# Find a point on the curve

Question:

Find a point on the curve $y=3 x^{2}+4$ at which the tangent is perpendicular to the line whose slope is $-\frac{1}{6}$.

Solution:

Given:

The curve $y=3 x^{2}+4$ and the Slope of the tangent is $\frac{-1}{6}$

$y=3 x^{2}+4$

Differentiating the above w.r.t $x$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=2 \times 3 \mathrm{x}^{2-1}+0$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=6 \times \ldots(1)$

Since, tangent is perpendicular to the line,

$\therefore$ The Slope of the normal $=\frac{-1}{\text { The Slope of the tangent }}$

i.e, $\frac{-1}{6}=\frac{-1}{6 x}$

$\Rightarrow \frac{1}{6}=\frac{1}{6 x}$

$\Rightarrow x=1$

Substituting $x=1$ in $y=3 x^{2}+4$

$\Rightarrow y=3(1)^{2}+4$

$\Rightarrow y=3+4$

$\Rightarrow y=7$

Thus, the required point is $(1,7)$.