Answer:
Option B
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
$\Rightarrow$ $\frac{bx}{ab}+\frac{ay}{ab}= \frac{2ab}{ab}$
$\Rightarrow$ $\frac{x}{a}+\frac{y}{b}=2$
