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

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Question

$\frac{x^{2}}{4}+\frac{y^{2}}{1}=1 \Rightarrow a=2 \& b=1, \quad$ Let $P(2 \cos 	heta, \sin 	heta)$

Equation of tangent at $P$ is $(\cos 	he

Vedclass · Accepted Answer

$(\sqrt{3}, 0)$

Vedclass · Answer

$\frac{x^{2}}{4}+\frac{y^{2}}{1}=1 \Rightarrow a=2 \& b=1, \quad$ Let $P(2 \cos 	heta, \sin 	heta)$

Equation of tangent at $P$ is $(\cos 	heta) x+2 \sin 	heta y=2$

$\mathrm{B}\left(-2, \frac{1+\cos 	heta}{\sin 	heta}ight)$

$C\left(2, \frac{1-\cos 	heta}{\sin 	heta}ight)$

$B\left(-2, \cot \frac{	heta}{2}ight)$

$C\left(2, 	an \frac{	heta}{2}ight)$

Equation of circle is

$(x+2)(x-2)+\left(y-\cot \frac{	heta}{2}ight)\left(y-	an \frac{	heta}{2}ight)=0$

$x^{2}-4+y^{2}-\left(	an \frac{	heta}{2}+\cot \frac{	heta}{2}ight) y+1=0$

so, $(\sqrt{3}, 0)$ satisfying equation $(1)$

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

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