Question

The equation of the normal to the curve $y^4=a{x}^3$ at $(a,a)$ is

• A. $x+2y=3a$
• B. $3x–4y+a=0$
• C. $4x+3y=7a$
• D. $4x–3y=0$

Answer 1

The correct option is C

$4x+3y=7a$

Explanation of the correct option.

Step $1$: Find the slope.

Given: $y^4=a{x}^3$

Differentiate with respect to $x$

$\Rightarrow$$4y^{3}y\text{'}=3a{x}^2$

$\Rightarrow$ $y\text{'}=\frac{3a{x}^2}{4y^3}$

Slope of the tangent at $\left(a,a\right)$,

$\Rightarrow$ $y\text{'}=\frac{3a{a}^2}{4a^3}$

$\Rightarrow$ $y\text{'}=\frac{3}{4}$

Since the slope of the normal is $-\frac{1}{m}=-\frac{4}{3}$.

Step $2$: Find the equation of the line.

We know that the line passing through point $\left(x_1,y_1\right)$ is given by

$y-y_1=m\left(x-x_1\right)$

Therefore, equation of the line passing through $\left(a,a\right)$ is,

$y-a=\left(-\frac{4}{3}\right)\left(x-a\right)$

$\Rightarrow$$3y-3a=-4x+4a$

$\Rightarrow$$4x+3y=3a+4a$

$\Rightarrow$ $4x+3y=7a$

Hence, Option $\left(\mathrm{C}\right)$ is the correct answer.