Question

What is the equation of the ellipse with foci at (5, 0), (-5, 0) and vertices (9, 0), (-9, 0)?
Can you help explain it to me as well? :S

Answer 1

Okay, do you know about the general equations of an ellipse?

Answer 2

The vertices are on the x axis so the major axis is horizontal and a = 9. The focal points are (c,0) and (-c,0)

Answer 3

yes .

Answer 4

So c = 5. We know that a^2-b^2=c^2 or 81-b^x=25

Answer 5

So b^2=56 and b=2sqrt14

Answer 6

\[\frac{x^2}{81}+\frac{y^2}{56}=1\]

Answer 7

Thanks I understand it better :)

What is the equation of the ellipse with foci (0, 4), (0, -4) and vertices (0, 8), (0, -8)?

so for this one it would be x^2/16 + y^2/64?

Answer 8

x^2/16 + y^2/64 = 1

Answer 9

I got 48 for b^2. How did you get 16?

Answer 10

a=8, c=4
a^2-b^2=c^2
8^2-b^2=4^2

Answer 11

Solve that for b^2

Answer 12

64 - b^2 = 16

Answer 13

yes. Finish it

Answer 14

it would be 48.

Answer 15

yes.

Answer 16

so the answer is x^2/48 + y^2/16?

Answer 17

And:
\[\frac{x^2}{48}+\frac{y^2}{64}=1\]

Answer 18

I have one last one..can you check if its correct or not?

Answer 19

yes

Answer 20

What is the equation of the ellipse with co-vertices (-20, 0), (20, 0) and foci (0, -8), (0, 8)?

-20^2 + b^2 = 0 ?

Answer 21

Co-vertices are (-20,0) and (20,0) so that means that b = 20
Focal points are (0,8) and (0,-8) so that means that c = 8

Answer 22

a^2-20^2=8^2

Answer 23

oh ok

Answer 24

a^2 - 400 = 64

Answer 25

yes

Answer 26

so the answer is x^2/400 + y^2/64 = 1?

Answer 27

\[\frac{x^2}{400}+\frac{y^2}{464}=1\]

Answer 28

a^2=464

Answer 29

Oh okay i get it.

Answer 30

Good for you.

Answer 31

Thanks for everything :)

Answer 32

yw