Answer
$(\frac{1}{8}, \pm \frac{1}{16})$
Work Step by Step
The slope of the lines $4x-3y=1$ is $\frac{4}{3}$ as we can rearrange the equation to be $y = \frac{4}{3}x - \frac{1}{3}$.
We use implicit differentiation on $y^2 = 2x^3$:
$2y * \frac{dy}{dx} = 6x^2$
$\frac{dy}{dx} = \frac{3x^2}{y}$
We then solve for all $(x,y)$ such that $\frac{dy}{dx} = -\frac{3}{4}$ in order to be perpendicular to the line with slope $\frac{4}{3}$:
$\frac{dy}{dx} = \frac{3x^2}{y} = -\frac{3}{4}$
$-12x^2 = 3y$
$y=-4x^2$
$y^2 = 16x^4$
From the equation, $y^2 = 2x^3$,
$2x^3 = 16x^4$
$x = \frac{1}{8}$ and $y = -4 (\frac{1}{8})^2 \Rightarrow y = \pm\frac{1}{16}$
Thus, the point is $(\frac{1}{8}, \pm \frac{1}{16})$
