Question

How to find the foci of an ellipse?
(x-7)^2/4 + (y+3)^2/ 16 = 1

Answer 1

\[\large \frac{ (x-7)^2 }{ 4 } + \frac{ (y+3)^2 }{ 16 } = 1\]

Answer 2

The equation of ellipse in conics form is:\[\Large \frac{ (x - h)^2 }{ a^2 } + \frac{ (y - k)^2 }{ b^2 } = 1\]where (h.k) is the center; a and b are semi-major and semi-minor axes.
The distance c from the center to either foci is given by the formula:
c^2 = a^ - b^2
Compare the general conics equation to the given equation and you can identify h, k, a, and b. Compute c. Find the point on either side of the center at distance c.

Note: If a > b, the ellipse will be elongated along the x-axis
If a < b, the ellipse will be elongated along the y-axis.
The focus will be on the longer axis.

Answer 3

Add an absolute sign around |a^2-b^2| above.

Answer 4

Okay.
so..
c^2 = 16 - 4
c^2 = 12
c = 2√3

Answer 5

(h,k) = (7,-3)

Answer 6

yes. a = 2, b = 4. So the ellipse will be elongated along the y axis and the foci will be located along the y axis. (7, -3 +/- 2sqrt(3)) will be the coordinates of the foci.

Answer 7

foci not on the y axis but on a line parallel to the y axis.

Answer 8

Alright, thanks.

Answer 9

you are welcome.