Question

Tangents are drawn from points on the hyperbola $\frac{x^2}{4}-\frac{y^2}{9}=1$ to circle $x^2+y^2=4$. 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. $(x^2+y^2{)}^2=16\left(\frac{x^2}{4}-\frac{y^2}{9}\right)$
• D. $(x^2+y^2{)}^2=9\left(\frac{x^2}{9}-\frac{y^2}{4}\right)$

Answer 1

Answer: (C)

$(x^2+y^2{)}^2=16\left(\frac{x^2}{4}-\frac{y^2}{9}\right)$

A point on the hyperbola is $(2\sec \theta ,3\tan \theta )$

Thus equation of chord of contact at this point is, $T=0\Rightarrow 2\sec \theta x+3\tan \theta y=4$ ...(1)

If $(x_1,y_1)$ is mid-point of chord then its equation is

$T=S_1\Rightarrow x{x}_1+y{y}_1=x_1^2+y_1^2$ ...(2)

(1) and (2) are identical

$\frac{2\sec \theta }{x_1}=\frac{3\tan \theta }{y_1}=\frac{4}{x_1^2+y_1^2}$

$\Rightarrow \sec \theta =\frac{4x_1}{2(x_1^2+y_1^2)}$

And $\tan \theta =\frac{4y_1}{3(x_1^2+y_1^2)}$

Eliminating $\theta$,

$1=\frac{16x_1^2}{4(x_1^2+y_1^2{)}^2}-\frac{16y^2}{9(x_1^2+y_1^2{)}^2}$

$\Rightarrow \frac{(x_1^2+y_1^2{)}^2}{16}=\frac{x_1^2}{4}-\frac{y_1^2}{9}$

Hence locus is, $(x^2+y^2{)}^2=16\left(\frac{x^2}{4}-\frac{y^2}{9}\right)$