Question

Using a directrix of y = -2 and a focus of (2, 6), what quadratic function is created?

Answer 1

I think it is a parabola

Answer 2

my answer choices are all in this format

Answer 3

and I know that it'll be - 2 at the end from the directrix

Answer 4

actually now that I'm looking at all the choices, the first fraction part has to be either 1/8 or -1/8, but I don't remember how to find that part

Answer 5

if I write your parabola as below:
\[y=ax ^{2}+bx+c\]
then coefficients a, b, c are related to the coordinates of focus and to the equation of directrix, by the subsequent formulas:
\[-\frac{ b }{ 2a }=2\]
\[\frac{ 1-\Delta }{ 4a }=6\]
\[-\frac{ 1+\Delta }{ 4a }=-2\]
where:
\[\Delta =b ^{2}-4ac\]

Answer 6

how can I find the values for a, b, and c? I don't think I can put them in the form of y=ax^2 + bx + c

Answer 7

don't worry after we found a, b and c we will get the parabola in your form
Now I'm finding a, b and c

Answer 8

ok

Answer 9

from the preceding formulas, we get:
\[a=\frac{ 1 }{ 16 },b=-\frac{ 1 }{ 4 },c=\frac{ 9 }{ 4 }\]
so our parabola is:
\[y=\frac{ 1 }{ 16 }x ^{2}-\frac{ 1 }{ 4 }x+\frac{ 9 }{ 4 }=\frac{ 1 }{ 16 }(x ^{2}-4x+36)\]

Answer 10

\[=\frac{ 1 }{ 16 }(x ^{2}-4x+4+32)=\frac{ 1 }{ 16 }[(x-2)^{2}+32]=\]

Answer 11

\[\frac{ 1 }{ 16 }[(x-2)^{2}]+\frac{ 32 }{ 2 }=\frac{ 1 }{ 16 }[(x-2)]^{2}+2\]

Answer 12

so your parabola is:
\[y=\frac{ 1 }{ 16 }[(x-2)]^{2}+2\]

Answer 13

ok thank you! :)

Answer 14

thank you! :)

Answer 15

I'm sorry, but I don't know a simpler method!

Answer 16

lol that's ok