Finding the slope at a point $P(x_1,y_1)$ on a parabola

Question

Given a point $P(x_1,y_1)$ on the graph of a parabola $y^2=4px$, prove that the slope at point P is $$\frac{y_1}{2x_1}$$

Answer 1

How 'bout this: from $y^2 = 4px$, we have, by implicit differentiation, $2yy' = 4p$; dividing the latter equation by the former, we get $\frac{2y'}{y} = \frac{1}{x}$, or $y' = \frac{y}{2x}$. And I leave it to you to re-insert the subscripts. Usual caveats in re. case $x= 0$ apply. Cheers.

Answer 2

Hint: split the parabola in the branches

$$\mathcal B_1:~ y=\sqrt{4px},$$

$$\mathcal B_2:~ y=-\sqrt{4px},$$

for all $x$ s.t. $px\geq 0$ (I presume $p$ is a constant).

Can you compute the derivatives of the parabola at $(x_1,y_1)$ in both cases

$$(x_1,y_1)\in\mathcal B_1$$

and

$$(x_1,y_1)\in\mathcal B_2?$$