Question

What is the equation of a parabola with the given vertex and focus?
vertex: (-2,5) focus: (-2,6)

Answer 1

@iGreen

Answer 2

I don't understand the focus part. The only thing I know about the focus is it's equidistant to the vertex as the directrix is to the vertex

Answer 3

I know that vertex form is y=a(x-h)^2+k but how do I find a?

Answer 4

I'm not sure either..

Answer 5

ohhhh wait I think I know. would I plug in the x and y from the focus into the vertex form to find a?

Answer 6

@iGreen

Answer 7

first off, do you know what this looks like? if so, it is real easy
if not, it is extra confusing

Answer 8

no...

Answer 9

I just have those two coordinates

Answer 10

yes i know that
the question is, given those two points, the vertex and the focus, do you know what the parabola looks like? i.e. does it open left, right, up, or down? we need that first

Answer 11

I have no clue. I'm assuming it opens up because the focus is one unit above the vertex and the directrix would have to be one unit below... but my reasoning could be wrong :/

Answer 12

because the vertex is below the focus, this means it opens up
and yes, the directrix is one unit below the vertex, so it is \(y=4\) but we don't need that for the equation, now we are done in one step

Answer 13

general form is
\[4p(y-k)=(x-h)^2\] where the vertex is \((h,k)\)
here \(p=1\) the distance between the focus and the directrix, so \(4p=4\) and \((h,k)=(-2,5)\)

Answer 14

you see once you know what it looks like, it is real easy
plug them in and get
\[4(y-5)=(x+2)^2\]

Answer 15

okay, so then you'd distribute the 4 then just solve for y?

Answer 16

Oh and do (x+2)(x+2)

Answer 17

i would leave it like it is, since that is a general form, but if you need to solve for \(y\) go head
i would not square it though

Answer 18

why not?

Answer 19

\[4y-20=(x+2)^2\\
4y=(x+2)^2+20\\
y=\frac{1}{4}(x+2)^2+5\]

Answer 20

it depends entirely on the form you want to put it in

Answer 21

it didn't specify...

Answer 22

if you want standard for \\[y=ax^2+bx+c\] then go ahead
if you want "vertex form" then the last line above

Answer 23

i would leave it as
\[4(y-5)=(x+2)^2\]and be done with it

Answer 24

okay. I just thought I needed to go from vertex form to get to standard form... but okay. thanks for your help :)