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- anonymous

Derive the equation of the parabola with a focus at (−5, 5) and a directrix of y = −1

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- anonymous

Derive the equation of the parabola with a focus at (−5, 5) and a directrix of y = −1

- chestercat

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- campbell_st

have you any method to use for this question...?

- anonymous

not really no

- campbell_st

so what methods do you know..
1. find the focal length and then find the vertex...
2. use the distance formula..

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- anonymous

number 1

- campbell_st

ok.... so start by plotting the given information
|dw:1437680444218:dw|
so this tells me the parabola is concave up... does that make sense...

- anonymous

yes

- campbell_st

ok... so the line of symmetry is x = -5
and the vertex is a units below the focus...
where a = the focal length
so |dw:1437680607256:dw|

- anonymous

so the length between the focus and the directrix is 6

- campbell_st

the distance between the focus and directrix is 2a
the vertex is midway between the vertex and directrix on the line of symmetry

- campbell_st

that's great
so
2a = 6
therefore the focal length is a = 3
is that ok..?

- anonymous

yes

- campbell_st

|dw:1437680777548:dw|

- campbell_st

so where do you think the vertex is located..?

- anonymous

woud it be at (-5,2)

- campbell_st

perfect....
now to get the correct equation I use a standard form
\[(x - h)^2 = 4a(y - k)\]
(h, k) is the vertex and a is the focal length
so the equation is
\[(x + 5)^2 = 4 \times 3 (y - 2)\]
so this can be simplified and rewritten in the form you need.

- campbell_st

hope it all makes sense

- anonymous

Thank you, it really does help

- campbell_st

if you need it rewritten I'd use
\[y = \frac{1}{12}(x + 5)^2 + 2\]

- anonymous

That's exactly what I thought it was. Thank you so much

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