Question 30463: (M) Question:
Find the center, vertices, and foci for the ellipse:
4x^2+16y^2=64
(A) center (0,0) vertices (0, +4) foci (0, +3.5)
(B) center (2, 4) vertices (4, 0) and (4,8) foci (4, -1.5) and (4, 5.5)
(C) center (0,0) vertices (+4, 0) foci (+3.5,0)
(D) center (4,2) vertices (0,2) and (4,2) foci (0.5,2) and (7.5,2)
Please be specific as to whcih letter is correct
Answer by sdmmadam@yahoo.com(530)   (Show Source):
You can put this solution on YOUR website! 4x^2+16y^2=64
Dividing by 64
x^2/16 +y^2/4 = 1
That is x^2/(4^2) +y^2/(2^2) = 1 ----(1)
which is in the standard form x^2/a^2 +y^2/b^2 = 1 (with a>b)
The major axis is the x-axis
and the minor axis is the y-axis
Hence Centre C= (0,0)
semi-major length = a = 4 and
semi-minor length = b = 2
The vertices are A(4,0) and A'(-4,0) on the major axis
and B(0,2) and B'(0,-2) on the minor axis.
To find the eccentricity e we use the formula
b^2 = a^2(1-e^2)
4 = 16(1-e^2)
1 = 4(1-e^2) (dividing by 4)
1 = 4 - 4e^2
4e^2 = 4-1
4e^2 = 3
4e^2 = 4-1
4e^2 = 3
e^2= 3/4
Therefore e = [sqrt(3)]/2 (taking the positive sqrt as e > 0)
The foci are given by S(ae,0) and S'(-ae,0)
And (ae) = 4X(rt(3)]/2 = 2(sqrt(3))
Therefore the foci are
S(ae,0)= S(2(rt3),0) and S'(-ae,0)=(-2(rt3),0)
Remark: We are asked to give center, vertices(notice the plural)
and the foci(notice the plural) and in the answer set only one vertex and one focus given along with the center
The choice (C) is close to the answer
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