Question

Find the center, vertices, and foci of the ellipse with equation x^2/100 + y^2/64= 1.

Answer 1

haven't done ellipises yet

Answer 2

mooor, do you have your ellipse formula? does it resemble what you have?

Answer 3

no :/

Answer 4

@Luigi0210

Answer 5

http://www2.seminolestate.edu/lvosbury/CalculusII_Folder/EllipseEx1.gif :)

Answer 6

ah I'm still a bit confused :/

Answer 7

What jdoe pretty much showed in the image
Confused about what?

Answer 8

well, check 100 and 64, the denominators, can you get a square root off them?

Answer 9

\[\frac{x^2}{100}+\frac{y^2}{64}=1 \implies\\
\frac{(x^2-0)}{10^2}+\frac{(y^2-0)}{8^2}=1 \\
center\ of\ the\ ellipse\ is\ at\ (h,k)\ from\ the\ formula \\
\\
foci = \sqrt={a^2+b^2} \implies \sqrt{10^2+8^2}\]

Answer 10

\[foci = \sqrt{a^2+b^2} \implies \sqrt{10^2+8^2}\]

Answer 11

\[\frac{x^2}{100}+\frac{y^2}{64}=1 \implies\\
\frac{(x-0)^2}{10^2}+\frac{(y-0)^2}{8^2}=1 \\
center\ of\ the\ ellipse\ is\ at\ (h,k)\ from\ the\ formula \\
\\
foci = \sqrt{a^2+b^2} \implies \sqrt{10^2+8^2}\]
there, better :)

Answer 12

\[distance\ from\ center = \sqrt{a^2+b^2} \implies \sqrt{10^2+8^2}\\
foci = (h\ or\ k) + \sqrt{10^2+8^2}\]

Answer 13

once you find your center (h,k), move THAT MUCH to find your foci, moving always over the major axis, of course the major axis will be the one with the bigger number(denominator)

Answer 14

so from theese 4 choices:
Center: (0, 0); Vertices: (-10, 0), (10, 0); Foci: (-8, 0), (8, 0)
Center: (0, 0); Vertices: (0, -10), (0, 10); Foci: (0, -8), (0, 8)
Center: (0, 0); Vertices: (-10, 0), (10, 0); Foci: (-6, 0), (6, 0)
Center: (0, 0); Vertices: (0, -10), (0, 10); Foci: (0, -6), (0, 6)

it would be the first, correct? :)

Answer 15

so from theese 4 choices:
Center: (0, 0); Vertices: (-10, 0), (10, 0); Foci: (-8, 0), (8, 0)
Center: (0, 0); Vertices: (0, -10), (0, 10); Foci: (0, -8), (0, 8)
Center: (0, 0); Vertices: (-10, 0), (10, 0); Foci: (-6, 0), (6, 0)
Center: (0, 0); Vertices: (0, -10), (0, 10); Foci: (0, -6), (0, 6)

it would be the first, correct? :)

Answer 16

so from theese 4 choices:
Center: (0, 0); Vertices: (-10, 0), (10, 0); Foci: (-8, 0), (8, 0)
Center: (0, 0); Vertices: (0, -10), (0, 10); Foci: (0, -8), (0, 8)
Center: (0, 0); Vertices: (-10, 0), (10, 0); Foci: (-6, 0), (6, 0)
Center: (0, 0); Vertices: (0, -10), (0, 10); Foci: (0, -6), (0, 6)

it would be the first, correct? :)

Answer 17

so from theese 4 choices:
Center: (0, 0); Vertices: (-10, 0), (10, 0); Foci: (-8, 0), (8, 0)
Center: (0, 0); Vertices: (0, -10), (0, 10); Foci: (0, -8), (0, 8)
Center: (0, 0); Vertices: (-10, 0), (10, 0); Foci: (-6, 0), (6, 0)
Center: (0, 0); Vertices: (0, -10), (0, 10); Foci: (0, -6), (0, 6)

it would be the first, correct? :)

Answer 18

@e.mccormick

Answer 19

@e.mccormick

Answer 20

@009infinity

Answer 21

@009infinity

Answer 22

OK, so you have the work and just need to onfirm at this point, right?

Answer 23

I'm looking for a better description of the foci formula for you. I think that is wrong in that answer, but want to be sure.

Answer 24

When you find the center, which is just \((0,0)\) in this case, you need to find out how far to go out from the center to be at the foci. This takes the form of:
(Radius of Major)\(^2\) - (Radius of Minor)\(^2\) = (Focal Distance)\(^2\)
This is most commony shortened to:
\(a^2-b^2=c^2\)

Here is a reference to that:
http://www.dummies.com/how-to/content/how-to-graph-an-ellipse.html

So I am not sure where that one person got: \(\sqrt{10^2+8^2}\), that would put the foci at \(\pm\sqrt{164}\), whicht hey are not.