Question

Find the locus of the poles of normal chords of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$.

• A. $a^2{x}^2-b^2{y}^2=a^4+b^4$
• B. $a^6{y}^2-b^6{x}^2=(a^2+b^2)x^2{y}^2$
• C. $a^2{y}^2-b^2{x}^2=(a^2+b^2)xy$
• D. none of these

Answer 1

The correct option is B $a^6{y}^2-b^6{x}^2=(a^2+b^2)x^2{y}^2$

Suppose $(h,k)$ is the pole of a normal chord of the
given hyperbola.
The equation of the polar of $(h,k)$ w.r.t the given hyperbola is
$\frac{xh}{a^2}-\frac{yk}{b^2}=1$ ...(1)
Also the equation of a normal chord of the given hyperbola is
$ax\cos \theta +by\cot \theta =a^2+b^2$ ...(2)
If (1) and (2) are the equation of the same straight line.
then comparing the coeff of like term in (1) and (2), we have
$\frac{h}{a^{2}a\cos \theta }=-\frac{k}{b^{2}b\cot \theta }=\frac{1}{a^2+b^2}$
$\therefore \sec \theta =\frac{a^3}{(a^2+b^2)h},\tan \theta =\frac{-b^3}{(a^2+b^2)h}$
Using ${\sec }^2\theta -{\tan }^2\theta =1$
we get $\frac{a^6}{(a^2+b^2)h^2}-\frac{b^6}{(a^2+b^2)h^2}=1$
Therefore the locus is
$(\frac{a^6}{x^2}-\frac{b^6}{y^2})={(a^2+b^2)}^2$