For the curve y = 4x3 – 2x5, find all the points at which the tangen

1.	For the curve y = 4x3 – 2x5, find all the points at which the tangent passes through the origin.
Answer» y = 4x³ – 2x⁵ dy \(\frac{dy}{dx}\)= 12x² – 10x⁴dx Let (a, b) be the point on the curve at which the tangent passes through the origin. ∴ Equation of tangent is y – b = (12a² – 10a²) (x – a) but this passes through the origin ∴ 0 – b = (12a² – 10a⁴) (-a) b = 12a³ – 10a⁵ ….(1) Also from the equation b = 4a³ – 2a⁵ …. (2) from (1) and (2) 12a³ – 10a⁵ = 4a3 – 2a⁵ 8a³ = 8a⁵ a^{3 (}1 – a²) = 0 ⇒ a = 0, a = + 1 when a = 0, b = 0, (0, 0) a = 1,b= 12(1)- 10(1) = 2, (1,2) a = -1, b = 12 (-1) -10 (-1) = -2, (-1, -2) Hence the required points are (0,0), (1,2), (-1,-2).

For the curve y = 4x3 – 2x5, find all the points at which the tangent passes through the origin.

Answer»

y = 4x³ – 2x⁵ dy

\(\frac{dy}{dx}\)= 12x² – 10x⁴dx

Let (a, b) be the point on the curve at which the tangent passes through the origin.

∴ Equation of tangent is

y – b = (12a² – 10a²) (x – a)

but this passes through the origin

∴ 0 – b = (12a² – 10a⁴) (-a)

b = 12a³ – 10a⁵ ….(1)

Also from the equation b = 4a³ – 2a⁵ …. (2)

from (1) and (2) 12a³ – 10a⁵ = 4a3 – 2a⁵ 8a³ = 8a⁵

a^{3 (}1 – a²) = 0 ⇒ a = 0, a = + 1

when a = 0, b = 0, (0, 0)

a = 1,b= 12(1)- 10(1) = 2, (1,2)

a = -1, b = 12 (-1) -10 (-1) = -2, (-1, -2)

Hence the required points are (0,0), (1,2), (-1,-2).