Question

An equation for the line that passes through $(10,-1)$ and is perpendicular to $y=\frac{x^2}{4}-2$ is

• A. $4x+y=39$
• B. $2x+y=19$
• C. $x+y=9$
• D. $x+2y=8$

Answer 1

The correct option is D $x+2y=8$

Given equation of parabola is
$y=\frac{x^2}{4}-2$
$\Rightarrow \frac{dy}{dx}=\frac{x}{2}$
Slope of tangent to parabola at $(x_1,y_1)=\frac{x_1}{2}$
Therefore, slope of normal at $(x_1,y_1)=-\frac{2}{x_1}$
Also, slope of normal $=\frac{y_1+1}{x_1-10}$
$\therefore \frac{y_1+1}{x_1-10}=-\frac{2}{x_1}$
$\Rightarrow x_1{y}_1+x_1=-2x_1+20$
$\Rightarrow x_1{y}_1+3x_1=20$
Substituting $y_1=\frac{x_1^2-8}{4}$ (from the given equation)
$x_1(\frac{x_1^2-8}{4}+3)=20$
$\Rightarrow x_1(x_1^2+4)=80$
$\Rightarrow x_1^3+4x_1-80=0$.
which has one root $x_1=4$
Hence, $x_1=4;y_1=2$
$\therefore P=(4,2)$
Therefore, equation of $PA$ is
$y+1=-\frac{1}{2}(x-10)$
$\Rightarrow 2y+2=-x+10$
$\Rightarrow x+2y-8=0$