Let the major axis of an ellipse be $AB$, let $O$ be its center, and let $F$ be one of its foci. $P$ is a point on the ellipse, and $CD$ a chord through $O$, such that $CD$ is parallel to the tangent of the ellipse at $P$. $PF$ and $CD$ intersect at $Q$. Compare the lengths of $PQ$ and $OA$.
Problem
Source: Taiwan NMO 2006
Tags: conics, ellipse, analytic geometry, graphing lines, slope, geometry, geometric transformation
Virgil Nicula
22.03.2006 20:02
Let $b^2x^2+a^2y^2=a^2b^2$ be the equation of the ellipse $\Sigma$, where $b^2+c^2=a^2$ and let $F(\epsilon c,0)$ be a focus, where $\epsilon ^2=1$. Let $P(au,bv)\in \Sigma$ be a mobile point, where $u^2+v^2=1$. Then the slope of the tangent $PP$ in the point $P\in \Sigma$ to the ellipse $\Sigma$ is $m_{PP}=-\frac{bu}{av}$. The equation of the line $CD$ is $y=-\frac{bu}{av}\cdot x$ and the equation of the line $CF$ is $y=\frac{bv}{au-\epsilon c}\cdot (x-\epsilon c)$. Thus, for the intersection point $Q\in CD\cap PF$ we have $x_Q=\frac{\epsilon acv^2}{a-\epsilon cu}$, $y_Q=-\frac{\epsilon bcuv}{a-\epsilon cu}$ and $\left(\frac{a-\epsilon cu}{a}\right)^2\cdot PQ^2=$ $\left[u(a-\epsilon cu)-\epsilon cv^2\right]^2+b^2v^2=$ $(au-\epsilon c)^2+b^2v^2=$ $a^2u^2+b^2(1-u^2)+c^2-2\epsilon acu=$ $(a^2-b^2)u^2+a^2-2\epsilon acu=$ $c^2u^2+a^2-2\epsilon acu=$ $(a-\epsilon cu)^2\Longrightarrow PQ=a$, i.e. $PQ=OA$.
nsato
23.03.2006 09:41
Let the foci be $F_1$ and $F_2$, and let $Q_1 = PF_1 \cap CD$ and $Q_2 = PF_2 \cap CD$. Let $M$, $R_1$, and $R_2$ be the projections of $P$, $F_1$, and $F_2$ onto $CD$, respectively. First, we claim that triangles $PMQ_1$ and $PMQ_2$ are congruent. We are given that $CD$ is parallel to the tangent at $P$, so by the reflection property of ellipses, $\angle MPQ_1 = \angle MPQ_2$. Also, $\angle PMQ_1 = \angle PMQ_2 = 90^\circ$, and both triangles share side $PM$, so they are congruent. Therefore, $PQ_1 = PQ_2$. Next, we claim that triangles $F_1 R_1 Q_1$ and $F_2 Q_2 R_2$ are congruent. Since $CD$ passes through the centre of the ellipse, foci $F_1$ and $F_2$ are equidistant from $CD$, i.e. $F_1 R_1 = F_2 R_2$. By the argument above, $\angle R_1 F_1 Q_1 = \angle R_2 F_2 Q_2$, and $\angle F_1 R_1 Q_1 = \angle F_2 R_2 Q_2 = 90^\circ$, so the two triangles are congruent. Therefore, $F_1 Q_1 = F_2 Q_2$. But, $PF_1 + PF_2 = AB$, and \begin{eqnarray*} PF_1 + PF_2 &=& F_1 Q_1 + Q_1 P + PQ_2 - Q_2 F_2 \\ &=& Q_1 P + PQ_2 \\ &=& 2 PQ_1 = 2PQ_2, \end{eqnarray*} so $PQ_1 = PQ_2 = AB/2 = OA$.