OpenStudy (anonymous):

@johnweldon1993

4 years ago
OpenStudy (anonymous):

Find the center, vertices, and foci of the ellipse with equation

4 years ago
OpenStudy (anonymous):

i know the center is (0,0)

4 years ago
OpenStudy (johnweldon1993):

Was the picture cut off on the left? or is that information irrelevant?

4 years ago
OpenStudy (anonymous):

nope its irrelevant, its what i wrote before it

4 years ago
OpenStudy (anonymous):

(0, -25), (0, 25) for the vertices right??

4 years ago
OpenStudy (anonymous):

there arnt any vectors here (; @SolomonZelman

4 years ago
OpenStudy (solomonzelman):

No, there are not:) I can see an ellipse, \(\Large\color{black}{ \frac{x^2}{225}+\frac{y^2}{625}=1\Huge\color{white}{ \rm │ }}\) \(\Large\color{black}{ \frac{x^2}{(±15)^2}+\frac{y^2}{(±25)^2}=1\Huge\color{white}{ \rm │ }}\)

4 years ago
OpenStudy (anonymous):

so this is definetly the foci

4 years ago
OpenStudy (solomonzelman):

vertices are the x intercepts. foci are the y intercepts (in this case they are intercepts, because the center is the origin)

4 years ago
OpenStudy (anonymous):

and this is the vertices (0, -25), (0, 25)

4 years ago
OpenStudy (anonymous):

wait foci are y? i did them the opposite

4 years ago
OpenStudy (solomonzelman):

no, close... `(25,0)` and `(-25,0)`.

4 years ago
OpenStudy (solomonzelman):

And foci are the `Y`s

4 years ago
OpenStudy (anonymous):

Vertices: (-25, 0), (25, 0); Foci: (0, -15), (0, 15) correct?

4 years ago
OpenStudy (solomonzelman):

Yes, and the asymptotes are y=±15/25 = ±3/5

4 years ago
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