Question

The point (s) on the curve $y^3+3x^2=12y,$ where the tangent is vertical (i.e., parallel to the y-axis), is / true

• 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 $(\pm \frac{4}{\sqrt{3}},2)$

We require the point where $\frac{dx}{dy}=0$
Or
$3y^2.dy+6x.dx=12dy$
Or
$dy(3y^2-12)=-6xdx$
Or
$\frac{-dx}{dy}=\frac{3y^2-12}{6}$
$=0$
Hence
$3y^2-12=0$
$y^2-4=0$
$y=\pm 2$
Now $y=-2$ does not satisfy the equation the curve.
Substituting in the equation of the curve, y=2,
$8+3x^2=24$
$3x^2=16$
$x=\pm \frac{4}{\sqrt{3}}$.