Question

Find the points on the curve 2a²y = x³ − 3ax² where the tangent is parallel to x-axis ?

Answer 1

Let (x₁, y₁) represent the required points.
The slope of the x-axis is 0.
Here,

\[2 a^2 y = x^3 - 3a x^2 \]

\[\text { Since, the point lies on the curve } . \]

\[\text { Hence }, 2 a^2 y_1 = {x_1}^3 - 3a {x_1}^2 . . . \left( 1 \right)\]

\[\text { Now }, 2 a^2 y = x^3 - 3a x^2 \]

\[ \text { On differentiating both sides w.r.t.x, we get }\]

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

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

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

\[\text { Given }:\]

\[\text { Slope of the tangent at }\left( x_1 , y_1 \right)= \text { Slope of the x-axis }\]

\[ \Rightarrow \frac{3 {x_1}^2 - 6a x_1}{2 a^2} = 0\]

\[ \Rightarrow 3 {x_1}^2 - 6a x_1 = 0\]

\[ \Rightarrow x_1 \left( 3 x_1 - 6a \right) = 0\]

\[ \Rightarrow x_1 = 0 \text { or }x_1 = 2a\]

\[\text { Also}, \]

\[2 a^2 y_1 = 0 \text { or }2 a^2 y_1 = 8 a^3 - 12 a^3 [\text { From eq. } (1)]\]

\[ \Rightarrow y_1 = 0 \text { or } y_1 = - 2a\]

\[\text { Thus, the required points are}\left( 0, 0 \right)\text { and }\left( 2a, - 2a \right).\]

Answer 2

The given equation of the curve is

`2a^2y = x^3 - 3ax^2`    ............(i)

Differentiating with respect to x , we get

`2a^2dy/dx = 3x^2 - 6ax`

∴ `"Slope"   m_1 = dy/dx = 1/(2a^2)[3x^2 - 6ax]`  ..........(ii)

Also , 

Slope `m_2 = dy/dx = tanθ`

= tan0° = 0     [∵ Slope is parallel to x-axis]

∴ m₁ - m₂

⇒ `1/(2a^2)[3x^2 - 6ax] = 0`

⇒ 3x[x - 2a] = 0

⇒ x = 0 or 2a

∴ From (i)

y = 0 or -2a

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