Tangents to Curves

A review of some ideas, relevant to the calculus, from high school

plane geometry

Straightedge and Compass

• The physical tools for drawing the figures that Plane Geometry investigates are:– The unmarked ruler (i.e., a ‘straightedge’)– The compass (used for drawing of circles)

Lines and Circles

• Given any two distinct points, we can use our straightedge to draw a unique straight line that passes through both of the points

• Given any fixed point in the plane, and any fixed distance, we can use our compass to draw a unique circle having the point as its center and the distance as its radius

The ‘perpendicular bisector’

• Given any two points P and Q, we can draw a line through the midpoint M that makes a right-angle with segment PQ

P QM

Tangent-line to a Circle

• Given a circle, and any point on it, we can draw a straight line through the point that will be tangent to this circle

How do we do it?

• Step 1: Draw the line through C and T

C

T

How? (continued)

• Step 2: Draw a circle about T that passes through C, and let D denote the other end of that circle’s diameter

C

T

D

How? (contunued)

• Step 3: Construct the straight line which is the perpendicular bisector of segment CD

C

T

D

tangent-line

Proof that it’s a tangent

• Any other point S on the dotted line will be too far from C to lie on the shaded circle (because CS is the hypotenuse of ΔCTS)

C

T

D

S

Tangent-line to a parabola

• Given a parabola, and any point on it, we can draw a straight line through the point that will be tangent to this parabola

directrix

axisfocus

parabola

How do we do it?

• Step 1: Drop a perpendicular from T to the parabola’s directrix; denote its foot by A

directrix

axisfocus

parabola

TA

F

How? (continued)

• Step 2: Locate the midpoint M of the line-segment joining A to the focus F

directrix

axisfocus

parabola

TA

F

M

How? (continued)

• Step 3: Construct the line through M and T (it will be the parabola’s tangent-line at T, even if it doesn’t look like it in this picture)

directrix

axisfocus

parabola

TA

F

M

tangent-line

Proof that it’s a tangent

• Observe that line MT is the perpendicular bisector of segment AF (because ΔAFT will be an isosceles triangle)

directrix

axisfocus

parabola

TA

F

M

tangent-line

TF = TA because T ison the parabola

Proof (continued)

• So every other point S that lies on the line through points M and T will not be at equal distances from the focus and the directrix

directrix

axisfocus

parabola

T

A

F

M

SB SB < SA

since SA is hypotenuse of right-triangle ΔSAB SA = SFbecause SA lies on AF’s perpendictlar bisector

Therefore: SB < SF

Tangent to an ellipse

• Given an ellipse, and any point on it, we can draw a straight line through the point that will be tangent to this ellipse

F1 F2

How do we do it?

• Step 1: Draw a line through the point T and through one of the two foci, say F1

F1 F2

T

How? (continued)

• Step 2: Draw a circle about T that passes through F2, and let D denote the other end of that circle’s diameter

F1 F2

TD

How? (continued)

• Step 3: Locate the midpoint M of the line-segment joining F2 and D

F1 F2

TD

M

How? (continued)

• Step 4: Construct the line through M and T (it will be the ellipse’s tangent-line at T, even if it doesn’t look like it in this picture)

F1 F2

TD

M

tangent-line

Proof that it’s a tangent

• Observe that line MT is the perpendicular bisector of segment DF2 (because ΔTDF2 will be an isosceles triangle)

F1 F2

TD

M

tangent-line

Proof (continued)

• So every other point S that lies on the line through points M and T will not obey the ellipse requirement for sum-of-distances

F1 F2

TD

M

tangent-line

S

SF1 + SF2 > TF1 + TF2 (because SF2 = SD and TF2 = TD )

Why are these ideas relevant?

• When we encounter some other methods that purport to produce tangent-lines to these curves, we will now have a reliable way to check that they really do work!