Question

The locus of the pole with respect to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ of any tangent to the circle, whose diameter is the line joining the foci, is

• A. $x^2+y^2=a^2-b^2$
• B. $\frac{x^2}{a^4}+\frac{y^2}{b^4}=\frac{1}{a^2+b^2}$
• C. $\frac{x^2}{a^4}+\frac{y^2}{b^4}=1$
• D. $\frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{1}{a^2+b^2}$

Answer 1

The correct option is B $\frac{x^2}{a^4}+\frac{y^2}{b^4}=\frac{1}{a^2+b^2}$

Let the pole be $P(x_1,y_1)$.
Poiar of $P$ wrt $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $\frac{x{x}_1}{a^2}-\frac{y{y}_1}{b^2}=1$
$\Rightarrow \left(b^2{x}_1\right)x-\left(a^2{y}_1\right)y=a^2{b}^2$ ----(1)
Now, Above eqaution is tangent to $x^2+y^2=a^2{e}^2=a^2+b^2$
A line $lx+my+n=0$ is tangent to $x^2+y^2=r^2$ is $n^2=r^2(l^2+m^2)$
$\Rightarrow a^4{b}^4=(a^2+b^2)(b^4{x}_1^2+a^4{y}_1^2)$
$\Rightarrow \frac{x_1^2}{a^4}+\frac{y_1^2}{b^4}=\frac{1}{a^2+b^2}$
$\therefore$ Required locus is $\frac{x^2}{a^4}+\frac{y^2}{b^4}=\frac{1}{a^2+b^2}$
Hence, option B.