Question

A point on the curve $2y^3 + x^2 = 12y$ at which the tangent to the curve is vertical is

• A. $\left(\sqrt{2}, \sqrt[4]{128}\right)$
• B. $\left( \sqrt[4]{128}, \sqrt{2}\right)$
• C. $\left( {2}, \sqrt[4]{128}\right)$
• D. $\left(\sqrt[4]{128} , 2\right)$

Answer 1

Answer: (B)

$\left( \sqrt[4]{128}, \sqrt{2}\right)$

Answer 2

The given curve is $2y^3 + x^2 = 12y$
$\Rightarrow \, \, \, \frac{dx}{dx} = \frac{x}{3(2-y^2)}$
For vertical tangents, we have $\frac{dy}{dx} =\frac{1}{0} \Rightarrow 2-y^{2} =0$
$\Rightarrow y\pm\sqrt{2}$
For $y=\sqrt{2} ,x^{2} =12y -2y^{3} =8\sqrt{2}$
$\, \, \, \, \, x=\left(8\sqrt{2}\right)^{\frac{2}{2}. \frac{1}{2}} =\left(64 \times2\right)^{\frac{1}{4} } =\left(128\right)^{\frac{1}{4}}$
For $y=-\sqrt{2} ,x^{2} =-2\sqrt{2}\left(6-2\right) =-8\sqrt{2}$
$x=\left(-8 \sqrt{2}\right)^{\frac{1}{2}} =\left(128\right)^{\frac{1}{4}}$
$\therefore \, \, \, \, \,$ Required point $\left(_{\left(128\right)} \frac{1}{4}, _{\sqrt{2}}\right)$