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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 directrix parabola 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 focus axis directrix parabola How? (continued) • Step 2: Locate the midpoint M of the linesegment joining A to the focus F T A M F focus axis directrix parabola 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) T A tangent-line M F focus axis directrix parabola Proof that it’s a tangent • Observe that line MT is the perpendicular bisector of segment AF (because ΔAFT will be an isosceles triangle) T A tangent-line TF = TA because T is on the parabola M F focus axis directrix 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 SB < SA B since SA is hypotenuse A of right-triangle ΔSAB SA = SF because SA lies on AF’s perpendictlar bisector S T M F focus axis Therefore: SB < SF 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 F1 F2 D How? (continued) • Step 3: Locate the midpoint M of the linesegment 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!