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If a tangent to the ellipse $x^{2}+4 y^{2}=4$ meets the tangents at the extremities of its major axis at $B$ and $\mathrm{C}$ , then the circle with $\mathrm{BC}$ as diameter passes through the point
Option: 1 $(\sqrt{3}, 0)$
Option: 2 $(\sqrt{2}, 0)$
Option: 3 $(1,1)$
Option: 4 $(-1,1)$

Answers (1)

best_answer

$\frac{x^{2}}{4}+\frac{y^{2}}{1}= 1\Rightarrow a^{2}= 4,b^{2}= 1$

Let point of Ellipse be $P\left ( 2\cos \theta ,\sin \theta \right )$

Equation of tangent : T = 0

$\frac{x}{4}\cdot 2\cos \theta +y\cdot \sin \theta = 1$

For point C, put x = 2 in this equation

$y= \frac{1-\cos \theta }{\sin \theta }= \tan \frac{\theta }{2}$

$C\left ( 2,\tan \frac{\theta }{2} \right )$ $\\Similarly \: B\left ( -2,\cot \frac{\theta }{2} \right )\, \\\\Equation\; of\, circle\, with\, BC\, as\, diameter$
$\left ( x+2 \right )\left ( x-2 \right )+\left ( y-\cot \frac{\theta }{2} \right )\left ( y-\tan \frac{\theta }{2} \right )= 0$
Cheking options, 1st option satisfies it

Posted by

Kuldeep Maurya

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