Question

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

Answer 1

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

Let $(x_1,y_1)$ be the required points.

slope of x axis is o.

as point lies on the curve

$\therefore 2a^2{y}_1=x_1^3-3a{x}_1^2$ ...(1)

Differentiating,

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

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

slope of taangent at $(x_1,y_1)\Rightarrow \frac{dy}{dx}=\frac{3x_1^2-6a{x}_1}{2a^2}$

Now$\frac{3x_1^2-6a{x}_1}{2a^2}=0$

$\Rightarrow 3x_1^2+6a{x}_1=0$

$\Rightarrow x_1(3x_1-6a)=0$

$x_1=0$ or $x_1=2a$

Also, $2a^2{y}_1=0$ or $2a^2{y}_1=8a^3-12a^3$

$\Rightarrow y_1=0$ or $y_1=-2a$ $\therefore points(0,0)(2a,-20)$