Question:
Find the point on the curve $y=3 x^{2}+4$ at which the tangent is perpendicular to the line whose slop is $-\frac{1}{6}$.
Solution:
Let $\left(x_{1}, y_{1}\right)$ be the required point.
Slope of the given line $=\frac{-1}{6}$
$\therefore$ Slope of the line perpendicular to it $=6$
Since, the point lies on the curve.
Hence, $y_{1}=3 x_{1}^{2}+4$
Now, $y=3 x^{2}+4$
$\therefore \frac{d y}{d x}=6 x$
Now,
Slope of the tangent at $\left(x_{1}, y_{1}\right)=\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}=6 x_{1}$
Slope of the tangent at $\left(x_{1}, y_{1}\right)=$ Slope of the given line [Given]
$\therefore 6 x_{1}=6$
$\Rightarrow x_{1}=1$
and
$y_{1}=3 x_{1}^{2}+4=3+4=7$
Thus, the required point is $(1,7)$