Question

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

Answer 1

$4y=x^2-8$
Differentiating w.r. to $x$
$4\frac{dy}{dx}=2x$
${\frac{dy}{dx}|}_{(x_1,y_1)}=\frac{x_1}{2}$
Slope of Normal $=-\frac{2}{x_1}$
But 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 $y_1=\frac{x_1^2-8}{4}$
$x_1(\frac{x_1^2-8}{4}+3)=20$
$\Rightarrow x_1(x_1^2-8+12)=80$
$\Rightarrow x_1(x_1^2+4)=80$
$\Rightarrow x_1^3+4x_1-80=0$
$\Rightarrow (x_1-4)(x_1^2+4x_1+20)=0$
Hence $x_1=4;y_1=2$
$P=(4,2)$, $A(10,-1)$
Equation of $PA$ is:
$y+1=-\frac{1}{2}(x-10)$
$\Rightarrow 2y+2=-x+10$
$\therefore x+2y-8=0$