Question

--Two questions for a fan please help--
1.What is the equation of the ellipse with foci (0, 3), (0, -3) and co-vertices (1, 0), (-1, 0)

2.What is the equation of the ellipse with vertices at (-25, 0), (25, 0) and co-vertices (0, -15), (0, 15)

Answer 1

what is a co-vertex? is that in line with the focus or perp to them?

Answer 2

from the points given, it looks like perps

Answer 3

do we agree that these are centered at the origin?

Answer 4

yes they are centered at the origin

Answer 5

then can you give me the general setup for an ellipse equation, them we can just fill in the parts as we find them

Answer 6

x^2 over b^2 + y^2 over b^2=1

Answer 7

allowing for a mistype, lets say x^2 over a^2, but the rest looks good

Answer 8

okay

Answer 9

the next step would be to plot the points in a relative fashion ...

.What is the equation of the ellipse with foci (0, 3), (0, -3) and co-vertices (1, 0), (-1, 0)


f 0, 3

v-1,0 v 1,0

f 0,-3


hmmm, so its like:

Answer 10

\[\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\]
when y=0, x=1 so we have
\[\frac{1^2}{a^2}=1\]
a^2 is simple to determine from then

Answer 11

the pythag thrm (will look odd tho) will determine the b^2 for us:

a^2 + c^2 = b^2

Answer 12

c is the distance from center to focus so:

1^2 + 3^2 = b^2 seems to fit, does that make sense?

Answer 13

yes

Answer 14

what does our final result look like thn?

Answer 15

Well I already knew that you the Pythagorean thrm to solve these types of problems, my online class taught me that. The issue is that I did not know how to set up the problems and I still don't. But thank you for your time.

Answer 16

the setup is just a matter finding a center, and finding the distances from that center to the vertexes

Answer 17

plotting helps see what kind of shape takes place, whether its longer on along the x or y lines

Answer 18

.What is the equation of the ellipse with

foci (0, 3), (0, -3) and co-vertices (1, 0), (-1, 0)

given that x^2/a^2 + y^2/b^2 = 1, we are given that when y=0, x=1 or -1 and can determine that:

1^2/a^2 + 0^2/b^2 = 1, which simplfies to 1/a^2 = 1, so a^2 has to be 1

the rest is pythag to find b^2