# For the curve y = 4x^{3} - 2x^{5} find all the points at which the tangent passes through the origin

**Solution:**

The equation of the given curve is y = 4x^{3} - 2x^{5}

Therefore,

dy/dx = 12x^{2} - 10x^{4}

Hence, the slope of the tangent at the point (x, y) is 12x^{2} - 10x^{4}

Thus, the equation of the tangent at (x, y) is given by,

Y - y = (12x^{2} - 10x^{4})(X - x)

When the tangent passes through the origin (0, 0),

X = Y = 0.

Therefore,

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

y = 12x^{3} - 10x^{5}

Also, we have y = 4x^{3} - 2x^{5}

Hence,

∴ 12x^{3} - 10x^{5} = 4x^{3} - 2x^{5}

⇒ 8x^{5} - 8x^{3} = 0

⇒ x^{5} - x^{3} = 0

⇒ x^{3} (x^{2} - 1) = 0

⇒ x = 0, ± 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

Thus, the points are (0, 0) , (1, 2) and (- 1, - 2)

NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.3 Question 18

## For the curve y = 4x^{3} - 2x^{5} find all the points at which the tangent passes through the origin

**Summary:**

For the given curve y = 4x^{3} - 2x^{5}, the points at which the tangent passes through the origin are (0, 0) , (1, 2) and (- 1, - 2)

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