Question

Tangents are drawn from the points on a tangent of the hyperbola $x^2-y^2=a^2$ to the parabola $y^2=4ax$. If all the chords of contact pass through a fixed point $Q$, then the locus of the point $Q$ for different tangents on the hyperbola is

• A. $\frac{x^2}{a^2}+\frac{y^2}{4a^2}=1$
• B. $\frac{x^2}{a^2}+\frac{y^2}{3a^2}=1$
• C. $\frac{x^2}{a^2}-\frac{y^2}{4a^2}=1$
• D. $\frac{x^2}{a^2}-\frac{y^2}{3a^2}=1$

Answer 1

The correct option is A $\frac{x^2}{a^2}+\frac{y^2}{4a^2}=1$

Tangent at a point $(a\sec \theta ,a\tan \theta )$ on the hyperbola $x^2-y^2=a^2$ is
$x\sec \theta -y\tan \theta =a\cdots \left(1\right)$
Any point on $\left(1\right)$ will be of the form
$(t,\frac{t\sec \theta -a}{\tan \theta })$
Equation of chord of contact of the point w.r.t parbola $y^2=4ax$ will be
$y\left(\frac{t\sec \theta -a}{\tan \theta }\right)-2a(x+t)=0$
$\Rightarrow (\frac{-ay}{\tan \theta }-2ax)+t(\frac{y\sec \theta }{\tan \theta }-2a)=0\cdots \left(2\right)$
Here $\left(2\right)$ represents family of straight lines each member of which passes through the point of intersection $Q(\alpha ,\beta )$ of straight lines
$-\frac{ay}{\tan \theta }-2ax=0\text{and}\frac{y\sec \theta }{\tan \theta }-2a=0$
$\Rightarrow (\alpha ,\beta )\equiv (-a\cos \theta ,2a\sin \theta )$
$\Rightarrow \frac{{\alpha }^2}{a^2}+\frac{{\beta }^2}{4a^2}=1$
So locus of $Q$ is
$\frac{x^2}{a^2}+\frac{y^2}{4a^2}=1$