Question

The points on the curve $y^3+3x^2=12y$ where the tangent is vertical (parallel to the$y-$axis), are

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

Solve for the points on the curve $y^3+3x^2=12y$

$y^3+3x^2=12y$

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

Finding the first derivative.

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

For vertical tangent, $\frac{dx}{dy}=0$

$$\begin{gathered} 12–3y^2=0 \\ y=\pm 2 \end{gathered}$$

Put $y=2,x=\pm \frac{4}{\sqrt{3}}$

There is no real solution for $3x^2=–16$

The required point is $\left(\pm \frac{4}{\sqrt{3}},2\right)$

Hence, option(D) is the correct answer.