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 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 axis focus parabola

How do we do it? Step 1: Drop a perpendicular from T to the parabola’s directrix; denote its foot by A directrix axis focus parabola T A F

How? (continued) Step 2: Locate the midpoint M of the line- segment joining A to the focus F directrix axis focus parabola T A 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 axis focus parabola T A 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 axis focus parabola T A F M tangent-line TF = TA because T is on 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 axis focus parabola T A F M S B SB < SA since SA is hypotenuse of right-triangle ΔSAB SA = SF because 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 F1F2

How do we do it? Step 1: Draw a line through the point T and through one of the two foci, say F1 F1F2 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 F1F2 T D

How? (continued) Step 3: Locate the midpoint M of the line- segment joining F2 and D F1F2 T D 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) F1F2 T D 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) F1F2 T D 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 F1F2 T D 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!