Question

write an equation of an ellipse with foci at (+-3,0) and vertices at (+-6,0)

Answer 1

lol everyone hates these conic section problems
they avoid them like the plague
first off, is it clear what it looks like?

Answer 2

we need that before we can start

Answer 3

yes

Answer 4

so you know that the larger number goes under the \(x^2\) term right?

Answer 5

general form is
\[\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\]
in your case it should be clear from the vertices an foci that the center is \((0,0)\) so it is just going to be
\[\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\]

Answer 6

thats it?

Answer 7

we need \(a\) and \(b\)

Answer 8

you are pretty much given \(a\) in the question
since the foci are \((-6,0)\) and \((6,0)\) you know \(a=6\) and so \(a^2=36\)

Answer 9

now we are here \[\frac{x^2}{36}+\frac{y^2}{b^2}=1\] but we still need \(b\)

Answer 10

and b?

Answer 11

\[a^2-c^2=b^2\] and in this case since the foci are \((-3,0)\) and \((3,0)\) you get \(c=3\) making
\[a^2-c^2=b^2\\
36-9=b^2\\
27=b^2\]

Answer 12

final answer, assuming i did not make a mistake, is
\[\frac{x^2}{36}+\frac{y^2}{27}=1\] want to check it ?

Answer 13

how do i check it

Answer 14

http://www.wolframalpha.com/input/?i=ellipse+%28x^2%29%2F36%2B%28y^2%29%2F27%3D1

Answer 15

looks like we got the foci and vertices correct right?
that is the easy check

Answer 16

@satellite73 : A typo again - in your previous post, you stated the following:
"since the foci are (−6,0) and (6,0) you know a=6 and so a^2=36 "
This should instead read as follows:
"since the vertices are (−6,0) and (6,0) you know a=6 and so a^2 =36"

I need to highlight these errors as someone new to this area (& wishes to learn it) might be confused with the comments you make. Thanks.