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For $0 < \theta < \frac{\pi}{2}$, four tangents are drawn at the four points $(\pm 3 \cos \theta, \pm 2 \sin \theta)$ to the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$. If $A(\theta)$ denotes the area of the quadrilateral formed by these four tangents, the minimum value of $A(\theta)$ is
$21$
$24$
$27$
$30$
Solution
(b)
We have,
Equation of ellipse
$\frac{x^2}{9}+\frac{y^2}{4}=1$
Equation of $\operatorname{tangent}$ at $(3 \cos \theta, 2 \sin \theta)$ is
$\frac{x}{3} \cos \theta+\frac{y}{2} \sin \theta=1$
Intercept of tangent is $(3 \sec \theta, 0)$ and $(0,2 \operatorname{cosec} \theta)$.
Area of quadrilateral
$=4 \times \cdot \frac{1}{2} 3 \sec \theta \times 2 \operatorname{cosec} \theta$
$=12 \sec \theta \operatorname{cosec} \theta=\frac{24}{\sin 2 \theta}$
Minimum area $=24$
