What is the center, foci, and the length of the major and minor axes for the ellipse whose equation is,

Question

16x^2 + 25y^2 + 32x - 150y = 159?

Answer 1

The equation of ellipse is 16x^2 + 25y^2 + 32x - 150y = 159.

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/a^2 + (y - k)^2/b^2 = 1 or (x - h)^2/b^2 + (y - k)^2/a^2 = 1 and the relation between a, b and c is c^2 = a^2 - b^2.

Write the equation 16x^2 + 25y^2 + 32x - 150y = 159 in standard form of the equation of an ellipse.

(16x^2 + 32x) + (25y^2 - 150y) = 159

16(x^2 + 2x) + 25(y^2 - 6y) = 159

 To change the expressions (x^2 + 2x) and 25(y^2 - 6y) into a perfect square trinomial add (half the x coefficient)² to each side of the expression.

 Here x coefficient = 2. so, (half the x coefficient)² = (2/2)²= 1.

Here y coefficient = 6. so, (half the x coefficient)² = (6/2)²= 9

Add 16(1) and 25(9) = 225 to each side.

16(x^2 + 2x + 1) + 25(y^2 - 6y + 9) = 159 + 16 + 225

16(x + 1)^2 + 25(y - 3)^2 = 400

[16(x + 1)^2]/400 + [25(y - 3)^2]/400 = 400/400

(x + 1)^2/25 + (y - 3)^2/16 = 1

(x + 1)^2/5^2 + (y - 3)^2/4^2 = 1.

Compare the equation (x + 1)^2/5^2 + (y - 3)^2/4^2 = 1 with (x - h)^2/a^2 + (y - k)^2/b^2 = 1.

Center = (h, k ) = (- 1, 3), a = 5 and b = 4.

The major axes of lengths 2a = 2(5) = 10.

The minor axes of lengths 2b = 2(4) = 8.

To find the value of c, substitute the values of a = 5 and b = 4 in the c^2 = a^2 - b^2.

c^2 = 25 - 16

c^2 = 3

c = ± 3

The foci of the ellipse is ( ± c, 0) = ( ± 3, 0).

The Center = (- 1, 3), Foci = ( ± 3, 0), the major axes of lengths is 10 and the minor axes of lengths is 8.