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:

\[ q = 2, \quad p = 3\sqrt{3} \]