Conic sections: Parabola - What is $p$?

Help my teacher says $$p$$ can't be negative because it's distance.

I watched TOCT's tutorial (The Organic Chemistry Tutor) in YouTube about parabola and he said "$$(x-h)^2 = 4p(y-k)$$ if $$p$$ is positive the parabola opens upward, if negative parabola opens downward. $$(y-k)^2 = 4p(x-h)$$ if $$p$$ is positive the parabola opens to the right, if $$p$$ negative parabola opens to the left."

So I tried answering the question my teacher gave me: "Find the equation of the parabola with Vertex$$(1,2)$$ and Focus$$(1, -8)$$"

I tried graphing the given points and found out the focus is below the vertex.

I used distance formula to find out $$p$$: $$p=\pm 10$$

Based on what I watched in YouTube if the parabola opens downwards, $$p$$ is negative: $$p=-10$$

But then my teacher said $$p$$ can't be negative so I get the absolute value so: $$p=10$$

I tried to get the equation of the parabola using $$(x-h)^2 = 4p(y-k)$$:

With $$p$$ positive $$(x-1)^2 = 4(10)(y-2)$$ $$(x-1)^2 = 40(y-2)$$

With $$p$$ negative $$(x-1)^2 = 4(-10)(y-2)$$ $$(x-1)^2 = -40(y-2)$$

I tried to find the directrix with the formula $$y=k-p$$:

With $$p$$ positive $$y=2-(+10)$$ $$y=2-10$$ $$y=-8$$

With $$p$$ negative $$y=2-(-10)$$ $$y=2+10$$ $$y=12$$

When I graphed the directrix I think the negative one makes more sense so I am soooo confused right now

I wanna know what is the correct formula. Which equation is the correct one? Which directrix is correct? What is $$p$$? I also want to know how to get the ELR (end of latus rectum which is $$2p$$ units away from the focus) Help me understand all of this better!!

You are right - since the vertex is above the focus, the parabola must open downwards. So $$p=-10$$ and the parabola has equation $$(x-1)^2 = -40(y-2)$$. Note that the right hand side is never negative because for all points on the parabola $$y \le 2$$ (because the parabola opens downwards). The directrix is on the opposite side of the vertex to the focus, and is the line $$y=12$$.
The latus rectum is the horizontal chord of the parabola that passes through the focus, so it is part of the line with equation $$y=-8$$. This intersects the parabola at the two points where
$$(x-1)^2 = -40(y-2) = -40(-8-2) = 400 \\ \Rightarrow x-1 = \pm20 \\ \Rightarrow x = -19 \text{ or } x=21$$
• For a downward opening parabola $p$ will be negative. The formula $(h \pm 2p, k + p)$ for the ends of the latus rectum is stil correct for negative $p$ - in your example the ends of the latus rectum are at $(1 \pm (-20), 2+(-10)) = (-19,-8) \text{ or } (21,-8)$. – gandalf61 Jan 29 '19 at 15:32