Asked by Kali
Ellipses
[(x+3)^2 / 12] + [(y-2)^2 / 16]
I Have to find the following:
Center:
Foci:
Vertices:
A:
B:
C:
Eccentricity:
[(x+3)^2 / 12] + [(y-2)^2 / 16]
I Have to find the following:
Center:
Foci:
Vertices:
A:
B:
C:
Eccentricity:
Answers
Answered by
Steve
check your text. The ellipse
(x-h)^2/a^2 + (y-k)^2/b^2 has
center: (h,k)
semi-axes a and b
c^2 = a^2-b^2
foci are at ±c away from the center, along the major axis.
eccentricity = c/a
So, plug in your numbers. What do you get?
(x-h)^2/a^2 + (y-k)^2/b^2 has
center: (h,k)
semi-axes a and b
c^2 = a^2-b^2
foci are at ±c away from the center, along the major axis.
eccentricity = c/a
So, plug in your numbers. What do you get?
Answered by
Reiny
And what have you got so far?
BTW, the ellipse should be an equation of the type
(x+3)^2/12 + (y-2)^2/16 = 1
BTW, the ellipse should be an equation of the type
(x+3)^2/12 + (y-2)^2/16 = 1
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