Find a point on the curve

Solution:

Given:

The curve $y^{2}=2 x^{3}$ and The Slope of tangent is 3

$y^{2}=2 x^{3}$

Differentiating the above w.r.t $x$

$\Rightarrow 2 y^{2}-1 \times \frac{d y}{d x}=2 \times 3 x^{3-1}$

$\Rightarrow y \frac{d y}{d x}=3 x^{2}$

$\Rightarrow \frac{d y}{d x}=\frac{3 x^{2}}{y}$

Since, The Slope of tangent is 3

$\frac{3 x^{2}}{y}=3$

$\Rightarrow \frac{x^{2}}{y}=1$

$\Rightarrow x^{2}=y$

Substituting $x^{2}=y$ in $y^{2}=2 x^{3}$

$\left(x^{2}\right)^{2}=2 x^{3}$

$x^{4}-2 x^{3}=0$

$x^{3}(x-2)=0$

$x^{3}=0$ or $(x-2)=0$

$x=0$ or $x=2$

If $x=0$

$\Rightarrow \frac{d y}{d x}=\frac{3(0)^{2}}{y}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\mathbf{0}$, which is not possible.

So we take $x=2$ and substitute it in $y^{2}=2 x^{3}$, we get

$y^{2}=2(2)^{3}$

$y^{2}=2 \times 8$

$y^{2}=16$

$y=4$

Thus, the required point is $(2,4)$