Question

Find the equation of the normal to the curve $y=|x^2-|x\left|\right|atx=-2.$

Answer 1

$=>y=|x^2-\left|x\right||...................\left(1\right)$

at $x=-2$, slope of tangent $=>\frac{dy}{dx}=|2x-\left|1\right||$

$=>\frac{dy}{dx}=|-4-1|=5(x=-2)$

Slope of normal $=>m=\frac{-1}{5}$

Now equation of normal be $y=-\frac{1}{5}x+c..........\left(2\right)$

at $x=-2$ in $\left(1\right)$, $y=2$

$=>$normal $\left(2\right)$ passes through $(-2,2)$, $=>2=-\frac{1}{5}(-2)+c$

$=>c=\frac{8}{5}$

So, normal becomes $5y+x-8=0.$