Question

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 :

• A. $(\sqrt{3},0)$
• B. $(\sqrt{2},0)$
• C. $(1,1)$
• D. $(-1,1)$

Answer 1

Answer: (A)

$\frac{x^2}{4}+\frac{y^2}{1}=1$
Equation of tangent is
$(\cos \theta )x+2\sin \theta y=2B\left(-2,\frac{1+\cos \theta }{\sin \theta }\right),C\left(2,\frac{1-\cos \theta }{\sin \theta }\right)$
$B\left(-2,\cot \frac{\theta }{2}\right)C\left(2,\tan \frac{\theta }{2}\right)$
Equation of circle is
$(x+2)(x-2)+\left(y-\cot \frac{\theta }{2}\right)\left(y-\tan \frac{\theta }{2}\right)=0$
$x^2-4+y^2-\left(\tan \frac{\theta }{2}+\cot \frac{\theta }{2}\right)y+1=0$
so, $(\sqrt{3},0)$ satisfying option (1)