Mathematics IV: Advanced Algebra

Chapter 3: Conic SectionsAt the end of this chapter, you should be able to:

identify the equations of the different conic sections; graph and identify the important parts of the different conic section; solve word problems involving applications of conic sections; and solve systems of nonlinear equations and inequalities.

Lesson 3.1: Introduction to Conic SectionsThe general equation of the second degree in two variables x and y, also called

quadratic equation in x and y, has the form

A x2+Bxy+C y2+Dx+Ey+F=0 ,

Where A ,B ,…, F are constants and A ,B ,C are not all zero.

In this chapter, we will only consider the case where B=0, hence, the equation

A x2+Cy2+Dx+Ey+F=0 .

As will be seen in the succeeding discussions, the following situations may arise depending on the values of the coefficients in Lesson 3.2:

1. If A=C, then the graph is either a circle, a point or the empty set.

2. In either of the following cases, the graph is a parabola:

(a) A=0, C≠0 and D≠0, or

(b) C=0, A≠0 and E≠0.

3. If AC>0, then the graph is either an ellipse, a point or the empty set.

4. If AC<0, then the graph is either a hyperbola or two intersecting lines.

The circle, parabola, ellipse and hyperbola are called conic sections, or simply conics. The point and a pair of intersecting lines are considered as degenerate conics.

The curves above are named conic sections because each can be illustrated as an intersection of a plane with a right circular cone of two nappes, each extending indefinitely far.

Although we have illustrated the different conic sections using the three – dimensional approach, in the succeeding sections, we will define each of them as a set of points in a plane satisfying certain geometric conditions.

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Figure 3.1 Circle

Figure 3.3 Parabola

Figure 3.5 A pair of lines

Figure 3.2 Ellipse

Figure 3.4 Hyperbola

Figure 3.6 A point

In Chapter, we have discussed circles. Before we proceed to the discussion of the other conic sections, we first study the concept of translation of axes.

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Lesson 3.2: Translation of Axes Let C (h , k ) be a point in the xy−¿ plane. To translate the axes to the point C (h , k )

means to construct another pair of axes, x '−¿ axis parallel to the x−¿ axis and y '−¿ axis parallel to the y−¿ axis, having the origin at C.

Figure 3.7

Thus, the plane has two sets of coordinate axes, and each point P in the plane has two pairs of coordinates: (x , y ) with respect to the original x−¿ and y−¿ axes, and (x ' , y ') with respect to the new x '−¿ and y '−¿ axes. The two pairs of coordinates are related according to the equations

x '=x−h, y '= y−k.

EXAMPLE 3.2.1Given the circle with equation x2+ y2−8 x−4 y+16=0 in the xy−¿ plane, translate the

axes to the center of this circle (name the new axes x ' and y ') and give it equation in terms of the new variables x ' and y '.

Solution: The standard form of the equation of this circle is( x−4 )2+( y−2 )2=4.

Therefore, its center is (4,2 ) .

Figure 3.8

When the axes are translated to the point (4,2), each point (x , y ) of the circle will have another pair of coordinates, (x ' , y '), with respect to the new set of axes, where

x '=x−4, y '= y−2.

Thus, an equation of the circle with respect to the variables x ' and y ' is

(x ' )2+ ( y ' )2=4.As shown in the previous example, translating the axes does not alter the shape of a given curve but possibly gives a simpler equation for the curve. In the succeeding lessons, though, we will apply translation of axes in the reverse manner. This will be illustrated in the following example.

EXAMPLE 3.2.2A circle has center at the point (−2,1) in the xy−¿ plane. The axes are translated to this

center so that an equation of the circle with respect to the new pair of coordinate axes is

( x ' )2+ ( y ' )2=9.

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Given an equation of the circle in terms of the variables x and y (with respect to the original coordinate axes).EXERCISES 3.2In exercises 1 – 4, a pair of equations is given. Draw a sketch of the graph of the first equation, then sketch the graph of the second equation using a suitable translation of axes.

1. y=|x|; y=|x−3|

2. y=|x|; y=|x+2|−13. y=√x ; y=√1+x−54. y=x2 ; y=( x−4 )2−2

In exercises 5 – 14, determine how many units and in what direction the graph of the first equation must be shifted to obtain the second equation.

5. x2+ y2=9 ; x2+ ( y−4 )2=9

6. x2+ y2=12 ; (x−2 )2+ ( y+3 )2=12

7. y2=8 x ; ( y−1 )2=8 ( x+1 )

8. x2=2 y ; ( x+5 )2=2 y−14

9. x2

9+ y2

16=1;

( x+3 )2

9+

( y−7 )2

16=1

10. x2

25− y2=1 ;

( x−5 )2

25− ( y−2 )2=1

11. ( x−2 )2+( y+3 )2=36 ; ( x−2 )2+ y2=36

12. ( x+7 )2

4+

( y−2 )2

25=1 ;

( x+3 )2

4+

( y+1 )2

25=1

13. x2+ y2+4 x+3=0 ;x2+ y2+6 x−6 y+17=0

14. y2+4 y−8 x+4=0 ; y2+10 y−8 x−7=0

In exercises 15 – 22, give the resulting equation if the following translations are applied to the given equations.

15. 4 units up; x3− y=−1

16. 5 units to the right; y+2=√−x

17. 3 units to the left; 12

units down; 2 y−2= ( x−3 )2

18. 14

units to the left; 2 units up; |4 x+1|=(3− y )3

19. 3 units to the left; 2 units up; x2+ y2=100

20. 2 units to the right; 3 units down; x2=12 y

21. 5 units to the left; 1 unit down; 4 x2+ y2+12 x−9 y=0

22. 2 units to the right; 5 units up; 5 y2−4 x2−30 y−32 x=99

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Lesson 3.3: ParabolasSuppose we have a line L and a point F not on L. Consider a point P whose distance

from L (the perpendicular distance) equals its distance from F. If we collect all such points P on the xy−¿ plane, we obtain a curve called parabola.

Figure 3.9 Figure 3.10

DEFINITION 3.3.1 A parabola is the set of all points in a plane equidistant from a fixed point F and a fixed

line L not containing F.

The point F in the definition above is called the focus of the parabola and the line L is its directrix. Throughout this section, we will assume that the line L is either horizontal or vertical.

The line through F perpendicular to the L is called the axis of symmetry, or simply axis, of the parabola. If we “fold” the plane along the axis, the parabola will be divided into two parts and these parts will coincide.

There is exactly one point of the parabola that lies on the axis of symmetry. It is called the vertex of the parabola. From the definition of a parabola, the vertex therefore is equidistant from the focus F and the directrix L. The distance from the vertex to the focus (or the directrix) is called the focal distance. Throughout this section, we shall denote the focal distance by c, where c>0.

Equation of a ParabolaA. Standard Form

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Consider a parabola with vertex at V (0,0) and focus at F (c ,0 ) , c>0. Since VF is horizontal, then the directrix L is the vertical line x=−c.

Figure 3.11

Let P(x , y) be any point of the parabola. If Q is the point on L fro which |PQ| is the distance from P ¿ L (so PQ is horizontal), then by definition of a parabola, we have

|PF|=|PQ|.

Note that for this equality to hold, the point P must be either in the first or the fourth quadrant. That is, the parabola must be opening to the right.

The point Q has coordinates (−c , y). Therefore, |PQ|=x+c. Also by the Distance Formula,

|PF|=√ ( x−c )2+ y2 .

Thus, we have

√(x−c)2+ y2=x+c .

Squaring both sides to remove the square root sign and simplifying, we have

(x−c)2+ y2=(x+c)2

x2−2cx+c2+ y2=x2+2cx+c2

y2=4cx .

This is an equation of the parabola with vertex at the origin and focus at (c ,0).

If, on the other hand, the focus at (−c ,0), where c>0, and the vertex is at the origin, then the parabola open to the left and its equation is y2=−4cx. This can be verified by going through the computation above, interchanging ( x−c ) and (x+c ).

Similarly, it can be verified that if the vertex of the parabola is the origin and the focus is (0 , c ), then the equation x2=4 cy. If the vertex is (0,0) and the focus is (0 ,−c), the equation is x2=−4cy.

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The four equations above are said to be in standard form. We summarize the details in the following table.

STANDARD FORM OF THE EQUATION OF A PARABOLA WITH VERTEX AT THE ORIGINFOCUS EQUATION PARABOLA OPENS(c ,0) y2=4cx To the right

(−c ,0) y2=−4cx To the left(0 , c ) x2=4 cy Upward

(0 ,−c) x2=−4cy Downward

Let us consider the general case where the vertex of the parabola is the point V (h ,k ) in the xy−¿ plane. Suppose the parabola opens upward. We wish to find an equation of the parabola.

First, translate the axes to the point(h , k ) and obtain a new set of axes, x ' and y '. Recall that if (x , y ) and (x ' , y ') are the coordinates of a point on the plane with respect to the original and the new set of axes, respectively, then x '=x−h and y '= y−k.

Figure 3.12

With respect to the axes x ' and y ', V has coordinates (0,0) and therefore the parabola has equation ( x ' )2=4cy '. Substituting x−h to x ' and y−k to y ', we obtain the desired equation in x and y,

(x−h)2=4c ( y−k ) .

This is the standard form of the equation of the parabola opening upward and having the vertex at (h , k ). Its directrix is the horizontal line y=k−c and its focus is the point (h , k+c).

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For the other three cases, parabola opening downward, to the right or to the left, the equations can also be derived by translation of axes. We give each of these equations together with the corresponding focus and directrix in the following table.

STANDARD FORM OF THE EQUATION OF A PARABOLA WITH VERTEX AT (h , k )PARABOLA OPENS EQUATION FOCUS DIRECTRIX

Upward (x−h)2=4c ( y−k ) (h , k+c) y=k−c

Downward (x−h)2=−4c ( y−k ) (h , k−c) y=k+cTo the right ( y−k )2=4 c (x−h) (h+c , k ) x=h−c

To the left ( y−k )2=−4 c (x−h) (h−c , k) x=h+c

B. General FormIf the square in any of the equation above are expanded and similar terms combined,

we obtain an equation in one of the following form.AXIS EQUATION

Vertical x2+Dx+Ey+F=0 ,whereE≠0Horizontal y2+Dx+Ey+F=0 ,where D≠0

Each of these equations is referred to as the general form of the equation of a parabola. Notice that exactly one of the variables x and y is squared. Also, if we multiply both sides of the equation by any nonzero constant other than 1, the graph of the resulting equation is the same parabola, although the coefficient of the squared term is not 1. Thus, the general form of the equation of a parabola with horizontal or vertical axis of symmetry is

Ax2+Cy2+Dx+Ey+F=0

Where exactly one of A and C is zero. When A=0, the axis is horizontal and the parabola open to the left or to the right. When C=0, the axis is vertical and the parabola opens upward or downward.

EXAMPLE 3.3.2Write an equation of the parabola with vertical axis of symmetry, vertex at the point

(2,3), and passing through the point (4,5).

EXAMPLE 3.3.3Find the vertex, focus and directrix of the parabola whose equation is

3 y2+4 x−24 y+44=0

EXAMPLE 3.3.4Find an equation of the parabola with horizontal axis and which contains the points

(2,0 ) ,(0 ,−1) and (0,7). Find its vertex, focal distance, focus and directrix.

To sketch a parabola if its equation is known, it suffices to plot three distinct points of the parabola then connect them through a smooth U – shaped curve. One of these three points

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must be the vertex. The other points can be obtained by assuming a value to the non – squared variable x or y, then solving for the values of the other variable from the resulting quadratic equation.

An example of pair of points of the parabola other than the vertex are the endpoints of the latus rectum. The latus rectum of a parabola is the line segment connecting two points of the parabola, perpendicular to the axis of symmetry and passing through the focus F. It can be verified using any of the previous standard form of the equation of a parabola that the length of the latus rectum is 4 c, where c is the focal distance.

Figure 3.14

EXAMPLE 3.3.5Sketch the parabola having the equation (x−2)2=12( y+3).

EXAMPLE 3.3.6Water issuing from the end of a horizontal pipe 6 meters (m) above the ground, forms a

parabolic curve, the vertex being at the end of the pipe. At the point 2m below the line of the pipe, the flow of the water has curve outward 3m beyond the vertical line through the end of the pipe. How far beyond this vertical line will the water strike the ground?

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4 c

EXERCISES 3.3For the parabola in exercises 1 – 16, find the (a) focal distance, and the coordinates of the (b) vertex, (c) focus, and (d) endpoints of the latus rectum. Also, find the equation of the (e) axis of symmetry and the (f) directrix. Finally, (g) sketch the graph.

1. x2=−8 y

2. x2=12 y

3. y2=16 x

4. y2=−6 x

5. ( y−3)2=12 (x+4)

6. (x+4 )2=4 ( y−2)

7. (x+2)2=−6( y−1)

8. ( y−1)2=−16 (x−2)

9. x2=−4 y+8

10. y2=3 x+12

11. x2+4 x−16 y+24=0

12. x2+2x− y=3=0

13. y2−8 y+6 x=0

14. y2−6 y−x+10=0

15. x2−2 x−2 y−1=0

16. y2+8 y+6 x+22=0

In exercises 7 – 34, find an equation of the parabola(s) having the given conditions.

17. Focus at (2,2), directrix at y=−4

18. Focus at (2,0), directrix at x=6.

19. Focus at (−3 ,−2), directrix at x=1

20. Focus at (−4 ,−4 ), directrix at y=−2

21. Vertex at (−3 ,−2), focus at (−3,2)

22. Vertex at (−1,3 ), focus at (3,3 )

23. Vertex at (−5 ,−2), directrix at y=−12

24. Vertex at (6,8), directrix at x=1

25. Vertex at (−1 ,−2 ), axis vertical, passes through (3,6 )

26. Axis horizontal, passes through (6,2 ) , (−3 ,−1 ) ,∧(−2,0)

27. Axis vertical, crosses the y−¿ axis at y=−3, and the x−¿ intercept are −6∧2

28. Opens downward and endpoints of latus rectum at (−6,1 )∧(2,1 )

29. Opens to the left and endpoints of the latus rectum at (−1 ,−5 )∧(−1,7 )

30. Vertex on the line 2 x+ y−5=0; focus on the line 3 x− y−21=0; directrix x=2

31. Vertex on the line 2 x−3 y+8=0; focus on the line 4 x+ y−1=0; directrix y=−1

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32. Length of the latus rectum is 8; directrix x=5; vertex on the line 5 x−4 y−7=0

33. Opens to the right, length of the latus rectum is 12, passing through (4 ,−1 ), and vertex on the line 2 x+ y−7=0

34. Focus (−1,3); passing through (3,6); axis parallel to the y−¿ axis.

35. A cable suspended between two poles which are 200 ft apart has a sag of 50 ft. If the cable hangs in a form of a parabola, find its equation with the origin at its lowest point.

36. The center cable of a suspension bridge forms a parabolic arc. The cable is suspended from the tops of the two support towers, which are 800 meters apart. The tops of the towers are 170 meters above the road and the lowest point of the cable is midway between the towers and ten meters above the road. Find the height of the cable above the road 100 feet from a tower.

37. A light reflector is in the shape of a paraboloid so that its cross – section is parabolic. If a light source is placed on the location of the focus of the reflector, the light rays will reflect off the surface, forming a beam of light parallel to the axis. Suppose that the light reflector is 12 inches in diameter. If the light source is located 1.5 inches from the vertex, determine the depth of the reflector so that the beam of the reflected light is parallel to the axis.

38. A searchlight is shaped like a paraboloid. If the light source is located 2 ft from the base along the axis of symmetry and the depth of the searchlight is 4 ft, what should the width of the opening be?

39. A satellite dish is shaped like a paraboloid. Signals that strike the surface of the dish are reflected to a single point which corresponds to the focus of the parabola. This point is where the receiver should be located. Suppose that the dish is 10 feet across and 3 feet deep at its center, at what position should the receiver be placed?

40. A satellite TV receiving dish is in the shape of a paraboloid. Find the location of the receiver, which is placed at the focus, if the dish is 6 ft across at its opening and 2 ft deep.

41. Water flowing from a suspended garden hose to the ground follows a parabolic path, with the hose’s opening as the vertex. Suppose this opening is 15 feet high and that the water strikes the ground a horizontal distance is 7 feet from where the opening is located. At what height is the horizontal distance of the water 1 foot from the opening?

42. A bridge is to be built in the shape of a parabolic arch and is to have a span of 100 ft. The height of the arch a distance of 40 ft from the center is to be 10 ft. Find the height of the arch at its center.

43. A parabolic arch has a height of 24 meters and a width of 32 meters at the base. If the vertex of the parabolic is at the top of the arch, at which height above the base is it 16 meters wide?

44. A water way for a boat ride in a recreational park has a parabolic cross – section. This water way has depth of 12 feet and a width of 6 feet at the surface. Boats with rectangular cross – section will be used and they are found to sink to a depth of 4 feet when placed on the

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water – filled water way. What is the maximum width that these boats can have so that they will float and move smoothly on the water way.

45. The endpoint of the latus rectum of a parabola are (4 ,−k ) and (4 , k ), where k>0. If the directrix of the parabola passes through the point (−2 ,10), find k and the equation of the parabola.

46. Find all values of b such that the parabola y=18x2+bx+ 9

2 will have a directrix having an

equation of y=2.

47. Find all points P of the parabola y2=8 x such that the perpendicular line segment from P to the directrix and the line segment from P to the focus are two sides of an equilateral triangle.

48. The locus of the midpoints of the segments joining the points of the parabola 4 cy=x2 and the focus is also a parabola. Find the vertex of the second parabola.

49. Show that the graph of an equation of the form

C y2+Dx+Ey+F=0 ,C≠0

(a) is a parabola if D≠0.

(b) is a horizontal line if D=0 and E2−4CF=0.

(c) is two horizontal lines if D=0 and E2−4CF>0.

(d) contains no point if D=0 and E2−4CF<0.

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Lesson 3.4: Ellipses

Definition 3.4.1

An ellipse is a set of all points in a plane that the sum of whose distances from two fixed points F1 and F2 is a constant. Each of the two fixed points F1and F2 is a focus (plural: foci) of the ellipse.

When F1 and F2 are the same point, the curve formed is a circle. Hence, we’ll assume throughout the discussion that the foci of an ellipse are two distinct points, thereby making a distinction between circles and ellipses.

According to the definition, if P1 and P2 are any two points of an ellipse with foci F1 and F2, then the sum |P1 F1|+|P1F2| of the distances of P1 from F1 and F2 is equal to the sum of |P2 F1|+|P2F2| of distances of P2 from F1 and F2. That is

|P1 F1|+|P1F2|=|P2 F1|+|P2F2|

Figure 3.18

The line through the foci of an ellipse is called its principal axis. In this section, we will only consider horizontal or vertical principal axis. The two points of the ellipse that lie on the principal axis are called vertices and the line segment joining them is callled the major axis. The midpoint of the major axis is called the center of the ellipse. It can be verified that the center is also the midpoint of the line segment whose endpoints are the foci. The distance from the center to a focus is called the focal distance

The line segment through the center, perpendicular to the major axis and whose endpoints are on the ellipse is called the minor axis. Its endpoints are called co – vertices.

Throughout this section, we will denote by a the distance from the center to a vertex (or half the length of the major axis), by b the distance from the center to a co – vertex (half the length of the minor axis), and by c the focal distance. Because F1 and F2 are distinct, then c>0. Likewise, a>0 and b>0.

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Lemma 3.4.2

For any point P of the ellipse with foci F1 and F2, then |P F1|+|PF2|=2aProof: The lemma says that the constant sum of distance mentioned in the definition of ellipse is equal to the length of the major axis. Let A be a vertex of the ellipse that is closer to F1 than to F2. If P is any point of the ellipse, then

|P F1|+|PF2|=|A F1|+|A F2|.

Thus, it suffices to show that |A F1|+|A F2|=2a. If A ' is the other vertex, then |A F1|=|A ' F2|, because C is the midpoint of both AA ' and F1F2. Thus,

|A F1|+|A F2|=|A' F2|+|A F2|=|AA '|=2a.

This show that |P F1|+|PF2|=2a.

Figure 3.20

Lemma 3.4.3

The constants a ,b , and c are related according to the equation a2=b2+c2.

Proof: Let B be a co – vertex of the ellipse with foci F1 and F2. From Lemma 3.4.2,

|B F1|+|B F2|=2a .

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a

b

c

If C is the center of the ellipse, then the triangles BC F1 and BC F2 ar congruent right triangle with right angle at C and legs of lengths b and c. Thus, |B F1|=|BF2|=√b2+c2 . Therefore,

2√b2+c2=2a .Dividing both sides by 2 and squaring both sides, we obtain b2+c2=a2 .

Because c>0, then the previous lemma implies that a>b. That is, the major axis is always longer than the minor axis.

Equations of an Ellipse

A. Standard Form

Consider an ellipse with center at the origin. Suppose the major axis is on the x−axis. Then the vertices are (−a ,0) and (a ,0), the foci are F1(−c ,0) and F2(c ,0), and the co –vertices are (0 ,−b) and (0 , b).

Figure 3.21

Let P(x , y) be any point of the ellipse. Then by Lemma 3.4.2

|P F1|+|PF2|=2a .

Using the distance formula, we have |P F1|=√ ( x+c )2+ y2 and |P F2|=√ ( x−c )2+ y2 . Therefore,

√ ( x+c )2+ y2+√( x−c )2+ y2=2a .

We want to have an equation free of radicals. To do this, first isolate one radical on the left hand side, then square both sides.

√ ( x+c )2+ y2=2a−√( x−c )2+ y2

( x+c )2+ y2=4 a2−4a√ ( x−c )2+ y2+( x−c )2+ y2 .

Expanding the squares and isolating the remaining radical, we have

x2+2xc+c2+ y2=4a2−4 a√( x−c )2+ y2+x2−2 xc+c2+ y2

4 a√ ( x−c )2+ y2=4 a2−4cx .

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Divide both sides by 4 and square both sides.

(a√ (x−c )2+ y2)=(a2−cx )2

a2 (x2−2 xc+c2+ y2)=a4−2a2cx+c2 x2

(a2−c2 ) x2+a2 y2=a4−a2 c2 .

Recall that a2−c2=b2, so substituting, we have

b2 x2+a2 y2=a2b2 .

Divide both sides by a2b2 to obtain

x2

a2+ y2

b2=1.

This is the standard form of the equation of an ellipse with center at the origin and major axis on the x−axis.

Using the same procedure, we can show that when the major axis of the ellipse is on the y−axis and the center is the origin, then an equation of the ellipse is

x2

b2+ y2

a2=1.

In general, when the center of the ellipse is the point (h , k ), then by translation of axes, similar to the derivation shown in the case of the parabola, an equation of the ellipse is

( x−h )2

a2+

( y−k )2

b2=1

when the major axis is horizontal, or

(x−h)2

b2+¿¿

when the major axis is vertical.

Standard Form of the Equation of an Ellipse

Center Major axis Equation

(0,0) horizontalx2

a2+ y2

b2=1

(0,0) verticalx2

b2+ y2

a2=1

(h , k ) horizontal(x−h)2

a2+

( y−k)2

b2=1

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(h , k ) vertical(x−h)2

b2+¿¿

B. General Form

If the square in any of the above equations are expanded and similar terms combined, we obtain the following general form of the equation of an ellipse

A x2+C y2+Dx+Ey+F=0

where AC>0. That is, either both A and C are positive or both are negative.

Example 3.4.4

Find the center and foci of the ellipse having the equation

4 x2+36 y2−24 x−144 y+171=0

Example 3.4.5

Find an equation of the ellipse with center at (1,1), a vertex at (1,3), and which passes through the origin.

Example 3.4.6

Determine whether the graph of the given equation is an ellipse, a point, or an empty set:

(a) x2+4 x+9 y2−18 y+13=0

(b) 16 x2−32 x+9 y2−72 y=−16

(c) 4 x2−16 x+9 y2+18 y+30=0

To draw a sketch of an ellipse, we begin by locating the center. Then we count a units from the center to both sides of the center along the major axis to obtain the vertices. Similarly, the co – vertices are obtained by counting b units alone the minor axis starting from the center. Then we connect the vertices and the co – vertices using a smooth curve that looks like an elongated circle.

Example 3.4.7

Sketch the ellipse having the equation ( x−1 )2

32+

( y−4 )2

42=1.

Example 3.4.8

The arch of a bridge over a two – lane highway is in the shape of a semi – ellipse. The distance from the base of the arch on one side of the road to the base on the other is 50 ft . The

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height of the arch at the center is 20 ft . Can a container van that is 14 ft high and 10 ftwide pass under the bridge without going over the center line of the bridge?

Exercises 3.4

For the ellipse in exercises 1 – 14, find the coordinates of the (a) center, (b) vertices, (c) foci, and (d) co – vertices. Finally, sketch the graph of the ellipse.

1. x2

16+ y2

49=1

2. x2

9+ y2

4=1

3. 4 y2+ x2=8

4. x2+9 y2=18

5. ( x−2 )2

25+

( y+1 )2

4=1

6. ( x+5 )2

4+

( y−6 )2

9=1

7. 25 ( x+2 )2+9 ( y−1 )2=225

8. 225 ( x+2 )2+289 ( y+3 )2=65025

9. 25 ( x+1 )2+169 ( y−2 )2=4225

10. 36 ( x+3 )2+4 y2−144=0

11. 16 x2+4 y2−32x+16 y−32=0

12. 2 x2+3 y2−8x+6 y+5=0

13. 4 x2+8 y2+4 x+24 y−13=0

14. 9 x2+4 y2−18x+16 y−11=0

In exercise 15 – 18, determine whether the graph of the given equation is an ellipse, a point, or an empty set.

15. 9 x2+4 y2−54 x−16 y+97=0

16. 9 x2+16 y2+90 x−160 y+481=0

17. 4 x2+ y2+16 x−14 y+61=0

18. x2+36 y2+6 x−72 y+45=0

In exercises 19 – 31, find an equation of the ellipse(s) satisfying the given conditions.

19. Center (0,0), length of minor axis 10, distance between foci 24.

20. Center (2,3), length of minor axis 8, foci (−2,3) and (6,3).

21. Center (5,4), major axis of length 16, minor axis of length 10.

22. Center (0,0), passing through (4,3) and (6,2).

23. Vertices (2,0) and (2,8); co – vertices (0,4 ) and (4,4).

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24. Vertices (4,3) and (4,9); focus (4,8).

25. Foci (3±√10 ,1); co – vertices (3 ,−2) and (3,4).

26. One vertex at (3,2), one co – vertex at (7,1).

27. One focus at (2,3), one vertex at (2,6), length of minor axis 18.

28. Center (0,0), one vertex at (3,0), one endpoint of the minor axis at (0,2).

29. Center at (1,2), one focus at (4,2), passing through (1,3).

30. Center (4 ,−1), one focus at (1 ,−1), and passing through(8,0).

31. Center (0,0), distance between foci 43

√33, passing through (2,1).

Solve problems in exercises 32 – 42.

32. A triangle in the plane has a perimeter of 12 units. If the two vertices lie at (−2,1) and (2,1), find the condition that must be satisfied by the third vertex.

33. A one – way road has an overpass in the form of a semi – ellipse, 15 feet high at the center, 40 feet wide. Assuming a truck is 12 feet wide, what is the tallest truck that can pass under the overpass.?

34. A bridge is built in the shape of a semi – ellipse. The bridge has a span of 120 ft and maximum height of 25 ft. Find the height of the arch at distances of 15ft and 40 ft from the center.

35. Suppose that a orbit of a planet is in the shape of an ellipse for which the distance between vertices is 1000 million kilometers and the distance between the foci is 600 million kilometers. Find an equation of the orbit.

36. A racetrack is in the shape of an ellipse 80 feet long and 40 feet wide. What is the width 10 feet from the side?

37. A whispering gallery is constructed such that its ceiling is a semi – ellipse. A person standing at the focus can whisper and be heard by another person standing at the other focus because all the sound waves that reach the ceiling from one focus are reflected to the other focus. How high will the ceiling be at the center of such a whispering gallery if the hall is 80 feet in length and foci are 60feet apart?

38. Mark, standing at one focus of a whispering gallery, is 6 meters from the nearest wall. His friend is standing at the other focus 100 meters away. Find the length of this whispering gallery and the height of its ceiling at the center.

39. A point moves in the coordinate plane so that the sum of its distances from (2,2) and (6,2) is equal to 16. Find the equation of the graph traced by the point.

40. The definition of the latus rectum for the ellipse is the same as for the parabola – a line segment through the focus, perpendicular to the major axis, and terminating on the curve. Find the equation of the ellipse centered at the origin where the distance between the two foci is √2 and the length of the latus rectum is 4.

19

41. Show that the length of each latus rectum for an ellipse is 2b2

a.

42. Find an equation of the set of points in a plane, each of whose distance from (2,0) is one – half its distance from the line x=8.

Lesson 3.5 Hyperbola

Recall that, to define an ellipse, we take the sum of the distances from two foci. For a hyperbola, instead of taking the sum, we take the difference. It will be seen that the equations of hyperbolas are, in some sense, similar to those of ellipses; the plus being replaced by minus. Despite the similarities, the ellipses and hyperbolas have completely different shapes.

Definition 3.5.1

A hyperbola is the set of all points in the plane that the absolute value of the difference of whose distances from two distinct fixed points F1 and F2 is a constant. These two fixed points are the foci of the hyperbola.

According the definition, if P1 and P2 are any two points of a hyperbola with foci F1 and F2, then

|P1 F1|−|P1F2|=|P2F1|−|P2 F2|.

20

Figure 3.25

The line through the foci is called the principal axis. Throughout this section, we will only consider vertical or horizontal principal axes. The two points of the hyperbola that lie on the principal axis are called vertices, and the line segment joining them is the transverse axis. The midpoint of the transverse axis is the center of the hyperbola. It can be shown that the center is also equidistant from the two foci.

Throughout this section, we shall denote by a the distance from the center to a vertex. It is also half the length of the transverse axis. The distance from the center to a focus will be denoted by c.

For the hyperbola, the vertices are closer to the center than the foci. That is, c>a.

Lemma 3.5.2

If P is any point of a hyperbola with foci F1 and F2, then |P F1−P F2|=2a .Proof: Let V 1 and V 2 be the vertices of the hyperbola. Essentially, the lemma says that the constant absolute value of differences mentioned in the definition of hyperbola is equal to the length of the transverse axis V 1V 2. Since V 1 is a point of the hyperbola, then |P F1|−|P F2|=|V 1F1|−|V 1F2|. Thus, it suffices to show that |V 1 F1|−|V 1F2|=2a. Now, if F1 is the focus that is on the same side of the center as V 1, then

|V 1 F2|=|V 1V 2|+|V 2F2|.

Since |V 2 F2|=|V 1F1|, then

|V 1 F2|=|V 1V 2|+|V 1F1|.

This implies that ||V 1F1|−|V 1 F2||=|V 1V 2|=2a .

Let b be the positive number such that b=√c2−a2 . We define the conjugate axis as the line segment whose length is 2b, perpendicular to the transverse axis and whose midpoint is the center of the hyperbola.

21

a

c

b

Figure 3.26

Take note that is possible that a>b or a<b, that is, the conjugate axis may be shorter or longer than the transverse axis.

Equation of Hyperbola

A. Standard Form

Suppose the center of the hyperbola is the origin and the transverse axis is on the x−axis. Then the foci are F1(−c ,0) and F2(c ,0). If P(x , y) is any point of the hyperbola, then, as shown earlier,

|P F1|−|P F2|=2a .

Using the distance formula, |P F1|=√ ( x+c )2+ y2 and |P F2|=√ ( x−c )2+ y2 . Substituting in the previous equation, we have

|√ ( x−c )2+ y2−√( x−c )2+ y2|=2a.Removing the absolute value sign and isolating one radical, we have

√ ( x+c )2+ y2=±2a+√( x−c )2+ y2 .

We want to obtain an equivalent equation free of radicals. Thus, squaring both sides and simplifying,

( x+c )2+ y2=4 a2±4a√ ( x−c )2+ y2+( x−c )2+ y2

2cx=4a2±4 a√( x−c )2+ y2−2cx

cx−a2=±a√ ( x−c )2+ y2 .

Squaring both sides, we have

c2 x2−2cx a2+a4=a2 ( x−c )2+a2 y2

c2 x2+a4=a2 x2+a2c2+a2 y2

(c2−a2 ) x2−a2 y2=a2 c2−a4 .

Recall that c2−a2=b2. Thus,

b2 x2−a2 y2=a2b2 .

Dividing both sides by a2b2, we obtain

x2

a2− y2

b2=1.

This is the standard form of the equation of the hyperbola with center at (0,0) and transverse axis on the x−axis.

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If the center of the hyperbola is the origin and the transverse axis is on the y−axis, then slight changes in the above derivation will yield the following equation.

y2

a2− x2

b2=1.

By translation of axes, then we obtain from above the following equations when the center is at point (h , k ). When the transverse axis is horizontal, the equation is

( x−h )2

a2−

( y−k )2

b2=1

And when the transverse axis is vertical, we have

( y−k )2

a2−

( x−h )2

b2=1.

Standard Form of the Equation of a Hyperbola

Center Major axis Equation

(0,0) horizontalx2

a2− y2

b2=1

(0,0) verticaly2

a2− x2

b2=1

(h , k ) horizontal(x−h)2

a2−

( y−k )2

b2=1

(h , k ) vertical ¿¿

B. General Form

Notice that the coefficients of the terms x2 and y2 in any of the above equations have different signs, one is positive and the other is negative. Thus, expanding the squares above and combining similar terms, we obtain the following general form of the equation of a hyperbola

A x2+C y2+Dx+Ey+ f=0

where AC<0.

Asymptotes of a Hyperbola

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A hyperbola is composed of two non – intersecting curves called its branches, each one containing a vertex and extending indefinitely to one side of the vertex it contains. These two curves approach two diagonal lines called asymptotes. If the hyperbola has center (h , k ), then the equations of the asymptotes are as follows:

1. when the transverse axis is horizontal (Figure 3.27)

y−k=±ba(x−h)

2. when the transverse axis is vertical (Figure 3.28)

y−k=±ab(x−h)

The derivation of these equations is beyond the scope of this book. The concept of limits in calculus is necessary in order to show that the branches of the hyperbola indeed approach these lines. At this point, we remark that a way to obtain such equations is to replaced by 0 the constant 1 on the right hand side of the standard form of the equation of the hyperbola.

Take note that the two asymptotes of the hyperbola intersects at the center of the hyperbola.

Figure 3.27 Figure 3.28

The asymptotes serve as a guide when sketching the hyperbola. Actually, these lines can be drawn even without getting their equations. Given the equation of the hyperbola, we start by locating its center. The direction of the transverse axis can be obtained from the given equation. Thus, along this direction (vertical or horizontal), we count a units from the center to both sides of the center. Then in the direction perpendicular to the transverse axis, we count b units from the center to both sides of the center. Next, draw a rectangle whose sides have these four points obtained as their midpoints. This rectangle is called the auxiliary rectangle. Connect each pair of opposite vertices of the auxiliary rectangle by a line. It can be verified that these two lines from have equations given above, and therefore, are the asymptotes of the hyperbola. Finally, we sketch the two branches of the hyperbola by drawing curves, starting from the vertices, approaching the asymptotes.

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y−k=( ba ) ( x−h )

y−k=−( ba ) ( x−h )

y−k=−( ab ) ( x−h )

y−k=( ab ) ( x−h )

(h . k )

Figure 3.29

Example 3.5.3

Sketch the hyperbola whose equation is

( x−2 )2

32−

( y+3 )2

42=1.

Example 3.5.4

The vertices of a hyperbola are (−5 ,−3) and (−5 ,−1) and the endpoints of the conjugate axis are (−7 ,−2) and (−3 ,−2). Find the equation of the hyperbola.

Example 3.5.5

Find the center, vertices, foci and asymptotes of the hyperbola having the equation

25 x2−9 y2−200 x+36 y+139=0

Example 3.5.6

Find an equation of the hyperbola that has a focus at (5,1) and the lines y−1=±2x as asymptotes.

Example 3.5.7

Determine whether the graph of the equation is a hyperbola or a pair of lines:

(a) 4 x2− y2+4 y=0

(b) x2−2 y2+4 y=2

Example 3.5.8

25

Points F1 and F2 are 1000 m apart, and it is determined from the sound of an explosion heard at these points at different times that the location of the explosion is 600 m closer to F1 than to F2. Show that the location of the explosion is a hyperbolic curve and give a Cartesian equation for this curve.

Exercises 3.5

For the hyperbolas in exercises 1 – 10, find the coordinates of the (a) center, (b) vertices, (c) foci. Also, (d) give the equations of the asymptotes and (e) sketch the graph.

1. x2

4− y2

25=1

2. y2

16− x2

9=1

3. 36 x2−81 y2=2916

4. x2− y2=4

5. 4 x2− y2−24 x−4 y+16=0

6. 9 x2−4 y2−36 x−16 y−16=0

7. 3 y2−2 x2−4 x−26=0

8. 9 x2−4 y2−90 x+189=0

9. 49 x2−4 x2−98 y+48 x−291=0

10. x2−2 y2−6 x−4 y+5=0

In exercises 11 – 14, determine whether the graph of the given equation is a hyperbola or pair of lines.

11. x2−4 y2−8 y−6 x−31=0

12. 9 y2−4 x2+8 x+18 y+5=013.

14x2− y2+6 y−2 x−5=0

14. 9 y2−16 x2−126 y−160x−535=0

In exercises 15 – 25, find an equation of the hyperbola satisfying the given conditions.

15. Center (0,0), transverse axis length of 6 is along the x−axis, a focus at (5,0).

26

16. Center (1 ,−2 ), transverse axis parallel to the x−axis and length of 6, conjugate axis of length 10.

17. Centered at the origin, passing through (−3 , 3 √52

) and (−32,0).

18. Center at (0,0), transverse axis along the y−axis, passing through points (5,3) and (−3,2).

19. Asymptotes are y−x−1=0 and y+x−1=0; y−intercepts are 3 and −1.

20. Conjugate axis along the y−axis, one vertex at (0,7), asymptotes are 6 x−5 y+30=0 and 6 x+5 y−30=0.

21. Foci (3±√13 ,2), passing through the point (0 , 3√52 −2).22. Foci at (−1,4 ) and (7,4 ), transverse axis of length

83.

23. Center (−3,2), transverse axis parallel to y−axis, passing through (1,7 ), the asymptotes are perpendicular to each other.

24. Asymptotes y=32x and y=

−32

x and passing through (4 ,√117 ) .

25. One focus at (0,25 ), Asymptotes intersecting at (0,5) and an asymptote passing through(24,13).

Solve the word problems in exercises 26 – 31.

26. One of the asymptotes of the hyperbola with horizontal transverse axis contains the points (−3 ,−6) and (9,0). Find the equation of the hyperbola if (−1,1) is one of the vertices of its auxiliary rectangle.

27. Two radio signaling stations at A and B lie on an east – west line, with A 100 miles west of B. A plane is flying west on a line 50 miles north of the line AB. Radio signals are sent (travelling at 0.2 miles/ μsec) simultaneously from A and B, and the one sent from B arrives at the plane 400 μsec before the one sent from A. Where is the plane?

28. Two stations that are 2 km apart receive a sound signal issued from a source located at point P(x , y). The station located at (0 ,−1) gets the signal 4 seconds earlier than the station at (0,1). Use 0.33 km/ sec as the speed of sound and find the equation of the hyperbola containing P.

29. Two stations that transmit radio signals are positioned 100 miles apart along a straight shore whereA is west of B. If the speed of each radio signal is 186, 000 miles per second,

(a) what time difference between the two signals should the ship be looking for if it wants to enter a harbor 20 miles west of station B?

(b) what is the approximate location of the ship if it is 80 miles offshore when the desired time difference is obtained?

27

30. Two LORAN stations that transmit radio signals are positioned 250 miles apart along a straight shore where A is west of B. If the speed of each radio signal is n186, 000 miles per second,

(a) what time difference between the two signals should the ship be looking for if it wants to enter a harbor 25 miles east of station A?

(b) what is the approximate location of the ship if it is 50 miles offshore when the desired time difference is obtained?

31. Suppose that the cost of producing product I is P150 less per unit at point A than at point B. The distance between A and B is 100 km. Suppose that the route of delivery of the product is a straight line, the delivery cost is P8 per unit per kilometer. Find the curve at any point of which the commodity can be supplied from either A or B at the same total cost. Take the points A and B at (−50,0) and (50,0), respectively.

32. Find the value(s) of k such that the graph of the equation

k (x2+ y2 )+x2− y2+x+ y=0 is

(a) a parabola.

(b) an ellipse.

(c) a hyperbola.

(d) a pair of intersecting lines.

Lesson 3.6: Systems of Linear and Quadratic Equations

In this section, we will solve systems involving linear and quadratic equations in two variables, say x and y. Solutions to such systems are pairs of values of x and y that will make all the equations in the system true. If x=a , y=b is a solution to a system of equations, then the point (a ,b) is a point of intersection of the graphs of the equations. Thus, if a system has no solution, then it means that the graphs do not intersect.

Example 3.6.1

Solve the system

{yy¿¿ x2

−2 x+8

Example 3.6.2

Solve the system

{ y=13

( x−3 )2−3

( x−3 )2+( y+2 )2=1

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Example 3.6.3

Sketch the graphs of the equations x2− y2−2 y=10 and x2−4 ( y+1 )2=6, and find their points of intersection.

Exercises 3.6

In exercises 1 – 15, solve the given system of equations, and make a rough sketch of the graph of each equation on a single coordinate system.

1. {x2− y2=21x+ y=7

2. {x2+ y2=2x− y=4

3. {x2+ y=1−4 x2x− y+10=0

4. {x2+ y2+6 y−x=−5x+ y+1=0

5. { 2x2−3 y2=618x2+3 y2=174

6. { x2− y2=19 x2+ y2=9

7. { ( x−2 )2+ y2=5x2+( y−1 )2=10

8. {x2− y2=4y=x2

9. { x2+ y2=9

x2

25+y2

16=1

10. { x2− y2=5x2+2 y2=17

11. { ( x−4 )2+( y−2 )2=9( x−4 )2

36+

( y−2 )2

9=1

29

12. {( y−2 )2=4 (x−4)( y−4 )2=(x−5)

13. { 12 ( x−6 )=( x−4 )2

( x−4 )2+( y−9 )2=36

14. { ( y−2 )2−3 x=9( y−2 )2+( x+1 )=4

15. {( x+1 )2

9+

( y−2 )2

25=1

(x+1 )2

9+

( y+1 )2

16=1

16. {4 ( x+4 )2−9 ( y−2 )2=369 ( x+4 )2+4 ( y−2 )2=36

17. { y=13

( x−3 )2−3

x2−6 x+ y2+2 y=−6

18. { x=12

( y+5 )2−2

y2+10 y+ (x−2 )2=−9

19. Find an equation of the common chord of the two circles x2+ y2−4 x−1=0 and x2+ y2−2 y−9=0.

20. The square of a certain number exceeds twice the square of another number by 18

. Also, the

sum of their squares is 516

. Find possible pairs of numbers that satisfy these conditions.

21. Prove that a hyperbola does not intersect its asymptotes.

Review for Chapter 3

1. Identify the graph of the conic section whose equation is given. Give the coordinates of its center, vertices, co – vertices and foci when applicable.

(a) 4 x2+ y2+16 x−6 y−39=0

(b) 4 x2−9 y2−16 x+54 y=101

(c) 9 x2−36 x−16 y2−64 y−28=0

2. Find an equation of the ellipse with center (2 ,−3), vertical principal axis, and which contains the points (5,13) and (−2,9).

3. Find an equation of the hyperbola with asymptotes y=2x−8 and y=−2x+4 and a vertex as V (3 ,−5). Graph the asymptotes and the hyperbola, indicating the center, foci, and vertices.

30

4. Find the points of intersection of the curves y2−4 x2+8x−2 y−7=0 and ( x+1 )2+ ( y−1 )2=9. Then sketch their graphs on the same Cartesian plane. Indicate the coordinates of the centers, and points of intersection.

5. A triangle in the plane has a perimeter of 12 units. If two vertices lie at (−2,0) and (2,0), sketch and identify the graph of the set of all possible locations of the third vertex.

6. Suppose a ,b>0. Sketch the parabolas

y2=4a2−4 ax

y2=4b2+4bx

Show that they have a common focus for any a and b.

7. A comet’s orbit is a parabola with the sun at the focus. When the comet is 100√2 million miles from the sun, the line from the sun to the comet makes an angle of 45 ° with the axis of the parabola. The comet comes closest to the sun when it is at the vertex of the parabola. What will be the minimum distance between the comet and the sun?

8. The graph of y=1x

is a hyperbola with coordinate axes as asymptotes. What translation of

y=1x

will produce the graph of y=8−4 x2x−3 ?

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