Algebra II-Please check
posted by Jodi .
Please help-What is the equation of the ellipse with foci (0,6), (0,-6) and co-vertices (2,0),(-2,0)
Please explain the steps because I have 5 to do for homework-Thank you-I'm really stuck on this
I thought the answer would be x^2/4 + y^2/40 = 1 but that can't be because the choices are:
x^2/1 + y^2/40 = 1 or x^2/1 + y^2/36
I'm really confused
The focus points show that the x-coordinate is constant @ zero, and the
y-coordinate varies from -6 to +6. Therefore, the focus points are on a vertical line. Which means we have a y-ellipse and a^2 is under y^2 in the Eq.
X^2/b^2 + Y^2/a^2 = 1.
b = +-2.
F^2 = a^2 - b^2,
F^2 = a^2 - 2^2 = 6^2,
a^2 - 4 = 36,
a^2 = 40,
Eq: X^2/4 + Y^2/40 = 1.
Check: Let Y = 0 and solve for X:
X^2/4 + 0/40 = 1,
X^2/4 = 1,
Multiply both sides by 4:
X^2 = 4,
X = +-2 = b = co-vertices = x-intercepts. In your 1st choice, the "1"
should be a 4.
1. "a" is always greater than b.
2. If this was an x-ellipse, a^2
would be located under X^2.
3. "a" is always on the major(longst)