Question

The locus of the poles of the chords of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ which subtend a right angle at the centre is

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

Answer 1

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

Let $P(x_1,y_1)$ be the pole.

Equation of hyperbola will be $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Then the equation of polar is $\frac{hx}{a^2}-\frac{ky}{b^2}=1$

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=\frac{hx}{a^2}-\frac{ky}{b^2}$

Since the lines are perpendicular then the coefficient of $x^2$ $+$ coefficient of $y^2$ ie equal to zero.

$\frac{1}{a^2}-\frac{h^2}{a^4}-\frac{1}{b^2}-\frac{k^2}{b^4}=0$

Hence,

$\frac{x^2}{a^4}+\frac{y^2}{b^4}=\frac{1}{a^2}-\frac{1}{b^2}$