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## PowerPoint Slideshow about 'Tangents to Curves' - Mercy

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### 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

Q

M

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? (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

tangent-line

C

T

D

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)

S

C

T

D

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

axis

focus

parabola

directrix

How do we do it?

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

T

A

F

axis

focus

parabola

directrix

How? (continued)

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

T

A

M

F

axis

focus

parabola

directrix

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)

tangent-line

T

A

M

F

axis

focus

parabola

directrix

Proof that it’s a tangent

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

tangent-line

T

A

TF = TA

because T is

on the parabola

M

F

axis

focus

parabola

directrix

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

SB < SA

since SA is hypotenuse

of right-triangle ΔSAB

SA = SF

because SA lies on AF’s

perpendictlar bisector

Therefore: SB < SF

B

S

A

T

M

F

axis

focus

directrix

parabola

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

T

F1

F2

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

T

D

F1

F2

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)

T

D

M

F1

F2

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)

T

D

M

F1

F2

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

S

T

D

M

F1

F2

tangent-line

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!

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