4x 2 16y 2 64

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Question 30463: (M) Question:
Find the eye, vertices, and foci for the ellipse:
4x^2+16y^two=64
(A) eye (0,0) vertices (0, +4) foci (0, +3.v)
(B) centre (2, 4) vertices (four, 0) and (iv,eight) foci (4, -1.v) and (iv, 5.v)
(C) centre (0,0) vertices (+4, 0) foci (+iii.v,0)
(D) center (4,2) vertices (0,2) and (4,ii) foci (0.5,ii) and (7.5,2)
Delight be specific as to whcih letter of the alphabet is correct
Answer by sdmmadam@yahoo.com(530) (Evidence Source):
You tin put this solution on YOUR website!
4x^2+16y^2=64
Dividing by 64
ten^2/sixteen +y^2/4 = one
That is x^2/(4^ii) +y^2/(2^ii) = 1 ----(1)
which is in the standard form x^2/a^2 +y^two/b^2 = 1 (with a>b)
The major axis is the x-axis
and the small-scale 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(four,0) and A'(-4,0) on the major axis
and B(0,2) and B'(0,-2) on the small-scale axis.
To find the eccentricity e we apply the formula
b^ii = a^2(1-e^2)
four = 16(ane-e^2)
1 = 4(1-eastward^2) (dividing by iv)
ane = 4 - 4e^two
4e^2 = 4-ane
4e^2 = three
4e^2 = four-i
4e^2 = 3
eastward^2= 3/four
Therefore east = [sqrt(3)]/2 (taking the positive sqrt as e > 0)
The foci are given by South(ae,0) and S'(-ae,0)
And (ae) = 4X(rt(three)]/two = two(sqrt(3))
Therefore the foci are
Southward(ae,0)= South(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 fix just 1 vertex and one focus given forth with the center
The option (C) is close to the answer