Question

The equation of the normal to the curve $y^4=a{x}^3$ at $\left(a,a\right)$ 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 for the correct option

Step 1: Solve for the slope of normal

Given curve is, $y^4=a{x}^3$ and the point is $\left(a,a\right)$

Differentiating it implicitly,

$\begin{array}{cc}& 4y^3\left(\frac{dy}{dx}\right)=3a{x}^2\\ \Rightarrow & \frac{dy}{dx}=\frac{3a{x}^2}{4y^3}\end{array}$

Where $\frac{dy}{dx}$ is the slope of the tangent

At $\left(a,a\right)$, $x=a$ and $y=a$. Substituting this in above equation,

$\begin{array}{cc}\Rightarrow & \frac{dy}{dx}=\frac{3a·a^2}{4a^3}\\ \Rightarrow & \frac{dy}{dx}=\frac{3}{4}\end{array}$

We know that,

Slope of normal $=$ $($Slope of tangent${)}^{-1}$

$\begin{array}{c}=-\frac{1}{{\displaystyle \frac{dy}{dx}}}\\ =-\frac{1}{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}\\ =-\frac{4}{3}\end{array}$

Step 2: Solve for the equation of normal

Equation of a line in point-slope form is

$y-y_1=m\left(x-x_1\right)$
where $\left(x_1,y_1\right)$ is the point the line passes through and $m$ is the slope.

For the normal of the given curve here, $\left(x_1,y_1\right)=\left(a,a\right)$ and $m=-\frac{4}{3}$.

Thus the equation of the normal is,

$\begin{array}{cc}& y-a=-\frac{4}{3}\left(x-a\right)\\ \Rightarrow & 3y-3a=4a-4x\\ \Rightarrow & 4x+3y=7a\end{array}$

Therefore, the equation of the normal to the given curve is $4x+3y=7a$.

Hence, option(C) is correct.