Question

From the points of the circle $x^2+y^2=a^2$, tangents are drawn to the hyperbola $x^2-y^2=a^2$, then the locus of the middle points of the chords of contact is

• A. $(x^2-y^2{)}^2=a^2(x^2+y^2)$
• B. $(x^2-y^2{)}^2=2a^2(x^2+y^2)$
• C. $(x^2+y^2{)}^2=a^2(x^2-y^2)$
• D. $2(x^2-y^2{)}^2=3a^2(x^2+y^2)$

Answer 1

The correct option is A $(x^2-y^2{)}^2=a^2(x^2+y^2)$

Equation of circle is $x^2+y^2=a^2$ ...(1)
Any point on the circle is $P(a\cos \theta ,a\sin \theta )$
Equation of the chord of contact of the tangent from the point
$P(a\cos \theta ,a\sin \theta )$
to the hyperbola $x^2-y^2=a^2$
is $x\cos \theta -y\sin \theta =a$ ...(2)
Let $m(x_1,y_1)$ be the mid point of the chord of contact (2)
Then its equation ${xx}_1-{yy}_1=x_1^2-y_1^2$ ...(3)
Equation (2) and (3) represent the same i.e chord of contact
Comparing the coeff. of the like terms in (2) and (3), we get
$\frac{\cos \theta }{x^1}=\frac{\sin \theta }{y^{\prime }}=\frac{a}{x_1^2-y_1^2}$
$\Rightarrow \cos \theta =\frac{a{x}_1}{(x_1^2-y_1^2)},\sin \theta =\frac{a{y}_1}{(x_1^2-y_1^2)}$
Now ${\cos }^2\theta +{\sin }^2\theta =1$
$\Rightarrow \frac{a^2{x}_1^2+a^2{y}_1^2}{{(x_1^2-y_1^2)}^2}=1\Rightarrow a^2(x_1^2+y_1^2)={(x_1^2-y_1^2)}^2$
Hence the locus of the mid point $m(x_1,y_1)$ is
$a^2(x^2+y^2)={(x^2-y^2)}^2$