Question

The Point (s) on the cure $y^3+{3x}^2=12y$ where the tangent is vertical (parallel to y-axis), is/are.

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

Answer 1

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

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

Thus differentiating w.r.t $x$

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

$\frac{dy}{dx}(y^2-4)+2x=0$

$\frac{dy}{dx}=\frac{-2x}{y^2-4}$

For tangent to be $\parallel$ to $y-$axis $\frac{dy}{dx}\to \infty$

Thus $y^2-4=0$

$y=\pm 2$

Putting $y=+2$ in curve we get $x=\pm \frac{4}{\sqrt{3}}$

Putting $y=-2$ in curve $x$ does not exit

Thus $y=-2$ not in range as $x^2<0$

Thus points where tangent is $\parallel$ to $y-$axis is $[\pm \frac{4}{\sqrt{3}},2]$ Answer.