Question

Find the foci of the ellipse whose equation is (y^2/16) + (x^2/12) = 1.

Answer 1

The equation of ellipse is y²/16 + x²/12 = 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 y² - term (16) is larger than the denominator of the x² - term (12), the major axis is vertical.

Compare the equation (x - 0)²/12 + (y - 0)²/16 = 1 with (x - h)²/b² + (y - k)²/a² = 1.

b² = 12, a² = 16, h = 0 and k = 0.

b = 2√3 and a = 4.

To find the value of c, substitute the value of a² = 16 and b² = 12 in c² = a² - b².

c² = 16 - 12

c² = 4

c = ± 2.

Center = (h, k ) = (0, 0).

Foci = (h, k ± c) = (0, 0 ± 2) = (0, 2) and (0, -2).

The major axes of lengths 2a = 2(4) = 8.

The minor axes of lengths 2b = 2(2√3) = 4√3.

 

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