Question

Using a directrix of y = -3 and a focus of (2, 1), what quadratic function is created?

Answer 1

you need the vertex first
do you know it ?

Answer 2

how can you find that

Answer 3

it is half way between the directrix and the focus

Answer 4

half way between \((2,1)\) and \((2,-3)\) is \((2,-1)\)

Answer 5

ohh okay gotcha

Answer 6

and the picture tells you it opens up. so the \(x\) term is squared
\[4p(y-k)=(x-h)^2\]

Answer 7

you have \((h,k)\) is \((2,-1)\) so far we have
\[4p(y+1)=(x-2)^2\] all we need is \(p\)

Answer 8

\(p\) is the distance between the focus and the vertex
the distance between \((2,-1)\) and \((2,1)\) is \(2\) so \(p=2\)

Answer 9

giving \[8(y+1)=(x-2)^2\]

Answer 10

and as usual we can cheat and check the answer
http://www.wolframalpha.com/input/?i=parabola+8%28y%2B1%29%3D%28x-2%29^2

Answer 11

ooh you need to solve this for \(y\) don't you?

Answer 12

i think so

Answer 13

we cheat for that too
http://www.wolframalpha.com/input/?i=+8%28y%2B1%29%3D%28x-2%29^2

Answer 14

not that it is that difficult to solve this for \(y\)
looks like you get y = x^2/8-x/2-1/2

Answer 15

hope the steps are clear ish

Answer 16

yes i got it thanks again