Answer :

Formula Used:


Equation of the tangent at (x1, y1) where slope is m is given by y – y1 = m(x – x1)


Equation of the normal at (x1, y1) where slope is m is given by


Let (x1, y1) be the required point on the given curve y = 4x3 – 2x5


y1 = 4x13 – 2x15 … (1)


Differentiating wrt x we get,




Equation of the tangent is y – y1 = (12x12 – 10x14)(x – x1)


This passes through the origin.


Hence, 0 – y1 = (12x12 – 10x14)(0 – x1)


y1 = 12x13 – 10x15 … (2)


Eq (1) - Eq (2),


0 = -8x13 + 8x15


x1 = 0 or x1 = ±1


When x1 = 0, from eq (2), y1 = 0


When x1 = 1, from eq (2),


y1 = 4 – 2 = 2


When x1 = -1, from eq (2),


y1 = -4 – (-2) = -2


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


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