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- Meehan98

Identify an equation in standard form for an ellipse with its center at the origin, a vertex at (0, 12), and a focus at (0, 8).

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- Meehan98

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- Meehan98

It appears that it's a horizontal ellipse so I know that the equation will be \[\frac{ x ^{2} }{ a ^{2} } + \frac{ y ^{2} }{ b ^{2} }=1\]

- anonymous

That is right.

- Meehan98

I just don't know how to find the other vertex in order to find b^2.

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- anonymous

Actually isn't it a verticle ellipse.

- anonymous

Because the vertex (furtherest point) is 12 up from the origin

- Meehan98

Oh yes, that's right!

- anonymous

So the semi major axis is 12, in the y direction

- anonymous

Therefore the equation is (y^2)/(12^2) + (x^2)/(b^2) = 1

- anonymous

Now for the b value I think you use the rule that for any point on the ellipse, the sum of the distance from both foci is 2a (a is the semi major axis)

- anonymous

Ill draw a graph to show

- Meehan98

So, you use the constant sum of an ellipse formula: d=PF1 + PF2

- anonymous

|dw:1441973292765:dw| in the picture i took the point to be the semi minor axis. Since the ellipse is symmertrical (both foci same distance) the distance from that point to each foci is a (semi major axis)

- anonymous

as they are same length and both add to 2a

- anonymous

Does that make sense?

- anonymous

Are you there?

- Meehan98

I'm sorry, just taking me little bit to understand. I'm used to strictly using formulas to get the solution.Okay, how do you know what point to pass through on the x-axis?

- anonymous

What do you mean exactly?

- anonymous

Btw you gotta understand really whats happening with everything in maths it makes things so much easier, and theres less things to remember.

- Meehan98

- Meehan98

This is the problem. I know that the denominator for y^2 is 144, but I don't know the denominator for x^2 because I'm not given thenother focus.

- Meehan98

Okay, I got it now! I used c^2=a^2-b^2 to get b^2= 80. So, my answer would be: (y^2/144) + (x^2/80)=1. Thanks for your help!

- anonymous

Both foci are in line with the centre and are the same distance from the centre. Therefore if you know one you can find the other.
If you take the point on the ellipse closest to the centre, the semi minor axis (the b value, last value we need) its not closer to one focus than the other. If the sum of the distances from any point to each foci is 2a, taking the closest point I just talked about, the distance from each focus to it is a (both distances are equal and must sum to 2a).
Using pythagoras theorem where f is the distance from centre to focus and a is the semi major axis and b is the semi minor axis,
(f^2) + (b^2) = (a^2) we know f=8 and a=12
therefore (8^2) + (b^2) = (12^2)
64 + (b^2) = 144
b^2 = 80
therefore (x^2)/(80) + (y^2)/(144) = 1

- anonymous

Oh i just read yours lol, yes you did eveything right, good job :)

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