What is the equation of the ellipse with co-vertices (-20, 0), (20, 0) and foci (0, -8), (0, 8)?
@DarkBlueChocobo come here to play, you are the right person to solve it. practice on other's problem is the best way to confirm your knowledge.
@juliamae10 since I saw you post 2 exactly- the- same problem on 2 different sections, I assume that you just want the answer. If it is so, why don't we let other student who needs practice work out on your problem, right? both you guys get what you want.
@Loser66 co vertices are they half of the vertices ?
@DarkBlueChocobo foci is about c, and it is on the vertical line, hence your main axis is y axix |dw:1431803368050:dw|
Or am I getting mixed up between hyperbolas and parabolas
co-vertices is a minor one, not the long one
so the length of the minor axis 20?
|dw:1431803420147:dw|
40, not 20
you find a by c^2 = a^2 -b^2
hence a^2 =??
does c^2 = 8^2? @Loser66
yup
b^2=20^2?
yup again
8^2=a^2-20^2
yup
need a^2
so -a^2=400-64 -a^2=336
hey
Did i mess up with -a^2?
\(8^2 = a^2 -400\\a^2 = 8^2 +400\) just +400 both sides
ohhhh that was simple: a=sqrt464
nope, let a^2 as it is. \(a^2 = 464\)
now, you have a^2, b^2, just plug all in formula for VERTICAL ellipse formula that is \(\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1\)
\[\frac{ x^2 }{ 20^2}+\frac{ y^2 }{ 464^2 }=1\]
\[\frac{ x^2 }{ 400}+\frac{ y^2 }{ 464 }=1\]
And that would be the equation
Remember, if it is HORIZONTAL ellipse, then the denominator of x term is a^2 if it is VERTICAL ellipse, the denominator of x term is b^2 They are different, that is why at the beginning, I have to define which one is the correct form by using foci and co-vertices.
yes, that is the required equation.
Yes if the number is bigger under y that means that is it vertical
|dw:1431804460345:dw|
if the bigger number is under x it is horizontal
ok, good job
Thank you for leading me to solve it
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