Question

The line $x\cos \alpha +y\sin \alpha =p$ touches the hyperbola
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ if

• A. $a^2{\cos }^2\alpha -b^2{\sin }^2\alpha =p^2$
• B. $a^2{\cos }^2\alpha -b^2{\sin }^2\alpha =p$
• C. $a^2{\cos }^2\alpha -b^2{\cos }^2\alpha =2p$
• D. $a^2{\cos }^2\alpha +b^2{\sin }^2\alpha =p$

Answer 1

The correct option is A

$a^2{\cos }^2\alpha -b^2{\sin }^2\alpha =p^2$

A given line $y=mx+c$ touches hyperbola when,

$c=\sqrt{a^2{m}^2-b^2}$ . . . (1)

Given line is,$y=-x\cot \alpha +p\text{cosec}\alpha ...\left(2\right)$
(1) and (2) gives

$p\text{cosec}\alpha =\sqrt{a^2{\cot }^2\alpha -b^2}$

$p^2{\text{cosec}}^2\alpha =a^2\frac{{\cos }^2\alpha }{{\sin }^2\alpha }-b^2$

$p^2=a^2{\cos }^2\alpha -b^2{\sin }^2\alpha$