Question

Find the standard form of the equation of the parabola with a focus at (0, -6) and a directrix at y = 6.

Answer 1

you know what this looks like? if you do, then we can do it quickly

Answer 2

no

Answer 3

would it just be a x^2 parabola but the starting point is shifted 6 to the down?

Answer 4

i meant down

Answer 5

no

Answer 6

then i dont know

Answer 7

lets draw a picture with the focus and the directrix only

Answer 8

so it isnt a parabola?

Answer 9

yes, it is
the vertex is half way between the focus and the directrix. so the vertex is the origin \((0,0)\)

Answer 10

ohh ok

Answer 11

but you should see from the picture that the parabola opens down, not up

Answer 12

since -6 that means opens down?

Answer 13

ok i see

Answer 14

but that does not make is \(y=-x^2\) we still have to be careful

Answer 15

y=-(x-6)^2?

Answer 16

general form is
\[4p(y-k)=(x-h)^2\] in your case \((h,k)\) is \((0,0)\) so we are at \[4py=x^2\]

Answer 17

ok

Answer 18

your answer would shift \(y=-x^2\) to the right 6 units, but that is not what we are trying to do
all you need now to finish
\[4py=x^2\] is find \(p\) which is the distance between the vertex and the focus
that distance is pretty clearly \(6\)

Answer 19

so it would be y=x^2/24?

Answer 20

don't forget the minus sign

Answer 21

yeah i forgot from the 24

Answer 22

\[-24y=x^2\] or
\[y=-\frac{x^2}{24}\]

Answer 23

want to check it is right?

Answer 24

sure

Answer 25

i wrote "parabola" and then the equation
it gives the focus, directrix, vertex, etc
http://www.wolframalpha.com/input/?i=parabola+-24y%3Dx^2

Answer 26

ohh wow thats convenient lol