Question

The tangent to the curve $2a^{2}y=x^3-3a{x}^2$ is parallel to the x-axis at the points

• A. $(0,0),(2a,-2a)$
• B. $(0,0),(-2a,2a)$
• C. $(0,0),(-2a,-2a)$
• D. $(2,2),(0,0)$

Answer 1

The correct option is A $(0,0),(2a,-2a)$

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

The slope of the $x-$axis is $0$

Here $2a^{2}y=x^3-3a{x}^2$, since the points lies on the curve we get

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

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

On differentiating both sides with respect to $x$, we get

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

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

Slope of the tangent at $(x_1,y_1)={\left(\frac{dy}{dx}\right)}_{(x_1,y_1)}=\frac{3x_1^2-6a{x}_1}{2a^2}$

It is given that slope of the tangent at $(x_1,y_1)=$ slope of the $x-$axis.

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

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

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

$\Rightarrow x_1=0$ or $x_1=2a$

Also, from $\left(1\right)$

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

$\Rightarrow y_1=0$ or $y_1=-2a$

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