Question

Find the point (s) on the curve $2a^{2}y=x^3-3a{x}^2$ where tangent is parallel to $x$-axis.

Answer 1

Tangent at a point on the curve parallel to $x$-axis.

So, slope of that tangent is zero. ( As, Slope's of parallel lines are same and Slope of X-axis is 0)

$2a^{2}y=x^3-3a{x}^2$

Differentiate w.r.t. $x$ ,

$2a^2\frac{dy}{dx}=3x^2-6ax$

$\frac{dy}{dx}=\frac{1}{2a^2}(3x^2-6ax)$

Now, slope of the tangent, $\frac{dy}{dx}=0$

$\Rightarrow 3x^2-6ax=0$

$3x(x-2a)=0$

$x=0,2a$

For $x=0$,

$2a^{2}y=(0{)}^3-3a(0{)}^2$

$y=0$

For $x=2a$,

$2a^{2}y=(2a{)}^3-3a(2a{)}^2$

$2a^{2}y=-4a^3$

$y=-2a$

So, the points are $(0,0)$ and $(2a,-2a)$