Practicing Success
The point(s) on the curve $y^3+3 x^2=12 y$ where the tangent is vertical, is (are) |
$\left( \pm \frac{4}{\sqrt{3}},-2\right)$ $\left( \pm \sqrt{\frac{\pi}{3}}, 1\right)$ (0, 0) $\left( \pm \frac{4}{\sqrt{3}}, 2\right)$ |
$\left( \pm \frac{4}{\sqrt{3}}, 2\right)$ |
We have, $y^3+3 x^2=12 y$ .....(i) $\Rightarrow 3 y^2 \frac{d y}{d x}+6 x=12 \frac{d y}{d x}$ [Diff. w.r.t. x] $\Rightarrow 3\left(y^2-4\right) \frac{d y}{d x}=-6 x \Rightarrow \frac{d y}{d x}=-\frac{2 x}{y^2-4}$ At point(s) where the tangent(s) is (are) vertical, $\frac{d y}{d x}$ is not defined. ∴ $y^2-4=0 \Rightarrow y= \pm 2$ From (i), we find that $y=2 \Rightarrow x= \pm \frac{4}{\sqrt{3}}$ and, $y=-2 \Rightarrow x^2=\frac{-16}{3}$, which is not possible. Hence, the required points are $( \pm 4 / \sqrt{3}, 2)$ |