Question

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

• A. $(2,-1)$
• B. $(6,7)$
• C. $(0,-2)$
• D. (4,2)

Answer 1

Answer: (D)

$\frac{\mathrm{dy}}{\mathrm{dx}}{|}_{\mathrm{P}}=\frac{2{\mathrm{x}}_1}{4}=\frac{{\mathrm{x}}_1}{2}\Rightarrow .$ 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 ${\mathrm{y}}_1=\frac{{\mathrm{x}}_1^2}{4}-2$
$\Rightarrow 4{\mathrm{y}}_1={\mathrm{x}}_1^2-8$