Question

Find the equation of the ellipse with the following properties.

The ellipse with foci at (0, 6) and (0, -6); y-intercepts (0, 8) and (0, -8).

Answer 1

please help anyone i am so lost i have been waiting 11 hours b4 reposting

Answer 2

from the graph, we can see pretty much what the MINOR AXIS is
it goes up to 8, then down to -8
so you should be able to tell what the "b" component is

it's noticeable the ellipse center is at the origin, due to the location of its foci
the MAJOR AXIS lies where the foci are
so is a horizontal traverse axis ellipse

Answer 3

so... what's the "a" component? well, we dunno
but we know the foci are at \(\bf (\pm 6, 0)\)

the distance from the center of the parabola, to either focus, has a distance of "c"
notice, the center of the parabola is the origin, (0, 0)
from there to a focus it goes 6 units, either left or right

thus c = 6
and keep in mind that that distance is also \(\bf c=\sqrt{a^2-b^2}\implies c^2=a^2-b^2\implies c^2+b^2=a^2\\ \quad \\
\implies \sqrt{c^2+b^2}=a\)

Answer 4

once you get "a" and "b", plug them in the ellipse equation \(\bf \cfrac{(x-h)^2}{a^2}+\cfrac{(y-k)^2}{b^2}=1\implies \cfrac{(x-0)^2}{a^2}+\cfrac{(y-0)^2}{b^2}=1\\ \quad \\\implies \cfrac{x^2}{a^2}+\cfrac{y^2}{b^2}=1\)