The points at which the tangent passes through the origin for the curve $y=4 x^{3}-2 x^{5}$ are

The equation of the given curve is $\begin{array}{l} y=4 x^{3}-2 x^{5} \\ \frac{\mathrm{dy}}{\mathrm{dx}}=12 x^{2}-10 x^{4} \end{array}$ Therefore, the slope of the tangent at point $(x, y)$ is $12 x^{2}-10 x^{4}$. The equation of the tangent at $(x, y)$ is given by $\mathrm{Y}-y=\left(12 x^{2}-10 x^{4}\right)(\mathrm{X}-x) \ldots \text { (i) }$ When, the tangent passes through the origin $(0,0)$, then $\mathrm{X}=\mathrm{Y}=0$ Therefore, eq. (i) reduce to $\begin{array}{l} -y=\left(12 x^{2}-10 x^{4}\right)(-x) \\ \Rightarrow y=12 x^{3}-10 x^{5} \end{array}$ Also, we have $y=4 x^{3}-2 x^{5}$ $\begin{array}{l} \therefore 12 x^{3}-10 x^{5} \\ =4 x^{3}-2 x^{5} \\ \Rightarrow 8 x^{5}-8 x^{3}=0 \\ \Rightarrow x^{5}-x^{3}=0 \\ \Rightarrow x^{3}\left(x^{2}-1\right)=0 \\ \Rightarrow x=0, \pm 1 \end{array}$ 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 require points are $(0,0),(1,2)$ and $(-1,-2)$.