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
$(x-1)^2 = -40(y-2) = -40(-8-2) = 400 \\ \Rightarrow x-1 = \pm20 \\ \Rightarrow x = -19 \text{ or } x=21$