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(C) The equation of the tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at point $(a \cos 	heta, b \sin 	heta)$ is $\frac{x \cos 	he

Vedclass · Accepted Answer

$9$

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(C) The equation of the tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at point $(a \cos 	heta, b \sin 	heta)$ is $\frac{x \cos 	heta}{a} + \frac{y \sin 	heta}{b} = 1$.Here,$a^2 = 27 \implies a = 3\sqrt{3}$ and $b^2 = 3 \implies b = \sqrt{3}$.The tangent equation is $\frac{x \cos 	heta}{3\sqrt{3}} + \frac{y \sin 	heta}{\sqrt{3}} = 1$.The coordinates of $A$ (x-intercept) are $(\frac{3\sqrt{3}}{\cos 	heta}, 0)$ and $B$ (y-intercept) are $(0, \frac{\sqrt{3}}{\sin 	heta})$.The area of triangle $OAB$ is $\Delta = \frac{1}{2} 	imes |x_A| 	imes |y_B| = \frac{1}{2} 	imes \frac{3\sqrt{3}}{\cos 	heta} 	imes \frac{\sqrt{3}}{\sin 	heta} = \frac{9}{2 \sin 	heta \cos 	heta} = \frac{9}{\sin 2	heta}$.For the area to be minimum,$\sin 2	heta$ must be maximum,i.e.,$\sin 2	heta = 1$.Thus,$\Delta_{\min} = 9$ sq. units.

If the tangent at a point on the ellipse $\frac{x^2}{27} + \frac{y^2}{3} = 1$ meets the coordinate axes at $A$ and $B,$ and $O$ is the origin,then the minimum area (in sq. units) of the triangle $OAB$ is

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