Question

Consider the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$. Let S (p, q) be a point in the first quadrant such that $\frac{p^2}{9}+\frac{q^2}{4}>1$. 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 Obe the center of the ellipse. If the area of the triangle △ORT is $\frac{3}{2}$ , then which of the following options is correct ?

• A. $q=2,p=3\sqrt3$
• B. $q=2,p=4\sqrt3$
• C. $0.5m,-5kg \ \frac{m}{s}$
• D. $-1m,5kg \ \frac{m}{s}$

Answer 1

The Correct Option is A

The equation of the ellipse is: 

$$\frac{x^2}{9} + \frac{y^2}{4} = 1 \quad (1)$$

Given $q = 2$, the area of $\triangle ORT$ is:

$$\text{Area} = \frac{1}{2} \times (OR) \times (OK) = \frac{3}{2}.$$ $$\Rightarrow \frac{1}{2} \times 3 \times k = \frac{3}{2}$$ $$\Rightarrow k = 1.$$

Let $T(h, -1)$ lie on the ellipse (1):

$$\frac{h^2}{9} + \frac{1}{4} = 1$$ $$\Rightarrow \frac{h^2}{9} = \frac{3}{4}$$ $$\Rightarrow h^2 = \frac{9}{4}$$ $$\Rightarrow h = \pm \frac{3\sqrt{3}}{2}.$$

The equation of the tangent at $T \left( \frac{3\sqrt{3}}{2}, -1 \right)$ is:

$$\frac{3\sqrt{3}}{2} \cdot \frac{x}{9} + \frac{-y}{4} = 1$$ $$\Rightarrow \frac{\sqrt{3}}{6} x - \frac{y}{4} = 1.$$

This line passes through $S(p, 2)$:

$$\frac{\sqrt{3}}{6} p - \frac{2}{4} = 1$$ $$\Rightarrow \frac{\sqrt{3}}{6} p = \frac{3}{2}$$ $$\Rightarrow p = \frac{9\sqrt{3}}{2} = 3\sqrt{3}.$$

Thus, $p = 3\sqrt{3}$ and $q = 2$.

Final Answer:

$$q = 2, \quad p = 3\sqrt{3}$$