Consider the ellipse . Let be a point in the first quadrant such that . Two tangents are drawn from 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 in the fourth quadrant. Let

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 $O$ be the center of the ellipse. If the area of the triangle $∆ O R T$ is $\frac{3}{2}$ , then which of the following options is correct? [2024]

Smart Achievers · Accepted Answer

(1)

$Given that \frac{p^{2}}{9} + \frac{q^{2}}{4} > 1 then S (p, q) lies outside the ellipse$

$SA is tangent \Rightarrow q = 2$

$S$

$∆ O R T$ $| \frac{1}{2} \times O R \times Q T |$ $= \frac{3}{2} \Rightarrow$ $| \frac{1}{2} \times 3 \times β |$ $= \frac{3}{2} \Rightarrow β = - 1$

$∴$ $\frac{α^{2}}{9} + \frac{β^{2}}{4} = 1 \Rightarrow \frac{α^{2}}{9} = 1 - \frac{1}{4} = \frac{3}{4}$

$\Rightarrow α^{2} = \frac{27}{4} \Rightarrow α = \frac{3 \sqrt{3}}{2}$

$Tangent at T$

$T = 0$

$\frac{x \cdot \frac{3 \sqrt{3}}{2}}{9} + \frac{y (- 1)}{4} = 1 |_{(p, 2)}$

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

$∴$ $p = 3 \sqrt{3}, q = 2$

Solution

1 $q = 2, p = 3 \sqrt{3}$

2 $q = 2, p = 4 \sqrt{3}$

3 $q = 1, p = 5 \sqrt{3}$

4 $q = 1, p = 6 \sqrt{3}$

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Solution

1 q=2, p=33

2 q=2, p=43

3 q=1, p=53

4 q=1, p=63

Ans. (1) Given that p29+q24>1 then S(p,q) lies outside the ellipse SA is tangent⇒q=2 Area of ∆ORT=32 ⇒|12×OR×QT|=32⇒|12×3×β|=32⇒β=-1 ∴ α29+β24=1⇒α29=1-14=34 ⇒α2=274⇒α=332 Tangent at T T=0 x·3329+y(-1)4=1|(p,2) ⇒p36-12=1⇒p36=32⇒p=33 ∴ p=33, q=2

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1 $q = 2, p = 3 \sqrt{3}$

2 $q = 2, p = 4 \sqrt{3}$

3 $q = 1, p = 5 \sqrt{3}$

4 $q = 1, p = 6 \sqrt{3}$