Q. A line $\mathrm{L}$ is perpendicular to the curve $\mathrm{y}=\frac{\mathrm{x}^{2}}{4}-2$ at its point $\mathrm{P}$ and passes through $(10,-1)$. The coordinates of the point $\mathrm{P}$ are
Application of Derivatives
Solution:
$ \frac{\mathrm{dy}}{\mathrm{dx}}|_{\mathrm{P}}=\frac{2 \mathrm{x}_{1}}{4}=\frac{\mathrm{x}_{1}}{2} \Rightarrow.$ slope of normal $=-\frac{2}{\mathrm{x}_{1}}$
$\Rightarrow -\frac{2}{\mathrm{x}_{1}}=\frac{\mathrm{y}+1}{\mathrm{x}_{1}-10} \Rightarrow 20-2 \mathrm{x}_{1}=\mathrm{x}_{1} \mathrm{y}_{1}+\mathrm{x}_{1}$
$\Rightarrow 3 \mathrm{x}_{1}+\mathrm{x}_{1} \mathrm{y}_{1}=20$
also $ \mathrm{y}_{1}=\frac{\mathrm{x}_{1}^{2}}{4}-2$
$\Rightarrow 4 \mathrm{y}_{1}=\mathrm{x}_{1}^{2}-8$
