Question

The point(s) on the curve $y^3+3x^2=12y$ where the tangent is vertical, is (are)

• A. $\left(\pm \frac{4}{\sqrt{3}},-2\right)$
• B. $\left(\pm \frac{\sqrt{11}}{3},1\right)$
• C. $\left(0,0\right)$
• D. $\left(\pm \frac{4}{\sqrt{3}},2\right)$

Answer 1

The correct option is D $\left(\pm \frac{4}{\sqrt{3}},2\right)$

The given equation is,

$y^3+3x^2=12y$ ------- ( 1 )

Differentiate above equation,

$3y^2\frac{dy}{dx}+6x=12\frac{dy}{dx}$

$\Rightarrow$ $(3y^2-12)\frac{dy}{dx}=-6x$

$\Rightarrow$ $\frac{dy}{dx}=\frac{-6x}{3y^2-12}$ ---------- ( 2 )

Let $(x_1,y_1)$ be the point on the curve where the tangent is vertical to it.

Now substituting the points $(x_1,y_1)$ in equation ( 2 ),

$\frac{dy}{dx}=\frac{-6x_1}{3y_1^2-12}$

Also given that the tangent at this point is verical to it.

That is the slope of the tangent is infinity at this point.

$\Rightarrow$ $3y_1^2-12=0$

$\Rightarrow$ $3y_1^2=12$

$\Rightarrow$ $y_1^2=4$

$\Rightarrow$ $y_1=\pm 2$

Substituting the value of $y_1$ in equation ( 1 ), we have,

$y_1^3+3x_1^2=12y_1$

$\Rightarrow$ $(2{)}^3+3x_1^2=12y_1$

$\Rightarrow$ $8+3x_1^2=12\left(2\right)$

$\Rightarrow$ $3x_1^2=24-8$

$\Rightarrow$ $3x_1^2=16$

$\Rightarrow$ $x_1^2=\frac{16}{3}$

$\therefore$ $x_1=\pm \frac{4}{\sqrt{3}}$

As the square root of a negative number is imaginary, we can neglect the value $y_1=-2$

$\therefore$ The required point on the curve is $(\pm \frac{4}{\sqrt{3}},2)$