SOLUTION: Find the standard form of the equation of the ellipse with the given characteristics. Center: (6, 7); a = 3c; foci: (2, 7), (10, 7)

Algebra ->  Equations -> SOLUTION: Find the standard form of the equation of the ellipse with the given characteristics. Center: (6, 7); a = 3c; foci: (2, 7), (10, 7)      Log On


   



Question 1136850: Find the standard form of the equation of the ellipse with the given characteristics. Center: (6, 7); a = 3c; foci: (2, 7), (10, 7)
Found 2 solutions by greenestamps, MathLover1:
Answer by greenestamps(13200) About Me  (Show Source):
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c is the distance from the center of the ellipse to each focus. The coordinates of the center and the two foci tell us c=4; that makes a, the semi-major axis, 3c = 12.

a, b, and c in an ellipse are related by c^2 = a^2-b^2; that gives us b, the semi-minor axis, equal to 8*sqrt(2).

The standard form of the equation of an ellipse with center (h,k), semi-major axis a, and semi-minor axis b is

%28x-h%29%5E2%2Fa%5E2%2B%28y-k%29%5E2%2Fb%5E2+=+1

You have all the numbers to fill in to write the equation.

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
Find the standard form of the equation of the ellipse with the given characteristics.
Center: (6, 7);
a+=+3c;
foci: (2, 7), (10, 7)

The equation of an ellipse is :
%28x-h%29%5E2%2Fa%5E2%2B%28y-k%29%5E2%2Fb%5E2=1 for a horizontally oriented ellipse and
%28x-h%29%5E2%2Fb%5E2%2B%28y-k%29%5E2%2Fa%5E2=1 for a vertically oriented ellipse.
(h,k) is the center and the distance+c from the center to the foci is given by a%5E2-b%5E2=c%5E2

a is the distance from the center to the vertices and b+is the distance from the center to the co-vertices.
The center of the ellipse is (6, 7)=>h=6 and k=7
a is the distance between the center and the vertices, so a=9
c is the distance between the center and the foci, and foci is at (2, 7), (10, 7)
distance from 6 to 2 is 4, and distance from 6 to 10 is 4
so c=4
since a+=+3c, we have a+=+3%2A4=>a+=+12
now find b
12%5E2-b%5E2=4%5E2
12%5E2-4%5E2=b%5E2
144-16=b%5E2
b%5E2=128
%28x-6%29%5E2%2F144%2B%28y-7%29%5E2%2F128=1 for a horizontally oriented ellipse