Question

The sides $AC$ and $AB$ of a $△ABC$ touches the conjugate hyperbola of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ at $C$ and $B.$ If the vertex $A$ lies on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$ then the side $BC$ always touches

• A. $\text{a parabola}$
• B. $\text{a circle}$
• C. $\text{a hyperbola}$
• D. $\text{an ellipse}$

Answer 1

The correct option is D $\text{an ellipse}$

Let the vertex $A$ be $(a\cos \theta ,b\sin \theta )$
Since $AC$ and $AB$ touches the hyperbola, $\frac{x^2}{a^2}-\frac{y^2}{b^2}=-1$ at $C,B.$
So, $BC$ is the chord of contact whose equation is $T=0$ i.e.,$\frac{x\cos \theta }{a}-\frac{y\sin \theta }{b}=-1$
$\Rightarrow \frac{x\cos \theta }{a}+\frac{y\sin \theta }{b}=1$
which can be written as
$\frac{x\cos (\pi -\theta )}{a}+\frac{y\sin (\pi -\theta )}{b}=1$
which is the equation of the tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at the point $\left(a\cos \right(\pi -\theta ),b\sin (\pi -\theta \left)\right).$

Hence, $BC$ touches the given ellipse.