Question

The point(s) on the curve $y^3+3x^2=12y$. where the normal is horizontal is(are)

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

Answer 1

Answer: (B)

$(\pm \frac{4}{\sqrt{3}},2)$

Slope of line which is horizontal is zero

$\tan \left(\theta \right)=\tan \left(0\right)=0$

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

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

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

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

$m_1{m}_2=-1$

$m_1\left(\frac{-6x}{3y^2-12}\right)=+1$

$m_1=\frac{3y^2-12}{6x}$ is slope of normal

Normal is horizontal

$m_1=0\Rightarrow 3y^2-12=0\Rightarrow y=\pm 2$

for $x=+2$

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

$(2{)}^3+3x^2=12(2)$

$3x^2=24-8\Rightarrow 3x^2=16$

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

for $x=-2$

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

$3x^2=-24+8$

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

Square can't be negative

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

Therefore points on the curve are $(\frac{4}{\sqrt{3}},2)$ and $(\frac{-4}{\sqrt{3}},2)$.