Show that the tangent line at ($x_{0}$, $y_{0}$) has y-intercept -$y_{0}$

Question

I've been wrapping my head around this problem for days and I'm not able to solve it. Could someone explain to me how to solve it?

Let p be a positive constant and consider the parabola $x^2=4py$ with vertex at the origin and focus at the point (0, p). (a) Show that the tangent line at ($x_{0}$, $y_{0}$) has y-intercept -$y_{0}$.

What I did this far is:

1. I used the slope formula (i) $y - y_{0} = m(x - x_{0})$
2. m = $\frac{dy}{dx} \frac{x^2}{4p} = \frac{x}{2p}$
3. Evaluated m when $x = x_{0}$

m = $\frac{x_{0}}{2p}$

3. Replaced m in (i)

$y = \frac{x_{0}}{2p}(x-x_{0}) + y_{0}$

4. In order to get the y-intercept, I evaluated $x = 0$

$y = \frac{x_{0}^2}{2p}+y_{0}$

What am I missing?

Answer 1

When $x = 0$, you'll get $\frac{-x_0^2}{2p} + y_0$. Since $(x_0,y_0)$ is a point on $x² = 4py$ we have $x_0^2 = 4py_0$.

The y-intercept is then $$\frac{-x_0^2}{2p} + y_0 = -2y_0 + y_0 = -y_0$$ as desired.

Answer 2

Since the tangent line is$$y=\frac{x_0}{2p}(x-x_0)+y_0,$$if you put $x=0$, then you get\begin{align}y&=-\frac{{x_0}^2}{2p}+y_0\\&=-\frac{{x_0}^2}{2p}+\frac{{x_0}^2}{4p}\\&=-\frac{{x_0}^2}{4p}\\&=-y_0.\end{align}