Ellipse
Tangent from external point — area of triangle
MJAT_TS7_P1
Grade 12
Question:
Consider the ellipse $\dfrac{x^2}{9}+\dfrac{y^2}{4}=1$. Let $S(p,q)$ be in the first quadrant with $\dfrac{p^2}{9}+\dfrac{q^2}{4}>1$. Two tangents from $S$: one at a minor axis endpoint, the other at $T$ in the fourth quadrant. $R$ = vertex with positive $x$-coordinate, $O$ = center. If area of $\triangle ORT=\dfrac{3}{2}$, then:
A) $q=2,\; p=3\sqrt{3}$
B) $q=2,\; p=4\sqrt{3}$
C) $q=1,\; p=5\sqrt{3}$
D) $q=1,\; p=6\sqrt{3}$
Step-by-Step Solution
Key Concept: Tangent at minor axis endpoint $(0,2)$: equation is $y=2$. The point $S(p,q)$ lies on this tangent so $q=2$. The tangent to ellipse at $(3\cos\theta,-2\sin\theta)$: $\frac{x\cos\theta}{3}-\frac{y\sin\theta}{2}=1$. $R=(3,0)$. Area of $\triangle ORT=\frac{1}{2}|3|\cdot|y_T|=\frac{3}{2}\Rightarrow|y_T|=1$.
$q=2$, $p=3\sqrt{3}$. Answer: **A**.
Correct Answer: A