Discussion Forum

If $x=a\left(t-\frac{1}{t}\right),y=a\left(t+\frac{1}{t}\right)$   , where t is the parameter , then $\frac{dy}{dx}$= ?

Answer:

Explanation:

Given , $x=a\left(t-\frac{1}{t}\right),y=a\left(t+\frac{1}{t}\right)$

Now,  $y^{2}-x^{2}=a^{2}\left[\left(t+\frac{1}{t}\right)^{2}-a^{2}\left(t-\frac{1}{t}\right)^{2}\right]$

 = $a^{2}\left[t^{2}+\frac{1}{t^{2}}+2-t^{2}-\frac{1}{t^{2}}+2\right]$

$\Rightarrow$  $y^{2}-x^{2}=4a^{2}$

On differentiating  both sides w.r.t 'x' , we get

    2y $\frac{dy}{dx}$-2x=0

$\Rightarrow$    $2\left( y\frac{dy}{dx}-x\right)=0$

$\therefore$     $\frac{dy}{dx}=\frac{x}{y}$