Question

The area ( in sq. units ) of the triangle formed by the tangent and normal drawn to the curves ${\left(\frac{x}{3}\right)}^n+{\left(\frac{y}{4}\right)}^n=2$ at $(3,4)$ and $X$-axis is

• A. $\frac{100}{3}$
• B. $48$
• C. $\frac{50}{3}$
• D. $144$

Answer 1

Answer: (C)

Equation of curve $\frac{x^n}{3^n}+\frac{y^n}{4^n}=2$
$\frac{n{x}^{n-1}}{3^n}+\frac{n{y}^{n-1}}{4^n}\frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}=\frac{-4^n{x}^{n-1}}{3^n{y}^{n-1}}$
${\left(\frac{dy}{dx}\right)}_{(3,4)}=\frac{-4^n\times 3^{n-1}}{3^n\times 4^{n-1}}=-\frac{4}{3}$
Slope of tangent is $-\frac{4}{3}$ and slope of normal $=\frac{3}{4}$
Equation of tangent of curve at $(3,4)$
$y-4=\frac{-4}{3}(x-3)$
$\Rightarrow 3y-12=-4x+12$
$4x+3y=24$
Equation of normal of $(3,4)$

Area of triangle
$ABC=\frac{1}{2}\times BC\times AM$
$=\frac{1}{2}\times \frac{25}{3}\times 4=\frac{50}{3}$