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Tangents to Curves

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# Tangents to Curves - PowerPoint PPT Presentation

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’)

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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 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

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

T

D

M

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!