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What is the center, foci, and the lengths of the major and minor axes for the ellipse,

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(x+3)^2/36 + (y-4)^2/9=1.
asked Mar 17, 2014 in ALGEBRA 2 by andrew Scholar

1 Answer

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The equation of ellipse is (x + 3)2/36 + (y - 4)2/9 = 1.

The standard form of the equation of an ellipse center (h, k) with major and minor axes of lengths 2a and 2b (where 0 < b < a) is (x - h)2/a2 + (y - k)2/b2 = 1 or (x - h)2/b2 + (y - k)2/a2 = 1.

The vertices and foci lie on the major axis, a and c units, respectively, from the center (h, k)  and the relation between a, b and c is c2 = a2 - b2.

Because the denominator of the x2 - term (36) is larger than the denominator of the y2 - term (9), the major axis is horizontal.

Compare the equation (x + 3)2/36 + (y - 4)2/9 = 1 with (x - h)^2/a^2 + (y - k)^2/b^2 = 1.

a2 = 36, b2 = 9, h = - 3 and k = 4

a = 6 and b = 3.

To find the value of c, substitute the value of a2 = 36 and b2 = 9 in c2 = a2 - b2.

c2 = 36 - 9

c2 = 27

c = ± 3√3.

Center = (h, k ) = (- 3, 4).

Foci = (h ± c, k ) = (- 3 ± 3√3, 4) = (- 3 + 3√3, 4) and (- 3 - 3√3, 4).

The major axes of lengths 2a = 2(6) = 12.

The minor axes of lengths 2b = 2(3) = 6.

 

answered Mar 27, 2014 by steve Scholar

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