Discussion Forum

The equation of tangent to the curve 

$\left(\frac{x}{a}\right)^{n}+\left(\frac{y}{b}\right)^{n}=2$  at the point (a,b)  is 

Answer:

Explanation:

We have,

$\left(\frac{x}{a}\right)^{n}+\left(\frac{y}{b}\right)^{n}=2$

on differentiating w.r.t x, we get

       $\frac{nx^{n-1}}{a^{n}}+\frac{ny^{n-1}}{b^{n}}\frac{dy}{dx}=0$

$\Rightarrow$        $\frac{dy}{dx}=\frac{-b^{n}x^{n-1}}{a^{n}y^{n-1}}$

$\Rightarrow$             $\left(\frac{dy}{dx}\right)_{(a,b)}=\frac{-b^{n}a^{n-1}}{a^{n}b^{n-1}}=\frac{-b}{a}$

 Equations of tangent (a,b) is 

  $y-b=\frac{-b}{a}(x-a)$

$\Rightarrow$      ay-ab=-bx+ab

$\Rightarrow$    bx+ay=2ab

$\Rightarrow$    $\frac{bx}{ab}+\frac{ay}{ab}= \frac{2ab}{ab}$

$\Rightarrow$   $\frac{x}{a}+\frac{y}{b}=2$