Question

The equation of the tangent to the curve ${\left(\frac{x}{a}\right)}^n+{\left(\frac{y}{b}\right)}^n=2$ at $(a,b)$ is

• A. $\frac{x}{a}+\frac{y}{b}=2$
• B. $\frac{x}{a}+\frac{y}{b}=\frac{1}{2}$
• C. $\frac{x}{b}-\frac{y}{a}=2$
• D. $ax+by=2$

Answer 1

The correct option is A $\frac{x}{a}+\frac{y}{b}=2$

The given curve is ${\left(\frac{x}{a}\right)}^n+{\left(\frac{y}{b}\right)}^n=2$
$\Rightarrow n{\left(\frac{x}{a}\right)}^{n-1}\cdot \frac{1}{a}+n{\left(\frac{y}{b}\right)}^{n-1}\cdot \frac{1}{b}\frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}=-\frac{b}{a}\frac{{\left(\frac{x}{a}\right)}^{n-1}}{{\left(\frac{y}{b}\right)}^{n-1}}$
At $(a,b)\frac{dy}{dx}=-\frac{b}{a}\frac{(1{)}^{n-1}}{(1{)}^{n-1}}=-\frac{b}{a}$
Therefore, the tangent at $(a,b)$ is
$y-b=-\frac{b}{a}(x-a)$
$\Rightarrow ay-ab=-bx+ab$
$\Rightarrow bx+ay=2ab$
$\Rightarrow \frac{x}{a}+\frac{y}{b}=2$.