If a tangent to the ellipse x^{2}+4y^{2}=4 me | vedclass

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(C) The equation of the ellipse is $\frac{x^{2}}{4} + \frac{y^{2}}{1} = 1$,where $a=2$ and $b=1$.Let the point of tangency be $P(2 \cos 	heta, \sin \

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$(\sqrt{3}, 0)$

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(C) The equation of the ellipse is $\frac{x^{2}}{4} + \frac{y^{2}}{1} = 1$,where $a=2$ and $b=1$.Let the point of tangency be $P(2 \cos 	heta, \sin 	heta)$.The equation of the tangent at $P$ is $\frac{x \cos 	heta}{2} + y \sin 	heta = 1$,or $x \cos 	heta + 2y \sin 	heta = 2$.The tangents at the extremities of the major axis are $x = -2$ and $x = 2$.For $B$,substitute $x = -2$: $-2 \cos 	heta + 2y \sin 	heta = 2 \Rightarrow y = \frac{1 + \cos 	heta}{\sin 	heta} = \cot \frac{	heta}{2}$. So,$B = (-2, \cot \frac{	heta}{2})$.For $C$,substitute $x = 2$: $2 \cos 	heta + 2y \sin 	heta = 2 \Rightarrow y = \frac{1 - \cos 	heta}{\sin 	heta} = 	an \frac{	heta}{2}$. So,$C = (2, 	an \frac{	heta}{2})$.The equation of the circle with $BC$ as diameter is $(x - x_B)(x - x_C) + (y - y_B)(y - y_C) = 0$.$(x + 2)(x - 2) + (y - \cot \frac{	heta}{2})(y - 	an \frac{	heta}{2}) = 0$$x^{2} - 4 + y^{2} - y(	an \frac{	heta}{2} + \cot \frac{	heta}{2}) + 	an \frac{	heta}{2} \cot \frac{	heta}{2} = 0$$x^{2} + y^{2} - y(	an \frac{	heta}{2} + \cot \frac{	heta}{2}) - 3 = 0$.Checking the point $(\sqrt{3}, 0)$: $(\sqrt{3})^{2} + 0^{2} - 0 - 3 = 3 - 3 = 0$. Thus,the circle passes through $(\sqrt{3}, 0)$.

If a tangent to the ellipse $x^{2}+4y^{2}=4$ meets the tangents at the extremities of its major axis at $B$ and $C$,then the circle with $BC$ as diameter passes through the point:

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