Question

At what point on the curve $x^3-8a^{2}y=0,$ the slope of the normal is $-\frac{2}{3}$.

• A. $\left(a,a\right)$
• B. $\left(2a,-a\right)$
• C. $\left(2a,a\right)$
• D. none of these

Answer 1

The correct option is C

$\left(2a,a\right)$

Explanation for correct option:

Step-1: Simplify the given data.

Given, $x^3-8a^{2}y=0,...........\left(i\right)$

$\Rightarrow$ $y=\frac{x^3}{8a^2}$

Step-2: Differentiate with respect to $x$.

$\Rightarrow$ $\frac{dy}{dx}=\frac{1}{8a^2}\times 3x^2$

$\Rightarrow$ $\frac{dy}{dx}=\frac{3x^2}{8a^2}$

Step-3: Finding slope of normal to the curve equate with the given.

Given, the slope of the normal is $-\frac{2}{3}$.

Slope of normal to the curve is $-\frac{1}{\left({\displaystyle \frac{dy}{dx}}\right)}$

$\Rightarrow$$-\frac{8a^2}{3x^2}=-\frac{2}{3}$

$\Rightarrow$$6x^2=24a^2$

$\Rightarrow$$x=2a$

Put value of $x$in equation $\left(i\right)$.

$\Rightarrow$$y=\frac{8a^3}{8a^2}$

$\Rightarrow$$y=a$

Therefore, required point is $\left(2a,a\right)$.

Hence, correct answer is option $\left(C\right)$