Answer
Proof is below.
Work Step by Step
We use implicit differentiation on $y^2 = kx$:
$2y *y' = k$
$y' = \frac{k}{2y}$
At $(x_0,y_0)$:
$y' = \frac{k}{2y_0}$
Creating the line using point-slope form with point $(x_0,y_0)$ and slope $\frac{k}{2y_0}$:
$(y-y_0) = \frac{k}{2y_0} * (x-x_0)$
$y y_0 - y_0^2 = \frac{k}{2}(x-x_0)$
$y y_0 = \frac{k}{2}(x-x_0) + y_0^2$
Using $y_0^2 = kx_0$:
$y y_0 = \frac{k}{2}(x-x_0) + kx_0$
$y y_0 = \frac{k}{2}(x+x_0) $
Thus, the equation is proven.
