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locus and parabola help (1 Viewer)

kaha167

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hey guys,
ive tried to answer this question but feel that im stuck. i can't find the focal length of the second parabola...?
parabola x^2 = 16y coordinates of the focus = S
A second parabola also has its focus at S. the vertex of this parabola is at (-2, 4)
Find the equation of the parabola:
i have gotten to:
(x - h)^2 = 4a(y-k)
(x+2)^2 = 4a(y-k)
i dont understand how to get a when they are on the same line [there is no gap between teh vertex and the focus]
thanks for the help
:)
 
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Focus of the first parabola is (0,4), so focal length of 2nd is 2 units, so substitute a=2 and V(-2,4)

so...

(y - k)2 = 4a(x - h)
(y - 4)2 = 4*2(x + 2)
(y - 4)2 = 8(x + 2)
 
Last edited:

kaha167

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well that makes me feel stupid. lol
thanks very much
 

hscishard

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hey guys,
ive tried to answer this question but feel that im stuck. i can't find the focal length of the second parabola...?
parabola x^2 = 16y coordinates of the focus = S
A second parabola also has its focus at S. the vertex of this parabola is at (-2, 4)
Find the equation of the parabola:
i have gotten to:
(x - h)^2 = 4a(y-k)
(x+2)^2 = 4a(y-k)
i dont understand how to get a when they are on the same line [there is no gap between teh vertex and the focus]
thanks for the help
:)
:S This is year 11 stuff. Are you catching up?

This one is a sideways parabola.
First step is to find S.
Since S if the focal point, and x^2=16y, a is 4. Therefore it's focus is at (0,4)

But the vertex of the second parabola is at (-2,4). Sketching will prove very useful here.

If you sketch the info so far, you'll notice the second parabola is a sideways parabola. We use (y-k)^2=4a(x=h)

The focal length can be found easily if you sketch it, it's 2. The vertex is (-2,4)
Therefore = (y-4)^2 = 8(x+2)
 

kaha167

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yeh i am.
thats why its frustrating. im trying to finish sheets off in a rush before tuesday and im just trying to get stuff done.
i know it, just not thinking.
thanks for the help
 

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