Question

Tangent drawn from point $(c,d)$ to the hyperbola $\frac{x^2}{25}-\frac{y^2}{16}=1$ make angles $\alpha$ and $\beta$ with the $x-$ axis. If $\tan \alpha \tan \beta =1$, then the value of $c^2-d^2$

Answer 1

Any tangent to hyperbola is
$y=mx\pm \sqrt{m^2{a}^2-b^2}$
Since, its passes through $(c,d)$
$\therefore d=mc\pm \sqrt{a^2{m}^2-b^2}$
$\Rightarrow (d-mc{)}^2=a^2{m}^2-b^2$
$\Rightarrow (c^2-a^2)m^2-2cdm+b^2+d^2=0$
Product of roots
$m_1{m}_2=\tan \alpha \tan \beta \Rightarrow \frac{d^2+b^2}{c^2-a^2}=1$
$\Rightarrow d^2+b^2=c^2-a^2$
$\Rightarrow c^2-d^2=a^2+b^2=25+16=41$