The locus of a variable point whose chord of contact w.r.t. the...

Given,

The equation of the given hyperbola is& $\dfrac{x2}{a2} - \dfrac{y2}{b2} = 1 ⋯ \left(1\right)$

Now let~ \left(h, k\right)$~be the pole of a chord~$PQ$~of the hyperbola</p><p>Then the straight line~$PQ$~is the polar of the point~ \left(h, k\right)$~w.r.t. the hyperbola and so its equation is

\dfrac{xh}{a2} - \dfrac{yk}{b2} =1 ⋯ \left(2\right)$</p><p>Now the center of the hyperbola is the origin, let it be point~$C$,</p><p>Now~making~ \left(1\right)$~homogeneous with the help of~ \left(2\right)$,~the combined equation of~$CP$~and~$CQ$~is</p><p>\dfrac{x2}{a2} - \dfrac{y2}{b2} =$\left(\dfrac{xh}{a2} - \dfrac{yk}{b2}\right)$2 ⋯ \left(3\right)$

Now the chord~$PQ$~subtends a right angle at~ \left(0, 0\right)

Therefore the lines~$CP$~and~$CQ$~given by~ \left(3\right)~are perpendicular and so in the equation~ \left(3\right), we have the coefficient of~$x^{2} +$coefficient of~$y^{2} = 0$

$⇒$\left(\dfrac{1}{a2} - \dfrac{h2}{a4}\right) + \left(\dfrac{- 1}{b2} - \dfrac{k2}{b4}\right) =0$</p><p>$⇒ \dfrac{h2}{a4} + \dfrac{k2}{b4} = \dfrac{1}{a2} - \dfrac{1}{b2}$</p><p>Thus locus is</p><p>$\dfrac{x2}{a4} + \dfrac{y2}{b4} = \dfrac{1}{a2} - \dfrac{1}{b2}$</p></blockquote>$