Mathematics - Hyperbola Question with Solution | TestHub

Let $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$... (i) be the hyperbola, then its conjugate hyperbola is

$\frac{y^2}{b^2}-\frac{x^2}{a^2}=1$........(ii) Let any point on (ii) be ( $(\mathrm{a}\tan \theta ,\mathrm{b}\sec \theta )$, then equation of the tangent to (ii) at this point is $\frac{\mathrm{y}\sec \theta }{\mathrm{b}}-\frac{\mathrm{x}\tan \theta }{\mathrm{a}}=1={\sec }^2\theta -{\tan }^2\theta$ i.e. $\frac{x\tan \theta }{a}-\frac{y\sec \theta }{b}-1=\frac{a^2{\tan }^2\theta }{a^2}-\frac{b^2{\sec }^2\theta }{a^2}-1$ which is the equation of the chord of (i) whose mid point is ( $a\tan \theta ,b\sec \theta$ ). Hence the result