# For the curve

Question:

For the curve $y=4 x^{3}-2 x^{5}$, find all the points at which the tangents passes through the origin.

Solution:

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

$\therefore \frac{d y}{d x}=12 x^{2}-10 x^{4}$

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

The equation of the tangent at (xy) is given by,

$Y-y=\left(12 x^{2}-10 x^{4}\right)(X-x)$   ....(1)

When the tangent passes through the origin (0, 0), then X = Y = 0.

Therefore, equation (1) reduces to:

$-y=\left(12 x^{2}-10 x^{4}\right)(-x)$

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

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

$\therefore 12 x^{3}-10 x^{5}=4 x^{3}-2 x^{5}$

$\Rightarrow 8 x^{5}-8 x^{3}=0$

$\Rightarrow x^{5}-x^{3}=0$

$\Rightarrow x^{3}\left(x^{2}-1\right)=0$

$\Rightarrow x=0, \pm 1$

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).