Question

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

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

Answer 1

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

$3y^2.y^{\prime }+6x=12y^{\prime }$
Or
$y^{\prime }(3y^2-12)=-6x$
Or
$y^{\prime }(y^2-4)=-2x$
Or
$y^{\prime }=\frac{-2x}{y^2-4}$
Hence for the tangent to be vertical
$y^2-4=0$
$y=\pm 2$
Substituting $y=2$, we get
$3x^2=12y-y^3$
$=24-8$
$=16$
Hence
$x=\frac{\pm 4}{\sqrt{3}}$
$y=-2$ gives
$3x^2=-16$
This is not possible.
Hence the required point is
$(\frac{\pm 4}{\sqrt{3}},2)$.