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(B) The equation of the ellipse is $\frac{x^2}{9}+\frac{y^2}{4}=1$.The equation of the tangent at the point $(3 \cos 	heta, 2 \sin 	heta)$ is $\frac

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$24$

Vedclass · Answer

(B) The equation of the ellipse is $\frac{x^2}{9}+\frac{y^2}{4}=1$.The equation of the tangent at the point $(3 \cos 	heta, 2 \sin 	heta)$ is $\frac{x}{3} \cos 	heta + \frac{y}{2} \sin 	heta = 1$.The intercepts of this tangent on the coordinate axes are $(3 \sec 	heta, 0)$ and $(0, 2 \operatorname{cosec} 	heta)$.The four tangents form a rhombus with vertices at $(\pm 3 \sec 	heta, 0)$ and $(0, \pm 2 \operatorname{cosec} 	heta)$.The area of this quadrilateral (rhombus) is given by $A(	heta) = 4 	imes \left( \frac{1}{2} 	imes |3 \sec 	heta| 	imes |2 \operatorname{cosec} 	heta| ight)$.$A(	heta) = 12 \sec 	heta \operatorname{cosec} 	heta = \frac{12}{\sin 	heta \cos 	heta} = \frac{24}{2 \sin 	heta \cos 	heta} = \frac{24}{\sin 2 	heta}$.Since $0 Therefore,the minimum value of $A(	heta) = \frac{24}{1} = 24$.

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

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