Question

Find the locus of the mid points of the chords of the circle $x^2+y^2=16$ which are tangents to the hyperbola $9x^2-16y^2=144$.

• A. $(x^2-y^2{)}^2=16x^2+9y^2$
• B. $(x^2+y^2{)}^2=16x^2+9y^2$
• C. $(x^2+y^2{)}^2=16x^2-9y^2$
• D. $(x^2-y^2{)}^2=16x^2-9y^2$

Answer 1

The correct option is C $(x^2+y^2{)}^2=16x^2-9y^2$

Given hyperbola may be written as, $\frac{x^2}{16}-\frac{y^2}{9}=1$
So equation of tangent to this curve at point $\theta$ is,
$\frac{x\sec \theta }{4}-\frac{y\tan \theta }{3}=1..\left(1\right)$

Now, let mid point of the chord of circle $x^2+y^2=16$ be $P(h,k)$
So equation of chord is given by,
$hx+ky=h^2+k^2.......\left(2\right)$

But both line (1) and (2) are same,
$\frac{\sec \theta }{4h}=\frac{-\tan \theta }{3k}=\frac{1}{h^2+k^2}$

$\Rightarrow \sec \theta =\frac{4h}{h^2+k^2},\tan \theta =\frac{-3k}{h^2+k^2}$

Eliminating $\theta$ we get,
${\left(\frac{4h}{h^2+k^2}\right)}^2-{\left(\frac{-3k}{h^2+k^2}\right)}^2=1$

Hence, required locus of $P(h,k)$ is given by, $16x^2-9y^2=(x^2+y^2{)}^2$