Question

The sides $AC$ and $AB$ of a $\Delta ABC$ touch the conjugate hyperbola of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. If the vertex $A$ lies on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, then the side $BC$ must touch :

• A. parabola
• B. circle
• C. hyperbola
• D. ellipse

Answer 1

Answer: (D)

ellipse

Let the vertex $A$ be $(a\cos \theta ,b\sin \theta )$
Since $AC$ and $AB$ touch the hyperbola
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=-1$
$BC$ is the chord of contact. Its equation is
$\frac{x\cos \theta }{a}-\frac{y\sin \theta }{b}=-1$
or $-\frac{x\cos \theta }{a}+\frac{y\sin \theta }{b}=1$
or $\frac{x\cos (\pi -\theta )}{a}+\frac{y\sin (\pi -\theta )}{b}=1$
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.