Find the point on the curve 2a2y = x3 – 3ax2 where the tangent is pa

1.	Find the point on the curve 2a2y = x3 – 3ax2 where the tangent is parallel to the x – axis.
Answer» Given: The curve is 2a²y = x³ – 3ax² Differentiating the above w.r.t x ⇒ 2a² x \(\frac{dy}{dx}\)= 3x^{3 – 1}– 3 x 2ax^{2 – 1} ⇒ 2a² \(\frac{dy}{dx}\)= 3x² – 6ax ⇒ \(\frac{dy}{dx}\) = \(\frac{3x^2-6ax}{2a^2}\)...(1) \(\therefore\frac{dy}{dx}\)= The Slope of the tangent = tanθ Since, the tangent is parallel to x – axis i.e, ⇒ \(\frac{dy}{dx}\) = tan(0) = 0 ...(2) \(\therefore\) tan(0) = 0 \(\therefore\frac{dy}{dx}\)= The Slope of the tangent = tanθ From (1) & (2),we get, ⇒ \(\frac{3x^2-6ax}{2a^2}\) = 0 ⇒ 3x² – 6ax = 0 ⇒ 3x(x – 2a) = 0 ⇒ 3x = 0 or (x – 2a) = 0 ⇒ x = 0 or x = 2a Substituting x = 0 or x = 2a in 2a²y = x³ – 3ax², when x = 0 ⇒ 2a²y = (0)³ – 3a(0)² ⇒ y = 0 when x = 2 ⇒ 2a²y = (2a)³– 3a(2a)² ⇒ 2a²y = 8a³ – 12a³ ⇒ 2a²y = – 4a³ ⇒ y = – 2a Thus, the required point is (0,0) & (2a, – 2a)

Find the point on the curve 2a2y = x3 – 3ax2 where the tangent is parallel to the x – axis.

Answer»

Given:

The curve is 2a²y = x³ – 3ax²

Differentiating the above w.r.t x

⇒ 2a² x \(\frac{dy}{dx}\)= 3x^{3 – 1}– 3 x 2ax^{2 – 1}

⇒ 2a² \(\frac{dy}{dx}\)= 3x² – 6ax

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

\(\therefore\frac{dy}{dx}\)= The Slope of the tangent = tanθ

Since, the tangent is parallel to x – axis

i.e,

⇒ \(\frac{dy}{dx}\) = tan(0) = 0 ...(2)

\(\therefore\) tan(0) = 0

\(\therefore\frac{dy}{dx}\)= The Slope of the tangent = tanθ

From (1) & (2),we get,

⇒ \(\frac{3x^2-6ax}{2a^2}\) = 0

⇒ 3x² – 6ax = 0

⇒ 3x(x – 2a) = 0

⇒ 3x = 0 or (x – 2a) = 0

⇒ x = 0 or x = 2a

Substituting x = 0 or x = 2a in 2a²y = x³ – 3ax²,

when x = 0

⇒ 2a²y = (0)³ – 3a(0)²

⇒ y = 0

when x = 2

⇒ 2a²y = (2a)³– 3a(2a)²

⇒ 2a²y = 8a³ – 12a³

⇒ 2a²y = – 4a³

⇒ y = – 2a

Thus, the required point is (0,0) & (2a, – 2a)