. For the curves y = 4x³ – 2x⁵ , find all points at which tangent passes through the origin ? . For the curves y = 4x ³ – 2x ⁵ , find all points at which tangent passes through the origin ?

. For the curves y = 4x ³ – 2x ⁵ , find all points at which tangent passes through the origin ?

The equation of the given curve is y = 4x3 − 2x5.

dy/dx = 12x^2 - 10 x^4
Therefore, the slope of the tangent at a point (x, y) is 12x2 − 10x4.
The equation of the tangent at (x, y) is given by,

Y -y =  (12x^2 - 10x^4) (X- x)       ---(1)
When  0the tangent passes through the origin (0, 0), then X = Y = 0.
Therefore, equation (1) reduces to:

-y=( 12x^2- 10x^4) (-x)

y = 12x^3 - 10x^5 

Also, we have 

y = 4x^3 -  2x^5

therefore 

12x^3 - 10x^5 = 4x^3 - 2x^5

8x^3 - 8x^5 = 0

x^3 - x^5 = 0

x^3 (1-x^2) = 0

x = 0 , plus, minus 1

when x = 0 ,y= 4 (0)^3 - 2(0)^5=0

When x = 1, y = 4 (1)3 − 2 (1)5 = 2.

When x = −1, y = 4 (−1)3 − 2 (−1)5 = −2.

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