I. ANALYTIC GEOMETRY

Analytic Geometry is a branch of algebra that is used to model geometric objects - points, (straight)

lines, and circles being the most basic of these. Analytic geometry is a great invention of Descartes

and Fermat.

In classical mathematics, analytic geometry, also known as coordinate geometry,

or Cartesian geometry, is the study of geometry using a coordinate and the principles

of algebra and analysis. Analytic geometry is widely used in physics and engineering, and is the

foundation of most modern fields of geometry, including algebraic, differential, discrete, and

computational geometry.

In plane analytic geometry, points are defined as ordered pairs of numbers, say, (x, y), while

the straight lines are in turn defined as the sets of points that satisfy linear equations. From the view

of analytic geometry, geometric axioms are derivable theorems. For example, for any two distinct

points (x1, y1) and (x2, y2), there is a single line ax + by + c = 0 that passes through these points. Its

coefficients a, b, c can be found (up to a constant factor) from the linear system of two equations

 ax1 + by1 + c = 0

ax2 + by2 + c = 0,

or directly from the determinant equation

 

However, no axiomatic theory may escape using undefined elements. In Set Theory that

underlies much of mathematics and, in particular, analytic geometry, the most fundamental notion

of set remains undefined.

Geometry of the three-dimensional space is modelled with triples of numbers (x, y, z) and a

3D linear equation ax + by + cz + d = 0 defines a plane. In general, analytic geometry provides a

convenient tool for working in higher dimensions.

Within the framework of analytic geometry one may (and does) model non-Euclidean geometries as

well. For example, in plane projective geometry a point is a triple of homogenous coordinates(x, y,

z), not all 0, such that

  (tx, ty, tz) = (x, y, z),

for all t ≠ 0, while a line is described by a homogeneous equation

  ax + bx + cz = 0.

In analytic geometry, conic sections are defined by second degree equations:

  ax² + bxy + cy² + dx + ey + f = 0.

That part of analytic geometry that deals mostly with linear equations is called Linear Algebra.

Cartesian analytic geometry is geometry in which the axes x = 0 and y = 0 are perpendicular.

The components of n-tuple x = (x1, ... xn) are known as its coordinates. When n = 2 or n = 3, the first

coordinates is called abscissa and the second ordinate.

II. DISTANCE BETWEEN TWO POINTS

The distance between two points, P, and P2, can be expressed in terms of their coordinates by using the Pythagorean Theorem. From your study of Mathematics, Volume 1, you should recall that this theorem is stated as follows:

In a right triangle, the square of the length of the hypotenuse (longest side) is equal to the sum of the squares of the lengths of the other two sides.

 

Let the coordinates of P, be (x),y,) and let those of P2 be (X2,Y2), as shown in figure 1-2. By the Pythagorean theorem,

Figure 1-2.-Distance between two points.

where P1N represents the distance between x, and x2, P2N represents the distance between y1 and y2, and d represents the distance from P1 to P2. We can express the length of P1N in terms of x, and x2 and the length of P2Nin terms of y1 and y2 as follows:

Although we have demonstrated the formula for the first quadrant only, it can be proven for all quadrants and all pairs of points.

EXAMPLE 1: In figure 1-2, x, = 2,x2 = 6, y, = 2, and y2 = 5. Find the length of d.

SOLUTION:

This result could have been foreseen by observing that triangle P1NP2 is a 3-4-5 triangle.

EXAMPLE 2: Find the distance between P1 (4,6) and P2 (10,4).

SOLUTION:

EXAMPLE 3: Find the distance between point (-1, -3) and the midpoint of the line segment joining (2, 4) and (4, 6). square root of.

We first find the coordinates of the midpoint M of the segment joining (2, 4) and (4, 6) 

M = [ (2 + 4) / 2 , (4 + 6) / 2 ] 

= (3, 5) 

We now use the distance formula to find the distance between the points (-1, -3) and (3, 5) 

D = sqrt [ (3 - (-1)) 2 + (5 - (-3)) 2 ] 

= sqrt (80) = 4 sqrt (5)

III. DIVISION OF A LINE SEGMENT

Many times you may need to find the coordinates of a point that is some known fraction of the distance between P1 and P2.

In figure 1-3, P is a point lying on the line joining P1 and P2 so that

If P should lie 1/4 of the way between P1 and P2, then k would equal 1/4.

Triangles P1MP and P1NP2 are similar.

Therefore,

 

Figure 1-3.-Division of a line segment.

Since   is the ratio that defines k1 then

Therefore,

P1M = k(PIN)

Refer again to figure 1-3 and observe that P1N is equal to X2- x,. Likewise, P1M is equal to x - x1. When you replace P1M and P1N with their equivalents in terms of x, the preceding equation becomes

By similar reasoning,

The x and y found as a result of the foregoing discussion are the coordinates of the desired point, whose distances from P1 and P2 are determined by the value of k.

EXAMPLE 1:  Given two points P1 and P2 in space find the point R dividing the line segment P1P2 in the ratio -2 : 1. 

Solution

If R divides P1P2 in the ratio -2 : 1 then   = -2 .

The position vector   is then equal to

EXAMPLE 2:  To illustrate the use of the formula for the division of a line segment in a certain ratio, consider the problem from classical geometry on the concurrency of the three medians of a triangle. LetABC be any triangle, and let D, E and F be the midpoints of the three sides of the triangle, as shown in the diagram below. The three medians are AD, BE, and CF, and the problem is to prove first that they are concurrent (that is, if G is the point of intersection ofAD and BE, then CF also passes through G), and then that the pointG divides each median in the ratio 2 : 1.

SolutionLet the position vectors of A, B, C, D, E, F and G relative to some origin O be a, b, c, d, e, f and g respectively. For simplicity, the origin O and the seven position vectors are not drawn on the diagram. The formula for the position vector of the midpoint of a line segment then allows us to write

Since G is defined to be the point of intersection of AD and BE, we seek an equation involving a, b, d, and e. This can be obtained by eliminating c from the first two equations above.

After multiplying by 2, rearranging and dividing by 3, we obtain

Now compare these expressions with the formula for the position vector (relative to O) of a point R on the line segment P1P2,

It is clear that   is the position vector of a point on AD, and   is a position vector of a point on BE. Since G is the point of intersection of these two lines, we see that

These equations also show that G divides both AD and BE in the ratio 2 : 1.

Finally, to show that CF also passes through G, observe that

This shows that G lies on CF and divides CF in the ratio 2 : 1, as required.

EXAMPLE 3: Find the coordinates of a point 1/4 of the way from P1(2,3) to P2(4,1).

SOLUTION:

Therefore, point P is   .

When the midpoint of a line segment is to be found, the value of  (Io is 1/2. Therefore,

such that

By similar reasoning,

IV. Slope of a Line

The slope of a straight line, parallel and perpendicular lines are all explored interactively using an applet. When the slope of the line is 0, you know that the line is horizontal and you know it's a vertical line when the slope of a line is undefined.

In the Figure below, the subscripts on point A, B and C indicate the fact that there are three points on the line. The change in y whether up or down is divided by the change in x going to the right, this is the 'rise over run' concept.

 

y = mx + b is the equation that represents the line and the slope of the line with respect to the  x-axis which is given by tan q = m. This is the slope-intercept form of the equation of a line. (m for slope? Seems to be the standard!)

When the slope passes through a point A(x1, y1) then y1 = mx1 + b or with subtraction y - y1 = m (x - x1)

You now have the slope-point form of the equation of a line.

You can also express the slope of a line with the coordinates of points on the line. For instance, in the above figure, A(x, y) and B(s, y) are on the line y= mx + b :

m = tan q =   therefore, you can use the following for the equation of the line AB:

The equations of lines with slope 2 through the points would be:

For (-2,1) the equation would be: 2x - y + 5 = 0.

For (-1, -1) the equation would be: 2x - y + 1 = 0

EXAMPLE 1: Find the slope of y = –2x + 3.

Graphing the line, it looks like this:   

Picking x = –1, I get y = –2(–1) + 3 = 2 + 3 = 5. Picking x = 2, get y = –2(2) + 3 = –4 + 3 = –1. Then the points (–1, 5) and (2, –1) are on the line y = –2x + 3. The slope of the line is then calculated as:

EXAMPLE 2: One line passes through the points (–4, 2) and (0, 3); another line passes through the points (–3, –2) and (3, 2). Are these lines parallel, perpendicular, or neither?

I'll find the slopes.

EXAMPLE 3: One line passes through the points (0, –4) and (–1, –7); another line passes through the points (3, 0) and (–3, 2). Are these lines parallel, perpendicular, or neither?

I'll find the values of the slopes. Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved

V. The Locus of a Moving Point

When a point moves in a plane according to some given conditions the path along which it moves is called a locus. (Plural of locus is loci.).

CONDITION 1 : A point P moves such that it is always m units from the point Q.

Locus formed: A circle with centre Q and radius m.

 

  

Example :

Construct the locus of a point P at a constant distance of 2 cm from a fixed point Q.

Solution:

Construct a circle with centre Q and radius 2 cm.

CONDITION 2 : A point P moves such that it is equidistant form two fixed pointsX and Y.

Locus formed: A perpendicular bisector of the line XY.

  

Example:

Construct the locus of point P moving equidistant from fixed points X and Y and XY= 6 cm.

Solution:

Construct a perpendicular bisector of the line XY.

CONDITION 3: A point P moves so that it is always m units from a straight lineAB.

Locus formed: A pair of parallel lines m units from AB.

 

  

Example:

Construct the locus of a point P that moves a constant distant of 2 cm from a straight line AB.

Solution:

Construct a pair of parallel lines 2 cm from AB.

CONDITION 4: A point P moves so that it is always equidistant from two intersecting lines AB and CD.

Locus formed: Angle bisectors of angles between lines AB and CD.

 

Example:

The following figure shows two straight lines AB and CD intersecting at point O. Construct the locus of point P such that it is always equidistant from AB and CD.

  

Solution:

Construct angles bisectors of angles between lines AB and CD.

INTERSECTION OF TWO LOCI

Sometimes you may be required to determine the locus of a point that satisfies two conditions. We could do this by constructing the locus for each of the conditions and then determine where the two loci intersect.

Example :

Given the line AB and the point Q, find one or more points that are 3 cm from ABand 5 cm from Q.

 

  

 

Solution:

Construct a pair of parallel lines 3 cm from line AB. Draw a circle with centre Q and radius 5 cm.

  

The points of intersections are indicated by points X and Y.

It means that the locus consists of the two points X and Y.

Example:

Given a square PQRS with sides 3 cm. Construct the locus of a point which is 2 cm from P and equidistant from PQ and PS. Mark the points as A and B.

Solution:

Construct a circle with centre P and radius 2 cm. Since PQRS is a square the diagonal PR would be the angle bisector of the angle formed by the lines PQ andPS. The diagonal when extended intersects the circle at points A and B

  

Note: A common mistake is to identify only one point when there could be another point which could be found by extending the construction lines or arcs; as in the above examples.

VI. Circle

Definition: A circle is the locus of all points equidistant from a central point.

Definitions Related to Circles

arc: a curved line that is part of the circumference of a circlechord: a line segment within a circle that touches 2 points on the circle.circumference: the distance around the circle.diameter: the longest distance from one end of a circle to the other.origin: the center of the circlepi ( ): A number, 3.141592..., equal to (the circumference) / (the diameter) of any circle.radius: distance from center of circle to any point on it.sector: is like a slice of pie (a circle wedge).tangent of circle: a line perpendicular to the radius that touches ONLY one point on the circle.

diameter = 2 x radius of circle

Circumference of Circle = PI x diameter = 2 PI x radius    where PI =   = 3.141592...

Area of Circle:

    area = PI r2 

Length of a Circular Arc: (with central angle  )    if the angle   is in degrees, then length =   x (PI/180) x r    if the angle   is in radians, then length = r x 

Area of Circle Sector: (with central angle  )    if the angle   is in degrees, then area = ( /360) PI r2

    if the angle   is in radians, then area = ( /2) r2

Equation of Circle: (cartesian coordinates)

  for a circle with center (j, k) and radius (r):    (x-j)2 + (y-k)2 = r2

Equation of Circle: (polar coordinates)    for a circle with center (0, 0):   r( ) = radius

    for a circle with center with polar coordinates: (c,  ) and radius a:      r2 - 2cr cos(  -  ) + c2 = a2

Equation of a Circle: (parametric coordinates)    for a circle with origin (j, k) and radius r:      x(t) = r cos(t) + j       y(t) = r sin(t) + k

EXAMPLE 1:  In the figure below, triangle ABC is a triangle inscribed inside the circle of center O and radius r = 10 cm. Find the lengths of AB and CB so that the area of the triangle is twice the shaded area. 

Solution to Problem :

If the center O is on AC then AC is a diameter of the circle and the triangle has a right angle at B (Thales's theorem). If At is the area of triangle ABC and As the shaded area then 

At = 2 As 

We also have. 

At + As = (1/2) area of circle = (1/2) Pi 10 2 = 50 Pi 

2 As + As = 50 Pi 

Which gives. 

As = 50 Pi / 3 

Since triangle ABC has a right angle, we now use the internal angle (to the triangle) A to write. 

sin(A) = CB / AC = CB / 20 which gives CB = 20 sin (A) 

and cos(A) = AB / AC = AB / 20 which gives AB = 20 cos (A) 

The area As might also be written as follows (using the identity sin(2A) = 2 sin (A) cos (A)). 

As = (1/2) * AB * CB = 200 cos (A) sin(A) = 100 sin (2A) = 50 Pi / 3 

The above equation gives. 

sin (2A) = 0.5 Pi / 3 

Use calculator to solve for 2A. Two solutions 

2A = 31.6 degrees (nearest tenth) or A = 15.8 degrees 

2A = 148.4 degrees (nearest tenth) or A = 74.2 degrees 

We now calculate the lengths of AB and CB. Two solutions 

first solution 

AB = 20 cos (15.8) = 19.24 (2 decimals) and CB = 20 sin (15.8) = 5.45 (2 decimals) 

second solution 

AB = 20 cos (74.2) = 5.45 (2 decimals) and CB = 20 sin (74.2) = 19.24 (2 decimals) 

The two solutions correspond to two congruent right triangles.

EXAMPLE 2:  The small square is inscribed inside the circle and the larger circle circumsrcibes the same circle. If A1 is the area of the large square and A2 is the area of the small square, what is the ration A1 / A2? 

Solution to Problem :

If x is the size of one side of the small square, then its area A2 is given by 

A2 = x 2 

The diagonal d of the small square is given by 

d 2 = x 2 + x 2 

d = x sqrt (2) 

d is also equal to the side of one side of the large square 

The area A1 of the large square A1 is given by. 

A1 = ( x sqrt (2) ) 2 = 2 x 2 

Hence 

A1 / A2 = 2 x 2 / x 2 = 2

EXAMPLE 3:  Given a circle with the radius of 5cm. Find the area of this circle. Take π as 3.14.

Step 1

The picture below shows the circle with the radius of 5cm.

Step 2To calculate the area, we can start with the formula for the area of a circle:

Step 3Since the radius is given as 5cm, we can substitute r with 5.  Similarly, we can substitute π with 3.14. After doing so, we can calculate for A, as shown below:

Step 4Now, the calculated number 78.5 only has meaning if we include the unit for it. Since the radius is in cm, the unit for the area will be cm2. Hence:

A = 78.5 cm2

VII. Parabola

The parabola has many applications in situations where:

Radiation needs to be concentrated at one point (e.g. radio telescopes, pay TV dishes, solar radiation collectors) or

Radiation needs to be transmitted from a single point into a wide parallel beam (e.g. headlight reflectors).

The parabola is defined as the locus of a point which moves so that it is always the same distance from a fixed point (called the focus) and a given line (called thedirectrix).

In the following graph,

The focus of the parabola is at (0, p). The directrix is the line y = -p.

The focal distance is |p| (Distance from the origin to the focus, and from the origin to the directrix. We take absolute value because distance is positive.)

The point (x, y) represents any point on the curve.

The distance d from any point (x, y) to the focus (0, p) is the same as the distance from (x, y) to the directrix.

The Formula for a Parabola - Vertical Axis

Adding to our diagram from above, we see that the distance d = y + p.

Now, using the distance formula on the general points (0, p) and (x, y), and equating it to our value d = y + p, we have

Squaring both sides gives:

(x − 0)2 + (y − p)2 = (y + p)2

Simplifying gives us the formula for a parabola:

x2 = 4py

In more familiar form, with "y = " on the left, we can write this as:

where p is the focal distance of the parabola.

Parabolas with Horizontal Axis

We can also have the situation where the axis of the parabola is horizontal:

In this case, we have the relation:

y2 = 4px

[In a relation, there are two or more values of y for each value of x. On the other hand, a function only has one value of y for each value of x.]

Parabola is a member of conic sections, along with hyperbola and ellipse. Parabola can be thought of as a limiting case of ellipse or hyperbola. Note that parabola is not a family of curves. The impression that some parabola are more curved is because we are looking at different scale of the curve. Similarly, part of a large circle appears to be a line may induce us to conclude that there are different shapes of circles.

Like ellipse and hyperbola, there are many ways to define parabola. A common definition defines it as the locus of points P such that the distance from a line (called the directrix) to P is equal to the distance from P to a fixed point F (called the focus). As a conics section, the eccentricity of Parabola is 1.

Tracing a Parabola Tracing a Parabola

The axis of a parabola is a line perpendicular to its directrix and passing its focus. Vertex of the parabola is the intersection of the parabola and its axis.

Formulas Parametric: {t, 1/4 t^2}, -∞ < t < ∞ Cartesian: y == 1/4 x^2.

For the given formulas, vertexes is at {0,0}, focus is at {0,1}.

Properties

Point and Tangent Construction

Let F be a given point and d be a give line. Let B := Point[d]. Let t := LineBisector[B,F]. Let b := Perpendicular[B,d]. Let P := Intersect[b,t] Since length[segment[B,P]]==length[segment[P,F]], P is a point on parabola. Further, t is the tangent at P.

Invariant under certain Dilation

Parabola have the property that when scaled (streching/shrinking) along a direction parallel or perpendicular to its axis, the curve remain unchanged. (For example, line also have this property, but circle do not. A streched line is still a line, but a streched circle is no longer a circle) When a parabola is streched along the directrix “a” units and along the axis by “b” units, the resulting curve is the original parabola scaled in both direction by “a^2/b”.

Given a parametrization of a parabola {xf[t], yf[t]} with vertex at Origin and focus along the y-axis, its focus is {0, xf[t]^2/(4 yf[t]) }.

Optical Property

A radiant point at the focus will reflect off the parabola into parallel lines. The figure shows three parabolas, two of which share a common focus.

         

above: Left, 3 parabolas with its reflection property. Right: A photo of a car's headlight (Honda Civic 2000). Parabola with a Moving Light Source 

Tangents of Parabola

Any set of tangents on the parabola will always cut a arbitrary tangent into the same proportion. That is, suppose you pick three tangents call them a, b, c. Now pick a arbitrary tangent x. Tagents a, b, c will cut x into segments with certain proportions. Now pick any other tangent x1, it will be cut into the same proportions. Thus, the envelope of lines with a positive constant sum of intercepts is a segment of parabola.

Tangents of parabola cutting other tangents into the same proportion.

 

A segment of parabola formed by envelope of lines. The left figure shows the line positions, the right figure is rotated to visually show that it coincides with parabola in the standard position. Note: this is not astroid, because the lines that forms astroid as Trammel of Archimedes do not have the same positions as this.

Evolute and Semicubic Parabola

The evolute of a parabola is the semicubic parabola.

Parabola and its normals. The envelope of the normals is the semicubic parabola.

Pedal

The pedal of a parabola with respect to its focus is a line; pedal with respect to its vertex is the cissoid of Diocles.

 

Inversion

The inversion of a parabola with respect to its focus is a cardioid; inversion with respect to its vertex is the cissoid of Diocles.

 

EXAMPLE 1:  What is the minimum value of the expression 2x2 – 20x + 17?

Solution:

        Consider the function y = 2x2 – 20x + 17. This function is defined by a second degree equation. This xo-efficient of its x2 term is positive. Hence the curve is a parabola opening up ward.`(-coefficient fo X term)/(2.coefficient fo X^2 term)` = `-b/(2a)` =` (-(-20))/(2(2))` = `20/4` = 5.               For x = 5, y = 2(5)2 – 20(5) + 17 = - 33. Therefore the minimum value of the expression 2x2- 20x + 17 for any value of x is – 33. This minimum value is assumed only when x = 5.  

EXAMPLE 2:  Find the coordinates of maximum point of the curve y = - 3x2 – 12x + 5, and locate the axis of symmetry.

Solution:        The curve is defined by a second degree equation. The coefficient of x2 term is negative.`(-coefficient fo X term)/(2.coefficient fo X^2 term)` = `-b/(2a)` = `(-(-12))/(2(-3))` = `12/-6` = -

2.               For x = -2, y = -3 (-2)2 – 12(- 2) + 5 = 17. Hence the coordinates of the vertex are (- 2, 17). The curve is symmetric with respect to the vertical line through its vertex, through the point (-2, 17), i.e., the line x = -2

EXAMPLE 3:  If a parabolic reflector is 16 cm in diameter and 4cm deep, find the focus.

 Solution:        let POQ be the vertical section of the reflector. Mid - point of PQ is M. Let OX be along OM and OY parallel to MP.Let the equation of the parabola be y2 = 4ax.The coordinates of P are (4, 8)(8)2= 4a (4) or a = 4Focus = (a, 0) = (4, 0).Focus coincides with M, the mid-point of PQ

VIII. ELLIPSE

In geometry, an ellipse (from Greek ἔλλειψις elleipsis, a "falling short") is a plane curve that

results from the intersection of a cone by a plane in a way that produces a closed curve. Circles are

special cases of ellipses, obtained when the cutting plane is orthogonal to the cone's axis. An ellipse

is also the locus of all points of the plane whose distances to two fixed points add to the same

constant.

Ellipses are closed curves and are the bounded case of the conic sections, the curves that result from

the intersection of a circular cone and a plane that does not pass through its apex; the other two

(open and unbounded) cases are parabolas and hyperbolas. Ellipses also arise as images of a circle

underparallel projection and the bounded cases of perspective projection, which are simply

intersections of the projective cone with the plane of projection. It is also the simplest Lissajous

figure, formed when the horizontal and vertical motions are sinusoids with the same frequency.

Elements of an ellipse

The ellipse and some of its mathematical properties.

An ellipse is a smooth closed curve which is symmetric about its horizontal and vertical axes. The

distance between antipodalpoints on the ellipse, or pairs of points whose midpoint is at the center of

the ellipse, is maximum along the major axis ortransverse diameter, and a minimum along the

perpendicular minor axis or conjugate diameter.[1]

The semimajor axis (denoted by a in the figure) and the semiminor axis (denoted by b in the

figure) are one half of the major and minor diameters, respectively. These are sometimes called

(especially in technical fields) the major and minor semi-axes,[2][3] the major and minor semiaxes,[4][5] or major radius and minor radius.[6][7][8][9]

The foci of the ellipse are two special points F1 and F2 on the ellipse's major axis and are equidistant

from the center point. The sum of the distances from any point P on the ellipse to those two foci is

constant and equal to the major diameter (PF1 + PF2 = 2a ). Each of these two points is called

a focus of the ellipse.

Refer to the lower Directrix section of this article for a second equivalent construction of an ellipse.

The eccentricity of an ellipse, usually denoted by ε or e, is the ratio of the distance between the two

foci, to the length of the major axis or e = 2f/2a = f/a. For an ellipse the eccentricity is between 0

and 1 (0<e<1). When the eccentricity is 0 the foci coincide with the center point and the figure is a

circle. As the eccentricity tends toward 1, the ellipse gets a more elongated shape. It tends towards a

line segment (see below) if the two foci remain a finite distance apart and a parabola if one focus is

kept fixed as the other is allowed to move arbitrarily far away.

The distance ae from a focal point to the centre is called the linear eccentricity of the ellipse

(f = ae).

Drawing ellipses

The pins-and-string method

Drawing an ellipse with two pins, a loop and a pen.

An ellipse can be drawn using two drawing pins, a length of string, and a pencil:

Push the pins into the paper at two points, which will become the ellipse's foci. Tie the string

into a loose loop around the two pins. Pull the loop taut with the pencil's tip, so as to form

a triangle. Move the pencil around, while keeping the string taut, and its tip will trace out an

ellipse. Using two pegs and a rope, this procedure is traditionally used by gardeners to

outline an elliptical flower bed; thus it is called the gardener's ellipse.

If the ellipse is to be inscribed within a specified rectangle, tangent to its four sides at their

midpoints, one must first determine the position of the foci and the length of the string loop:

Let A,B,C,D be the corners of the rectangle, in clockwise order, with A-B being one of the

long sides. Draw a circle centered on A, whose radius is the short side A-D. From

corner B draw a tangent to the circle. The length L of this tangent is the distance between the

foci. This length L can be calculated with the Pythagorean theorem. As the tangent is at a

right angles to the radius at the intersect of the tangent with the circle L equals square root

((A-B)squared - (A-D)squared) i.e. square root of the long side of the rectangle squared

minus short side squared. Draw a horizontal line through the center of the rectangle. This

will be the major axis of the ellipse. Place the foci on the major axis, at distance L/2 from the

center.

To adjust the length of the string loop, insert a pin at one focus, and the second pin at the

opposite side of the rectangle on the major axis. Loop the string around the two pins and

tie it taut. Move the second pin to the other focus. Then draw the ellipse as above; it

should fit snugly in the original rectangle. Unfortunately strings tend to be elastic so if

you push harder on the pencil stretching the string more you will get a bigger ellipse,

pushing less it will be smaller. It may take a few tries to push just hard enough to make

the ellipse fit the rectangle.

Other methods

Trammel of Archimedes(ellipsograph) animation

An ellipse can also be drawn using a ruler, a set square, and a pencil:

Draw two perpendicular lines M,N on the paper; these will be the major and minor axes of

the ellipse. Mark three points A, B, C on the ruler. A->C being the length of the major axis

and B->C the length of the minor axis. With one hand, move the ruler on the paper, turning

and sliding it so as to keep point Aalways on line N, and B on line M. With the other hand,

keep the pencil's tip on the paper, following point C of the ruler. The tip will trace out an

ellipse.

The trammel of Archimedes or ellipsograph is a mechanical device that implements

this principle. The ruler is replaced by a rod with a pencil holder (point C) at one

end, and two adjustable side pins (points A and B) that slide into two perpendicular

slots cut into a metal plate.[10] The mechanism can be used with a routerto cut ellipses

from board material. The mechanism is also used in a toy called the "nothing

grinder".

Approximations to ellipses

An ellipse of low eccentricity can be represented reasonably accurately by a circle

with its centre offset. With the exception of Mercury, all the planets have an orbit

whose minor axis differs from the major axis by less than half of one percent. To

draw the orbit with a pair of compasses the centre of the circle should be offset from

the focus by an amount equal to the eccentricity multiplied by the radius.

Mathematical definitions and properties

In Euclidean geometry

Definition

In Euclidean geometry, an ellipse is usually defined as the bounded case of a conic

section, or as the set of points such that the sum of the distances to two fixed points

is constant. The equivalence of these two definitions can be proved using

the Dandelin spheres.

Eccentricity

The eccentricity of the ellipse (commonly denoted as either e or ε) is

(where again a and b are one-half of the ellipse's major and minor axes

respectively) or, as expressed in terms using

the flattening factor 

The distance from the center to either focus is ae, or simply 

Directrix

Each focus F of the ellipse is associated with a line parallel to the minor axis called a directrix.

Refer to the illustration on the right. The distance from any point P on the ellipse to the focus F is a

constant fraction of that point's perpendicular distance to the directrix resulting in the

equality, e=PF/PD. The ratio of these two distances is the eccentricity of the ellipse. This property

(which can be proved using the Dandelin spheres) can be taken as another definition of the ellipse.

Besides the well known ratio e=f/a, it is also true that e=a/d.

Ellipse as hypotrochoid

An ellipse (in red) as a special case of the hypotrochoid with R=2r.

The ellipse is a special case of the hypotrochoid when R=2r.

Area

The area enclosed by an ellipse is πab, where (as before) a and b are one-half of the ellipse's major

and minor axes respectively.

If the ellipse is given by the implicit equation Ax2 + Bxy + Cy2 = 1, then the area is  .

Circumference

 C of an ellipse is:

where the function E is the complete elliptic integral of the second

kind.

The exact infinite series is:

or

For computational purposes a much faster series where the

denominators vanish at a rate   is given by:[11]

A good approximation is Ramanujan's:

or better approximation:

For the special case where the minor

axis is half the major axis, we can use:

or the better approximation

More generally, the arc length of a portion of the circumference, as a function of the angle

subtended, is given by an incomplete elliptic integral.

See also: Meridian arc#Meridian distance on the ellipsoid

The inverse function, the angle subtended as a function of the arc length, is given by the elliptic

functions.[citation needed]

Chords

The midpoints of a set of parallel chords of an ellipse are collinear.

EXAMPLE 1:  Given the following equation

                                        9x2 + 4y2 = 36

a) Find the x and y intercepts of the graph of the equation.

b) Find the coordinates of the foci.

c) Find the length of the major and minor axes.

d) Sketch the graph of the equation.

Solution

a) We first write the given equation in standard form by dividing both sides of the equation by 36

                                        9x2 / 36 + 4y2 / 36 = 1

                                        x2 / 4 + y2 / 9 = 1

                                        x2 / 22 + y2 / 32 = 1

We now identify the equation obtained with one of the standard equation in the review above and we can say that the given equation is that of an ellipse with a = 3 and b = 2 (NOTE: a >b) .

Set y = 0 in the equation obtained and find the x intercepts.

                                        x2 / 22  = 1

Solve for x.

                                        x2  = 22

                                        x = ± 2

Set x = 0 in the equation obtained and find the y intercepts.

                                     y2 / 32 = 1

Solve for y.

                                       y2  = 32

                                        y = ± 3

b) We need to find c first.

                                     c2 = a2 - b2

a and b were found in part a).

                                     c2 = 32 - 22

                                     c2 = 5

Solve for c.

                                    c = ± (5)1/2

The foci are    F1 (0 , (5)1/2) and  F2 (0 , -(5)1/2)

c) The major axis length  is given by  2 a = 6.

 The minor axis length  is given by  2 b = 4.

d) Locate the x and y intercepts, find extra points if needed and sketch.

                                         

EXAMPLE 2:  Find the length of major and minor axes ,the co ordinates of foci and vertices ,and the eccentricity of the ellipse 3x2 + 2y2 = 6. Also find the equation of directives of an ellipse.Solution: Deriving at ellipse problem related to length of axes and directive,please follow the below steps.3x2 + 2y2 = 6 x2/2 + y2/3 = 1 (by dividing by 6)Since b > a, the major axis lie along x-axis and the minor axis lie along y - axis and the minor axis is along x-axis Major axis = 2b = 2√3 so b =√3 Minor axis = 2a = 2√2 so a =√2C = √b2 - a2 = √3 - 2 = 1 e = c/b = 1/√3 

1. Length of Ellipse major axis = 2√32. Length of Ellipse minor axis = 2√2

3. The foci are (0,c) and (0,-c) => (0,1) and (0,-1)

4. The vertices (0,b)and (0,-b) => (0,√3)and (0,-√3)

5. Eccentricity = 1/√3

6. Equation of direction y = b/e and y = - b/e => y=3 and y= -3

EXAMPLE 3: Find the ellipse eccentricity centre, verticals ,foci ,directrices ,the length of latus rectum and the equation of the latus rectum of the equation of the latus rectum of the ellipse x 2 + 4y2 + 2x + 16y + 13 = 0.

Solution: (x2 + 2x + 1) + 4(y2 + 4y + 4) - 1 - 16 + 13 = 0(x + 1) 2 + 4(y + 2) 2 = 4(x + 1) 2/4 + (y+2) 2/1 = 1Let x = x + 1 and y = y + 1This is of the form( x2/a2) + (y2/b2) = 1 .Where a = 2 , b = 1 c = √(a2 - b2) = √3 ,e = c/a = √3/2Vertices are (x = ± 2, y = 0) => x + 1 = 2 or x + 1 = - 2 => x = 1 or - 3

And y = 0=> y + 2 = 0 so y = - 2So the vertices are (1,-2) and (-3,-2) .The cetre is mid point of two vertices (-1,-2)The foci are x = ± c, y = 0=> x = ± √3, y = 0 => x + 1 = ± √3 so, x = -1+√3 or -1-√3y + 2 = 0 => y = -2Foci(-1 + √3, -2) and (-1-√3, -2)The equation of the directrices are x = ± a/e = ± 4/√3=> x + 1 = ± 4/√3 so x = -1 – 4/√3 The length of latus rectum is 2b2/a = 2/2 =1The equation of the latus rectum are x = ae so x = -1-√3

IX. THE HYPERBOLA

Hyperbola

Cartesian equation: x2/a2 - y2/b2 = 1 or parametrically: x = a sec(t), y = b tan(t)

A hyperbola is a conic section with an eccentricity greater than 1.

The formulas

And

developed in the section concerning the ellipse were derived so that they are true for any value of eccentricity. Thus, they are true for the hyperbola as well as for an ellipse. Since e is greater than 1 for a hyperbola, then

Therefore c > a > d.

Figure 2-17.-The hyperbola.

According to this analysis, if the center of symmetry of a hyperbola is the origin, then the foci lies farther from the origin than the directrices. An inspection of figure 2-17 shows that the

curve never crosses the Y axis. Thus the solution for the value of b, the semiminor axis of the ellipse, yields no real value for b. In other words, b   is an imaginary number. This can easily be seen from the equation

since c > a   for a hyperbola.

However, we can square both sides of the the above equation, and since the square of an imaginary number is a negative real number we write

or

and, since c = ae ,

Now we can use this equation to obtain the equation of a hyperbola from the following equation, which was developed in the section on the ellipse:

and since

we have

This is a standard form for the equation of a hyperbola with its center, O, at the origin. The solution of this equation for y gives

which shows that y is imaginary only when x2 < a 2.  The curve, therefore, lies entirely beyond the two lines x = ± a   and crosses the X axis at V1 (a,0) and V2( - a,0), the vertices of the hyperbola.

The two straight lines

can be used to illustrate an interesting property of a hyperbola. The distance from the line bx - ay = 0 to the point (x1,y1) on the curve is given by

Since (x1,y1) is on the curve, its coordinates satisfy the equation

which may be written

or

Now substituting this value into equation (2.11) gives us

As the point (x1,y1) is chosen farther and farther from the center of the hyperbola, the absolute values for x, and y, will increase and the distance, d, will  approach zero. A similar result can easily be derived for the line bx + ay = 0.

The lines of equation (2.10), which are usually written

are called the asymptotes of the hyperbola. They are very important in tracing a curve and studying its properties. The

Figure 2-18.-Using asymptotes to sketch a hyperbola.

asymptotes of a hyperbola, figure 2-18, are the diagonals of the rectangle whose center is the center of the curve and whose sides are parallel and equal to the axes of the curve. The focal chord

of a hyperbola is equal to   .

Another definition of a hyperbola is the locus of all points in a plane such that the difference of their distances from two fixed points is constant. The fixed points are thefoci, and the constant difference is 2a.

The nomenclature of the hyperbola is slightly different from that of an ellipse. The transverse axis is of length 2a and is the distance between the intersections (vertices) of the hyperbola with its focal axis. The conjugate axis is of length 2b and is perpendicular to the transverse axis.

Whenever the foci are on the Y axis and the directrices are lines of the form y = ± k, where k is a constant, the equation of the hyperbola will read

This equation represents a hyperbola with its transverse axis on the Y axis. Its asymptotes are the lines by - ax = 0 and by +ax=0 or

The properties of the hyperbola most often used in analysis of the curve are the foci, directrices, length of the focal chord, and the equations of the asymptotes.

Figure 2-17 shows that the foci are given by the points F, (c,0) and FZ ( - c,0) when the equation of the hyperbola is in the form

If the equation were

the foci would be the points (0,c) and (0, -c). The value of c is either determined from the formula

or the formula

Figure 2-17 also shows that the directrices are the lines   or, in the case where the

hyperbolas open upward and downward,   . This was also given earlier in this

discussion as   .

EXAMPLE 1:  Given the following equation

                                        9x2 - 16y2 = 144

a) Find the x and y intercepts, if possible,  of the graph of the equation.

b) Find the coordinates of the foci.

c) Sketch the graph of the equation.

Solution

a) We first write the given equation in standard form by dividing both sides of the equation by 144

                                        9x2 / 144  -  16y2 / 144  =  1

                                        x2 / 16  -  y2 / 9  =  1

                                        x2 / 42  -  y2 / 32  =  1

We now compare the equation obtained with the standard equation (left) in the review above and we can say that the given equation is that of an hyperbola with a = 4 and b = 3.

Set y = 0 in the equation obtained and find the x intercepts.

                                        x2 / 42  = 1

Solve for x.

                                        x2  = 42

                                        x = ± 4

Set x = 0 in the equation obtained and find the y intercepts.

                                   - y2 / 32 = 1

NO y intercepts since the above equation does not have real solutions.

b) We need to find c first.

                                     c2 = a2 + b2

a and b were found in part a).

                                     c2 = 42 + 32

                                     c2 = 25

Solve for c.

                                    c = ± 5

The foci are    F1 (5 , 0) and  F2 (-5 , 0)

c) 

1 - Find the asymptotes y = - (b/a) x and y = (b/a) x and plot them.

                                        y = -(3/4) x  and y = (3/4) x

2 - plot the x intercepts

3 - Find extra points (if necessary)

set x = 6 and find y                9(6)2 - 16y2 = 144

                                                 - 16y2 = 144 - 324

                                                y2 = 45 / 4

 Solve for y                               y = 3(5)1/2 / 2         and            y = - 3(5)1/2 / 2          

so the points (6,  3(5)1/2 / 2) and    (6,  -3(5)1/2 / 2) are on the graph of the hyperbola.

Also because of the symmetry of the graph of the hyperbola, the points (-6,  3(5)1/2 / 2) and    

(-6,  -3(5)1/2 / 2) are also on the graph of the hyperbola.

                 

EXAMPLE 2:  Given the following equation

9x2 - 16y2 = 144

a) Find the x and y intercepts, if possible, of the graph of the equation.

b) Find the coordinates of the foci.

Solution a) We first write the given equation in standard form by dividing both

sides of the equation by 144

9x2 / 144 - 16y2 / 144 = 1

x2 / 16 - y2 / 9 = 1

x2 / 42 - y2 / 32 = 1

We now compare the equation obtained with the standard equation (left) in the review above and we can say that the given equation is that of an hyperbola with a = 4 and b = 3.

Set y = 0 in the equation obtained and find the x intercepts.

x2 / 42 = 1

Solve for x.

x2 = 42 

x = ± 4 

Set x = 0 in the equation obtained and find the y intercepts.

- y2 / 32 = 1

NO y intercepts since the above equation does not have real solutions.

b) We need to find c first.

c2 = a2 + b2

a and b were found in part a).

c2 = 42 + 32

c2 = 25

Solve for c.

c = ± 5

The foci are F1 (5 , 0) and F2 (-5 , 0)

EXAMPLE 3:  The foci  of a hyperbola are given by (-2,0) and (2,0)  directrices are x =3/2d x=-3/2.Find Major Axis.Solution:   The mid point of the foci are (0,0).  So centre of the hyperbola is (0,0).  Now coordinates offoci are (h-ae,0) and (h+ae,0)                    Since h =0 we get ae =2.   ...   1                    Directrices are given by  x=h+a/e and x =h-a/e or x =a/e or -a/e here                    a/e =3/2   Multiplying these two we get a^2 = 3 or a = root 3                     So 2a = 2√3 = Major axis.

Magnetism

   

I.INTRODUCTION

Magnetism, an aspect of electromagnetism, one of the fundamental forces of nature. Magnetic forces are produced by the motion of charged particles such as electrons, indicating the close relationship between electricity and magnetism. The unifying frame for these two forces is called electromagnetic theory (see Electromagnetic Radiation). The most familiar evidence of magnetism is the attractive or repulsive force observed to act between magnetic materials such as iron. More subtle effects of magnetism, however, are found in all matter. In recent times these effects have provided important clues to the atomic structure of matter.

The ancient Greeks, originally those near the city of Magnesia, and also the early Chinese knew about strange and rare stones (possibly chunks of iron ore struck by lightning) with the power to attract iron. A steel needle stroked with such a "lodestone" became "magnetic" as well, and around 1000 the Chinese found that such a needle, when freely suspended, pointed north-south.

    The magnetic compass soon spread to Europe. Columbus used it when he crossed the Atlantic ocean, noting not only that the needle deviated slightly from exact north (as indicated by the stars) but also that the deviation changed during the voyage. Around 1600 William Gilbert, physician to Queen Elizabeth I of England, proposed an explanation: the Earth itself was a giant magnet, with its magnetic poles some distance away from its geographic ones (i.e. near the points defining the axis around which the Earth turns).

II. HISTORY OF STUDY

The phenomenon of magnetism has been known of since ancient times. The mineral lodestone (see Magnetite), an oxide of iron that has the property of attracting iron objects, was known to the Greeks, Romans, and Chinese. When a piece of iron is stroked with lodestone, the iron itself acquires the same ability to attract other pieces of iron. The magnets thus produced are polarized—that is, each has two sides or ends called north-seeking and south-seeking poles. Like poles repel one another, and unlike poles attract.

The compass was first used for navigation in the West some time after AD1200. In the 13th century, important investigations of magnets were made by the French scholar Petrus Peregrinus. His discoveries stood for nearly 300 years, until the

English physicist and physician William Gilbert published his book Of Magnets, Magnetic Bodies, and the Great Magnet of the Earth in 1600. Gilbert applied scientific methods to the study of electricity and magnetism. He pointed out that the earth itself behaves like a giant magnet, and through a series of experiments, he investigated and disproved several incorrect notions about magnetism that were accepted as being true at the time. Subsequently, in 1750, the English geologist John Michell invented a balance that he used in the study of magnetic forces. He showed that the attraction and repulsion of magnets decrease as the squares of the distance from the respective poles increase. The French physicist Charles Augustin de Coulomb, who had measured the forces between electric charges, later verified Michell's observation with high precision.

III. ELECTROMAGNETIC THEORY

In the late 18th and early 19th centuries, the theories of electricity and magnetism were investigated simultaneously. In 1819 an important discovery was made by the Danish physicist Hans Christian Oersted, who found that a magnetic needle could be deflected by an electric current flowing through a wire. This discovery, which showed a connection between electricity and magnetism, was followed up by the French scientist André Marie Ampère, who studied the forces between wires carrying electric currents, and by the French physicist Dominique François Jean Arago, who magnetized a piece of iron by placing it near a current-carrying wire. In 1831 the English scientist Michael Faraday discovered that moving a magnet near a wire induces an electric current in that wire, the inverse effect to that found by Oersted: Oersted showed that an electric current creates a magnetic field, while Faraday showed that a magnetic field can be used to create an electric current. The full unification of the theories of electricity and magnetism was achieved by the English physicist James Clerk Maxwell, who predicted the existence of electromagnetic waves and identified light as an electromagnetic phenomenon.

Subsequent studies of magnetism were increasingly concerned with an understanding of the atomic and molecular origins of the magnetic properties of matter. In 1905 the French physicist Paul Langevin produced a theory regarding the temperature dependence of the magnetic properties of paramagnets (discussed below), which was based on the atomic structure of matter. This theory is an early example of the description of large-scale properties in terms of the properties of electrons and atoms. Langevin's theory was subsequently expanded by the French physicist Pierre Ernst Weiss, who postulated the existence of an internal, “molecular” magnetic field in materials such as iron. This concept, when combined with Langevin's theory, served to explain the properties of strongly magnetic materials such as lodestone.

After Weiss's theory, magnetic properties were explored in greater and greater detail. The theory of atomic structure of Danish physicist Niels Bohr, for example,

provided an understanding of the periodic table and showed why magnetism occurs in transition elements such as iron and the rare earth elements, or in compounds containing these elements. The American physicists Samuel Abraham Goudsmit and George Eugene Uhlenbeck showed in 1925 that the electron itself has spin and behaves like a small bar magnet. (At the atomic level, magnetism is measured in terms of magnetic moments—a magnetic moment is a vector quantity that depends on the strength and orientation of the magnetic field, and the configuration of the object that produces the magnetic field.) The German physicist Werner Heisenberg gave a detailed explanation for Weiss's molecular field in 1927, on the basis of the newly-developed quantum mechanics (see Quantum Theory). Other scientists then predicted many more complex atomic arrangements of magnetic moments, with diverse magnetic properties.

IV. The Magnetosphere

    On Earth one needs a sensitive needle to detect magnetic forces, and out in space they are usually much, much weaker. But beyond the dense atmosphere, such forces have a much bigger role, and a region exists around the Earth where they dominate the environment, a region known as the Earth's magnetosphere. That region contains a mix of electrically charged particles,

and electric and magnetic phenomena rather than gravity determine its structure. We call it the Earth's magnetosphere

    Only a few of the phenomena observed on the ground come from the magnetosphere: fluctuations of the magnetic field known as magnetic storms and sub storms, and the polar aurora or "northern lights," appearing in the night skies of places like Alaska and Norway. Satellites in space, however, sense much more: radiation belts, magnetic structures, fast streaming particles and processes which energize them. All these are described in the sections that follow.

But what is magnetism?

    Until 1821, only one kind of magnetism was known, the one produced by iron magnets. Then a Danish scientist, Hans Christian Oersted, while demonstrating to friends the flow of an electric current in a wire, noticed that the current caused a nearby compass needle to move. The new phenomenon was studied in France by Andre-Marie Ampere, who concluded that the nature of magnetism was quite different from what everyone had believed. It was basically a force between electric currents: two parallel currents in the same direction attract, in oposite directions repel. Iron magnets are a very special case, which Ampere was also able to explain.

What Oersted saw...

    In nature, magnetic fields are produced in the rarefied gas of space, in the glowing heat of sunspots and in the molten core of the Earth. Such magnetismmust be produced by electric currents, but finding how those currents are produced remains a major challenge.

V. Magnetic Field Lines    Michael Faraday, credited with fundamental discoveries on electricity and magnetism (an electric unit is named "Farad" in his honor), also proposed a widely used method for visualizing magnetic fields. Imagine a compass needle freely suspended in three dimensions, near a magnet or an electrical current. We can trace in space (in our imagination, at least!) the lines one obtains when one "follows the direction of the compass needle." Faraday called them lines of force, but the term field lines is now in common use.

Compass needles outlining field lines

Field lines of a bar magnet are commonly illustrated by iron filings sprinkled on a sheet of paper held over a magnet. Similarly, field lines of the Earth start near the south pole of the Earth, curve around in space and converge again near the north pole.

    However, in the Earth's magnetosphere, currents also flow through space and modify this pattern: on the side facing the Sun, field lines are compressed earthward, while on the night side they are pulled out into a very long "tail," like that of a comet. Near Earth, however, the lines remain very close to the "dipole pattern" of a bar magnet, so named because of its two poles.

 

Magnetic field lines from an idealized model.

    To Faraday field lines were mainly a method of displaying the structure of the magnetic force. In space research, however, they have a much broader significance, because electrons and ions tend to stay attached to them, like beads on a wire, even becoming trapped when conditions are right. Because of this attachment, they define an "easy direction" in the rarefied gas of space, like the grain in a piece of wood, a direction in which ions and electrons, as well as electric currents (and certain radio-type waves), can easily move; in contrast, motion from one line to another is more difficult.

A map of the magnetic field lines of the magnetosphere, like the one displayed above (from a mathematical model of the field), tells at a glance how different regions are linked and many other important properties.

VI. Electromagnetic Waves    Faraday not only viewed the space around a magnet as filled with field lines, but also developed an intuitive (and perhaps mystical) notion that such space was itself modified, even if it was a complete vacuum. His younger contemporary, the great Scottish physicist James Clerk Maxwell, placed this notion on a firm mathematical footing, including in it electrical forces as well as magnetic ones. Such a modified space is now known as an electromagnetic field.

    Today electromagnetic fields (and other types of field as well) are a cornerstone of physics. Their basic equations, derived by Maxwell, suggested that they could undergo wave motion, spreading with the speed of light, and Maxwell correctly guessed that this actually was light and that light was in fact an electromagnetic wave.

Heinrich Hertz in Germany, soon afterwards, produced such waves by electrical means, in the first laboratory demonstration of radio waves. Nowadays a wide variety of such waves is known, from radio (very long waves, relatively low

frequency) to microwaves, infra-red, visible light, ultra-violet, x-rays and gamma rays (very short waves, extremely high frequency).

    Radio waves produced in our magnetosphere are often modified by their environment and tell us about the particles trapped there. Other such waves have been detected from the magnetospheres of distant planets, the Sun and the distant universe. X-rays, too, are observed to come from such sources and are the signatures of high-energy electrons there.

VII. APPLICATIONS

Numerous applications of magnetism and of magnetic materials have arisen in the past 100 years. The electromagnet, for example, is the basis of the electric motor and the transformer. In more recent times, the development of new magnetic materials has also been important in the computer revolution. Computer memories can be fabricated using bubble domains. These domains are actually smaller regions of magnetization that are either parallel or antiparallel to the overall magnetization of the material. Depending on this direction, the bubble indicates either a one or a zero, thus serving as the units of the binary number system used in computers. Magnetic materials are also important constituents of tapes and disks on which data are stored.

In addition to the atomic-sized magnetic units used in computers, large, powerful magnets are crucial to a variety of modern technologies. Powerful magnetic fields are used in nuclear magnetic resonance imaging, an important diagnostic tool used by doctors. Superconducting magnets are used in today's most powerful particle accelerators to keep the accelerated particles focused and moving in a curved path. Scientists are developing magnetic levitation trains that use strong magnets to

enable trains to float above the tracks, reducing friction.

Republic of the Philippines

Philippine State College of Aeronautics

Piccio Garden, Villamor, Pasay City

PHYSICS:

MAGNETISM

Submitted by:

Bryan M. Almonidovar

Submitted to:

 Engr. Mark Gerald Dadivo

Republic of the Philippines

Philippine State College of Aeronautics

Piccio Garden, Villamor, Pasay City

Analytical Geometry

Submitted by:

Bryan M. Almonidovar

Submitted to:

 Engr. Mark Gerald Dadivo