Question

If the curve ${\left(\frac{x}{a}\right)}^n+{\left(\frac{y}{b}\right)}^n=2$ touches the straight line $\frac{x}{a}+\frac{y}{b}=2$, then find the value of $n$.

• A. $2$
• B. $3$
• C. $4$
• D. any real number

Answer 1

The correct option is D any real number

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

Differentiating both sides w.r.t $x,$ we get

$\frac{n}{a}{\left(\frac{x}{a}\right)}^{n-1}+\frac{b}{b}{\left(\frac{y}{b}\right)}^{n-1}\times \frac{dy}{dx}=0$

$\Rightarrow \frac{dy}{dx}=-\frac{n}{a}{\left(\frac{x}{a}\right)}^{n-1}\times \frac{b}{a}{\left(\frac{b}{y}\right)}^{n-1}$

$\therefore \frac{dy}{dx}$ at $(a,b)=\frac{b}{a}$

$\therefore$ Tangent is $y-b=-\frac{b}{a}(x-a)\Rightarrow bx+ay=2ab\Rightarrow \frac{x}{a}+\frac{y}{b}=2$

for all values of $n$ $(\because \frac{dy}{dx}$ is independent of $n)$