I have this MCQ.

Which of the following is the equation of the parabola with focus at $(1,2)$ and vertex at $(3,2)$ :

A. $ y^2 - 4y + 8x - 20 = 0 $

B. $ y^2 + y + 8x -20 = 0 $

C. $ y^2 + 4y + 8x - 20 = 0 $

D. $ y^2 - 4y + x - 20 = 0 $

E. $ y^2 - 4y + 8x + 20 = 0 $

Now, I know how to find equation of a parabola from focus and vertex, or from directrix and vertex, but that takes some time. I was wondering if there is a way to just sort of see which equation is the right one, because right from the focus and vertex we can see that the parabola will open on the left and its equation will be $y^2 = - 4ax$.

I'm not sure if it's actually possible though, but I think it would be cool if there was a way.

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    $\begingroup$ If you know how to do the work then do the work. $\endgroup$ – John Douma Mar 23 '19 at 0:03
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    $\begingroup$ For a parabola like $y=ax^2+bx+c$ the vertex is located at the point where the derivative vanishes. This is, when $2ax+b=0$, or what is the same $x=-\frac{b}{2a}$. In your case $x$ and $y$ are exchanged. You can see that only A,D, and E result in a vertex where $y=2$. Of those only A results in $x=3$ when $y=2$. Then you only need to check if A has focus at $(1,2)$. $\endgroup$ – user647486 Mar 23 '19 at 0:11
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    $\begingroup$ Why are people voting this question down? This genuinely perplexes me! $\endgroup$ – Theo Bendit Mar 23 '19 at 0:13
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    $\begingroup$ By the way, the equation wouldn't be $y^2 = -4ax$, it would be $(y-\color{red}{k})^2 = -4a(x-\color{red}{h})$, where $(h,k)$ is the vertex (which is known here). Also, $a$ is the focal length, which you can deduce since you are given the focus and vertex. So since you can easily work out $a$ and know $h,k$, you can put them into $(y-\color{red}{k})^2 = -4a(x-\color{red}{h})$, and then expand and rearrange to get the desired answer. $\endgroup$ – Minus One-Twelfth Mar 23 '19 at 0:18
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    $\begingroup$ The vertex tells you that A. is the only possibility. $\endgroup$ – herb steinberg Mar 23 '19 at 0:38

You don't need any particular formula here: just plug the coordinates of the vertex into the equation and you'll see that only in case A the point belongs to the parabola.

  • $\begingroup$ Jesus how did I forget to PITA $\endgroup$ – Arkilo Mar 23 '19 at 13:14

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