Question

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

• A. $\frac{x}{a}=-\frac{y}{b}$
• B. $\frac{x}{a}+\frac{y}{b}=2$
• C. $\frac{x^2}{a^2}-\frac{y^2}{b^2}=n$
• D. $\frac{x}{a}=\frac{y}{b}$

Answer 1

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

Given equation of the curve, ${\left(\frac{x}{a}\right)}^n+{\left(\frac{y}{b}\right)}^n=2$

Differente the above equation to find the slope of the curve

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

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

$\therefore$Slope at $(a,b)=\frac{-b}{a}{\left(\frac{a}{a}\right)}^{n-1}{\left(\frac{b}{b}\right)}^{n-1}=\frac{-b}{a}$

$\therefore$Equation of the tangent at $(a,b)$ is:

$(y-b)=\frac{-b}{a}(x-a)$

$bx+ay=2ab$

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