Question

Tangents are drawn from any point on the hyperbola $4x^2-9y^2=36$ to the circle $x^2+y^2-9=0$. If the locus of the midpoint of the chord of contact is $\left(\frac{x^2}{9}-\frac{y^2}{4}\right)=\lambda {\left(\frac{x^2+y^2}{9}\right)}^2$, then $\lambda$ is equal to

• A. 1
• B. 4
• C. 9
• D. 16

Answer 1

Answer: (A)

$\frac{x^2}{9}-\frac{y^2}{4}=1,x^2+y^2=9$
Let $P(3\sec \theta ,2\tan \theta )$
From $P$, tangents $PA$ and $PB$ are drawn to circle $x^2+y^2=9$
$\therefore$ Equation of chord of contact $AB$ w.r.t. $P$ is $T=0$
$3x\sec \theta +2y\tan \theta =9$....(i)
Let Mid point of $AB$ is $M(h,k)$
$\therefore$ Equation of chord w.r.t. M
$T=S_1$
$hx+ky-9=h^2+k^2-9$.....(ii)
Compare (i) and (ii)
$T=S_1$
$hx+ky-9=h^2+k^2-9$
$\frac{3\sec \theta }{h}=\frac{2\tan \theta }{k}=\frac{9}{h^2+k^2}\Rightarrow \sec \theta =\frac{h}{3}\left(\frac{9}{h^2+k^2}\right)$
$\text{ and }\tan \theta =\left(\frac{k}{2}\right)\left(\frac{9}{h^2+k^2}\right)$
$\text{ As, }{\sec }^2\theta -{\tan }^2\theta =1$

$\Rightarrow \left(\frac{h^2}{9}-\frac{k^2}{4}\right)={\left(\frac{h^2+k^2}{9}\right)}^2$
$\therefore$ compare with $\left(\frac{x^2}{9}-\frac{y^2}{4}\right)=\lambda {\left(\frac{x^2+y^2}{9}\right)}^2$
\Rightarrow \lambda=1 $