Question

Find the equation of normal to the curve $y=x^3-2x^2$ at $x=0$ .Find the co-ordinates of the point of intersection of the normal and the line $y=4.$

Answer 1

$y=x^3-2x^2$

$\frac{dy}{dx}=3x^2-4x$

$\frac{dy}{dx}{|}_{x=0}=0$ [$\because$ tangent is horizontal line]

$\therefore$ Slope of normal is $\infty$ mean normal is vertical line as $x=0$ normal is y-axis itself

Intersection of y-axis and $y=4$ is $(0,4)$.