Question

Find the point(s) on the curve $y^3+3x^2=12y$ where the tangent is vertical (parallel to $y-axis$).

Answer 1

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

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

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

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

If tangent is parallel to y_ axis , then

$\frac{dy}{dx}=\tan {90}^{\circ }$

$\Rightarrow \frac{dy}{dx}=\infty$

$\Rightarrow \frac{dy}{dx}=0$

$\Rightarrow \frac{(3y^2-12)}{-6x}=0$

$\Rightarrow 3y^2=12$

$\Rightarrow y^2=4$

$\Rightarrow y=\pm 2$

putting $y=2$ in (1)

${\left(2\right)}^3+3x^2=12\times 2$

$\Rightarrow 8+3x^2=24$

$\Rightarrow 3x^2=16$

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

$\Rightarrow x=\pm \frac{4}{\sqrt{3}}=\pm \frac{4\sqrt{3}}{\sqrt{3}}$

putting $y=-2$ in ...(1)

${\left(-2\right)}^3+3x^2=12(-2)$

$\Rightarrow -8+3x^2=-24$

$\Rightarrow 3x^2=-16$

$\Rightarrow x^2=\frac{-16}{3}$

Required points on curve are

$\left(\frac{4\sqrt{3}}{3},2\right)$ and $\left(\frac{-4\sqrt{3}}{3},2\right)$