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

The equation of the given hyperbola is
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ ..(1)
Let $(h,k)$ be the pole of a chord $PQ$ of the hyperbola
Then the straight line $PQ$ is the polar of the point $(h,k)$
w.r.t the hyperbola and so its equation is
$\frac{xh}{a^2}-\frac{yk}{b^2}=1$ ...(2)
The center of the hyperbola is the origin.
Let it be the point $C$, making (1) homogeneous with the help of (2),
the combined equation of $CP$ and $CQ$ is
$\frac{x^2}{a^2}-\frac{y^2}{b^2}={(\frac{xh}{a^2}-\frac{yk}{b^2})}^2$ ...(3)
Now the chord $PQ$ subtends a right angle at $(0,0)$
Therefore the lines $CP$ and $CQ$ given by (3)
are perpendicular and so in the equation (3)
We have
the coeff of $x^2$ + coeff of $y^2$ =0
$\Rightarrow (\frac{1}{a^2}-\frac{h^2}{a^4})+(\frac{-1}{b^2}-\frac{k^2}{b^4})=0$
$\Rightarrow \frac{h^2}{a^4}+\frac{k^2}{b^4}=\frac{1}{a^2}-\frac{1}{b^2}$
Thus locus is
$\frac{x^2}{a^4}+\frac{y^2}{b^4}=\frac{1}{a^2}-\frac{1}{b^2}$