Question

If tangents are drawn from points on the hyperbola $\frac{x^2}{4}-\frac{y^2}{9}=1$ to the circle $x^2+y^2=4,$ then the locus of the mid-point of the chord of contact is

• A. $x^2+y^2=\frac{x^2}{9}-\frac{y^2}{4}$
• B. $(x^2+y^2{)}^2=\frac{x^2}{4}-\frac{y^2}{9}$
• C. $(x^2+y^2{)}^2=16(\frac{x^2}{4}-\frac{y^2}{9})$
• D. $(x^2+y^2{)}^2=9(\frac{x^2}{9}-\frac{y^2}{4})$

Answer 1

The correct option is C $(x^2+y^2{)}^2=16(\frac{x^2}{4}-\frac{y^2}{9})$

Given hyperbola is $\frac{x^2}{4}-\frac{y^2}{9}=1$
Let any point on hyperbola be
$P(2\sec \theta ,3\tan \theta )$
Given circle is $x^2+y^2=4$
Chord of contact of to a given circle from this point is given by $T=0$
$\Rightarrow 2\sec \theta x+3\tan \theta y=4\cdots \left(1\right)$

If $M(h,k)$ is the mid point of the chord, then equation of chord of contact with $M$ as its mid point is given by
$T=S_1\Rightarrow x\left(h\right)+y\left(k\right)=h^2+k^2\cdots \left(2\right)$

Equation $\left(1\right)$ and $\left(2\right)$ represent the same chord of contact.
Hence, $\frac{2\sec \theta }{h}=\frac{3\tan \theta }{k}=\frac{4}{h^2+k^2}$
$\Rightarrow \sec \theta =\frac{4h}{2(h^2+k^2)},\tan \theta =\frac{4k}{3(h^2+k^2)}$

We know that,
${\sec }^2\theta -{\tan }^2\theta =1\Rightarrow \frac{16h^2}{4(h^2+k^2{)}^2}-\frac{16k^2}{9(h^2+k^2{)}^2}=1\Rightarrow \frac{h^2}{4}-\frac{k^2}{9}=\frac{(h^2+k^2{)}^2}{16}$

Hence, the locus of the mid point $M(h,k)$ of chord of contact is
$(x^2+y^2{)}^2=16(\frac{x^2}{4}-\frac{y^2}{9})$