Question

What are the foci of the following graph?

Answer 1

Last graph! =) @jim_thompson5910

Answer 2

The length of the major axis is 6 units (count from y = 1 to y = -5 to get 6). Half of that is 6/2 = 3 units.


The length of the minor axis is 2 units (count from x = -1 to x = 1 to get 2). Half of that is 2/2 = 1 unit.



The equation of an ellipse is (x-h)^2/a^2+(y-k)^2/b^2=1


So because the major axis is vertical, this means b = 3. So a = 1


The center of this ellipse is (0,-2). You can see this visually or you can find this by averaging the vertices and co-vertices.


So h = 0 and k = -2


So...


(x-h)^2/a^2+(y-k)^2/b^2=1


(x-0)^2/1^2+(y-(-2))^2/3^2=1


x^2/1+(y+2)^2/9=1


x^2+(y+2)^2/9=1


Therefore the equation is \[\Large x^2+\frac{(y+2)^2}{9}=1\]

Answer 3

One more sec while I find the foci

Answer 4

The foci lie along the major axis. So the x coordinates of both foci will be x = 0


The y coordinates are what we want.


To find the y coordinates, we need to know the distance the foci are from the center. They are equally spaced from the center and this distance will be known as 'c'.


It turns out that c^2 = b^2-a^2


So

c^2 = b^2-a^2


c^2 = 3^2-1^2


c^2 = 9-1


c^2 = 8


c = sqrt(8)


c = 2*sqrt(2)


So we're adding and subtracting 2*sqrt(2) to the y coordinate of the center to get the y coordinates of the foci


So the two foci are: \[\Large \left(0,-2+2\sqrt{2}\right) \text{ and } \left(0,-2-2\sqrt{2}\right)\]

Answer 5

Thank you very much for all your amazing help! =)

Answer 6

anytime