What is the equation of the quadratic graph with a focus of (8, −8) and a directrix of y = −6? and can someone show me how to solve problems like this?

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- anonymous

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- anonymous

i have two others like this
What is the equation of the quadratic graph with a focus of (4, −3) and a directrix of y = −6?
and
What is the equation of the quadratic graph with a focus of (1, 1) and a directrix of y = −1?
but if someone could just tell me how to solve these that'd be cool too

- campbell_st

ok... there are several ways to do this
I start by graphing things so I know what I'm looking at
|dw:1436903991982:dw|
so looking at things... the directrix is above the focus so the parabola is concave down.
so then I know the general form of the curve is
\[(x -h)^2 = -4a(y - k)\]
where (h, k) is the vertex and a is the focal length.

- anonymous

okay, I'm getting it so far

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## More answers

- campbell_st

|dw:1436904182134:dw|
What is the distance between the focus y = -6 and the y value in the focus y = -8
this is double to focal length or 2a... so if you can find it, then halve the answer you have the focal length

- anonymous

so the focal length would be 4?

- campbell_st

|dw:1436904321116:dw|

- campbell_st

close its y = -6 to y = -8
so the distance is 2 units
so
2a = 2 or a = 1
so the focal length a = 1
next the vertex is 1 unit above the focus on the line of symmetry
vertex is at (8, -8 + a)
where do you think the vertex is..?

- anonymous

(8, -7) ?

- campbell_st

great so substituting the vertex and focal length into the general form you would have
\[(x - 8)^2 = 4 \times 1 \times (y +7)\]
or
\[(x -8)^2 = 4(y + 7)\]
you may have to rewrite the equation with y as the subject

- campbell_st

which would require you to divide both sides on the equation by 4
then subtract 7 from both sides...

- anonymous

okay that lost me a little bit... my answer starts with f(x) as the subject which I'm pretty sure is the same as y but none of my options match what i got

- campbell_st

ok if you have
\[(x-8)^2 = 4(y + 7)\]
divide both sides by 4
\[y + 7 = \frac{1}{4}(x -8)^2 \]
now subtract 7 from both sides and change y to f(x)

- anonymous

so i got \[f(x) = \frac{ 1 }{ 4 } (x-8)^{2} + 7\] but my only options are \[f(x) = \frac{ 1 }{ 4 } (x-1)^{2}\] and \[f(x) = \frac{ 1 }{ 4 } (x-1)^{2} +1\]

- anonymous

so would it be the first one?

- anonymous

I'm sorry I'm a moron i was looking at the wrong question....

- anonymous

but would you mind walking me through the other two that i posted?

- campbell_st

oops I forgot the negative it should be
\[f(x) = -\frac{1}{4}(x -8)^2 - 7\]
as the parabola is concave down.
I think the answer choices you posted don't match the question you posted...

- anonymous

yeah i was looking at my options for the third question not the first...

- campbell_st

so there is a solution for your question..
if you need help with others in the same form, you'll need to post them as seperate questions...
always start by plotting the information you are given, that way you'll know the concavity of the parabola.
then work from there...

- anonymous

is the equation the same no matter which way the parabola is facing?

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