Question

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

• A. $a^{2}co{s}^2\alpha -b^{2}si{n}^2\alpha =p^2$
  
• B. $a^{2}co{s}^2\alpha -b^{2}si{n}^2\alpha =p$
  
• C. $a^{2}co{s}^2\alpha +b^{2}si{n}^2\alpha =p^2$
  
  
• D. $a^{2}co{s}^2\alpha +b^{2}si{n}^2\alpha =p$
  

Answer 1

The correct option is A $a^{2}co{s}^2\alpha -b^{2}si{n}^2\alpha =p^2$


The equation of line is $xcos\alpha +ysin\alpha =p$
$\Rightarrow y=-xcot\alpha +pcosec\alpha$
Now, $c^2=a^2{m}^2-b^2$
$\Rightarrow$ $p^{2}cosec\alpha =a^{2}co{t}^2\alpha -b^2$
$\Rightarrow$ $a^{2}co{s}^2\alpha -b^{2}si{n}^2\alpha =p^2$