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

Let $({x_1},{y_1})$ be the required point on the given curve
$y = 4{x^3} - 2{x^5}$ …(i)
Therefore, ${y_1} = 4{x_{_1}}^3 - 2{x_1}^5$

Differentiating (i) w. r. t. $x$, we get
$\cfrac{{dy}}{{dx}} = 12{x^2} - 10{x^4}$

So, ${\left( {\cfrac{{dy}}{{dx}}} \right)_{\left( {{x_1},{y_1}} \right)}} = 12{x_1}^2 - 10{x_1}^4$

Therefore the equation of the tangent at $\left( {{x_1},{y_1}} \right)$ is
$y - {y_1} = (12x_1^2 - 10{x_1}^4)(x - {x_1})$

Since, it passes through the origin,

Therefore, $0 - {y_1} = (12x_1^2 - 10{x_1}^4)(0 - {x_1})$
or ${y_1} = 12x_1^3 - 10x_1^5$ …(iii)

From (ii) and (iii), $4{x_1}^3 - 2x_1^5 = 12x_1^3 - 10{x_1}^5$
$\Rightarrow - 8x_1^3 + 8x_1^5 = 0 \Rightarrow 8x_1^3( - 1 + x_1^2) = 0 \Rightarrow {x_1} = 0,{x_1} = \pm 1$

When ${x_1} = 0$, then from(ii),${y_1} = 0$

When ${x_1} = 1$, then from(ii),${y_1} = 4(1) - 2(1) = 2$

When ${x_1} = - 1$, then from(ii),${y_1} = 4{( - 1)^3} - 2{( - 1)^5} = - 4 + 2 = - 2$

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