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

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

  • $\begingroup$ This is really helpful! I also like what you did for finding the LR I didn't know that because the formula that I got from my teacher for finding LR was (h+2p, k+p) and (h-2p, k+p) and that is only for parabola facing up, there's like 6 more for facing down, right, and left. Are these formulas even necessary and is my teacher correct about having p as positive? $\endgroup$ – user639467 Jan 29 '19 at 14:37
  • $\begingroup$ 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 $(1 \pm (-20), 2+(-10)) = (-19,-8) \text{ or } (21,-8)$. $\endgroup$ – gandalf61 Jan 29 '19 at 15:32

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