Ellipse and Tangent Lines
DAILY_CHALLENGE
Grade None
Question:
Consider the ellipse $\dfrac{x^2}{9}+\dfrac{y^2}{4}=1$. Let $S(p,q)$ be a point in the first quadrant such that $\dfrac{p^2}{9}+\dfrac{q^2}{4}>1$. Two tangents are drawn from $S$ to the ellipse, of which one meets the ellipse at one end point of the minor axis and the other meets the ellipse at a point $T$ in the fourth quadrant. Let $R$ be the vertex of the ellipse with positive $x$-coordinate and $O$ be the center of the ellipse. If the area of the triangle $\triangle ORT$ is $\dfrac{3}{2}$, then which of the following options is correct?
$q=2,\; p=3\sqrt{3}$
$q=2,\; p=4\sqrt{3}$
$q=1,\; p=5\sqrt{3}$
$q=1,\; p=6\sqrt{3}$
Step-by-Step Solution
Key Concept: Tangent at minor axis endpoint is horizontal; use area condition to find T; tangent at T gives S
Minor axis endpoints: $(0,\pm2)$. The tangent at $(0,2)$ is $y=2$. For $S(p,q)$ to lie on this tangent: $q=2$.
$R=(3,0)$, $O=(0,0)$. $T=(x_T,y_T)$ with $y_T<0$ on the ellipse.
Area of $\triangle ORT=\dfrac{1}{2}\cdot|OR|\cdot|y_T|=\dfrac{1}{2}\cdot3\cdot|y_T|=\dfrac{3}{2}\Rightarrow|y_T|=1\Rightarrow y_T=-1$.
$T$ on ellipse: $\dfrac{x_T^2}{9}+\dfrac{1}{4}=1\Rightarrow x_T^2=\dfrac{27}{4}\Rightarrow x_T=\dfrac{3\sqrt{3}}{2}$ (fourth quadrant).
Tangent at $T$: $\dfrac{(3\sqrt{3}/2)x}{9}+\dfrac{(-1)y}{4}=1\Rightarrow\dfrac{\sqrt{3}x}{6}-\dfrac{y}{4}=1$.
$S(p,2)$ on this tangent: $\dfrac{\sqrt{3}p}{6}-\dfrac{1}{2}=1\Rightarrow\dfrac{\sqrt{3}p}{6}=\dfrac{3}{2}\Rightarrow p=\dfrac{9}{\sqrt{3}}=3\sqrt{3}$.
Correct Answer: A