Discussion Forum

If 2x-y+1=0,  is a tangent to the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{16}=1$  , then which of the following CANNOT be sides of a right angled triangle?

Answer:

Explanation:

Tangent=2x-y+1=0

Hyperbola= $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{16}=1$

It point = $(a\sec\theta,4\tan\theta)$

$tangent=\frac{x\sec\theta}{a}-\frac{y\tan\theta}{4}=1$

  On comparing we get,

$\sec\theta=-2a$

$\tan\theta=-4$

$\Rightarrow$    $4a^{2}-16=1$

        $a=\frac{\sqrt{17}}{2}$

  Subsuitute the value of a in option (a).(b), (c)and (d)