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UCR MATH 133 - ANGENTS TO AN ELLPISE THROUGH AN EXTERNAL POINT

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TANGENTS TO AN ELLPISE THROUGH AN EXTERNAL POINTWe have noted that the concept of geometrical transformation was implicit in theclassical Greek idea of proof by superposition. Our purpose here is to show how suchtransformations can yield relatively simple proofs of some geometrical results.BACKGROUND MATERIAL. As the title suggests, we are interested in questions involvingtangent vectors to ellipses in R2. As such, we shall use the standard description of tangentsin terms of differential calculus, and we shall also use the representability of ellipses andother curves by parametric equations. The starting point is the following observation,which can be proved by direct calculation:PROPOSITION. Let y(t) be a parametrized curve defined on an open interval (c, d)with values in R2, and assume that y(t) has a continuous derivative y0(t). Suppose furtherthat we are given an affine transformation of R2defined by F (x) = Ax + b, where A isan invertible 2 × 2 matrix and b ∈ R2, and let z be the curve z(t) = Fy(t). Then wehave z0(t) = A y0(t).Verification is left as an exercise to the reader; one can use the standard coordinateforms of vectors and matrices in which vectors are viewed as 2 × 1 matrices, the vector whas coordinates (w1, w2), and the linear transformation associated to A sends this vectorto the one with coordinatesa1,1w1+ a1,2w2, a2,1w1+ a2,2w2.Here is a statement of the main result.THEOREM. Let Γ ⊂ R2be an ellipse, and let p ∈ R2be an external point. Then thereare two tangents to Γ which pass through p.Our proof will use the results in Section V.2 (pp. 80–83) of the following online linearalgebra notes:http://math.ucr.edu/∼res/math132/linalgnotes.pdfIn particular, these results imply that if Γ is an arbitrary ellipse in the coordinateplane, then there is an affine transformation (in fact, a rigid motion or Galilean transfor-mation) F such that F maps Γ to a standard ellipse Γ1of the formx2a2+y2b2= 1where a, b > 0. We have not been precise about the notion of external point for Γ, butone way to describing the external points of this ellipse is to say that they are the pointswhich map under F to the obvious external set of Γ1consisting of all points for whichx2a2+y2b2> 1 .1By the proposition, it is enough to prove the theorem for ellipses defined by the standardequations as above.One can take this another step further and reduce the proof to the case where Γ1isthe standard unit circle Γ0defined by x2+ y2= 1, for if G is the affine transformationsuch thatG(w1, w2) = G(w1/a1, w2/a2)then G maps the previously described Γ1to Γ0and sends external points for the curve Γ1into external points for Γ0.Thus we have reduced everything to a special case of a basic result in classical Eu-clidean geometry: If we are given a circle C and an external point p, then there are exactlytwo tangents from p to C.For the sake of completeness, here is a proof using vectors. It requires to pieces ofbackground information:(1) Suppose we are given ∆ABC in the plane such that the midpoint D of [AC] isequidistant from the three vertices. Then6ABC is a right angle.(2) If C is a circle and L is a line through the point p ∈ C, then L is tangent to Cat p if and only if L is perpendicular to the line joining p to the center of C.Both of these can be proven using vectors. In the first case, one starts with theequations D =12(A + C) and |D − A|2= |D − B|2= |D − C|2and uses them to showthat the dot product hA − B, C − Bi is equal to zero. In the second case, one has aparametrization x(t) such that |x|2= r2for some fixed r > 0, and simple differentiationshows that 2 x(t) · x0(t) = 0, so that direction vectors for the two lines — which are x0(t)for the tangent and x(t) for the line through the center — must be perpendicular.It follows (as in a classical Greek geometrical construction) that if the circle |x|2= r2meets the circle |x−p|2= |p|2in two points q1and q2, then the lines pxiare perpendicularto the lines 0xiby (1) above, and hence by (2) the lines pxiare tangent to the circle atthe points xi. To complete the proof, we need to show that the system of equations|w − p|2= |p|2, |w|2= r2has two solutions. Our assumption on p means that |p|2> r2, and hence if we write p =(a, b) then after some algebraic manipulation we obtain a system of two scalar equations:w21+ w22= r2, 2(a w1+ b w2) = r2We can solve the second equation for one of w1or w2in terms of the other coordinatebecause a and b are not both zero, and if we substitute the resulting expression into thefirst equation we obtain a quadratic equation in one of the coordinates. The condition|p|2= a2+ b2> r2implies that this equation has two real roots, and it follows that thereare exactly two points which lie on the two


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