Question

Find the center and vertices of the ellipse.

4x^2 + 16y^2 - 64x - 32y + 208 = 0

Answer 1

4X^2 + 16Y^2 - 64X - 32Y + 208 = 0.

Rearrange variables:
4X^2 - 64X + 16Y^2 - 32Y = -208,
Divide both sides by 4:
X^2 - 16X + 4Y^2 - 8Y = -52,
Complete the square:
X^2 - 16X + (-16/2)^2 + 4(Y^2 - 2Y +
(-2/2)^2) = -52 + 64 + 4 = 16,
Simplify:
X^2 - 16X + 64 + 4(Y^2 - 2Y + 1) = 16.
Write the perfect squares as binomials:
(X - 8)^2 + 4(Y - 1)^2 = 16,
Divide both sides by 16 and get:
(X - 8)^2 / 16 + (Y - 1)^2 / 4 = 1.

C(h , k) = C(8 , 1).

a^2 = 16,
a = +- 4.

b^2 = 4,
b = +-2.

Major Axis:

V1(X1 , Y1), C(8 , 1), V2(X2 , Y2).

h - X1 = a,
X1 = h - a,
X1 = 8 - 4 = 4.

Y1 = Y2 = K = 1.

X2 - h = a,
X2 = h + a,
X2 = 8 + 4 = 12.

Minor Axis:

V4(X4 , Y4)

C(8 , 1)

V3(X3 , Y3)

X3 = X4 = h = 8.

k - Y3 = b,
Y3 = k - b,
Y3 = 1 - 4 = -3.

Y4 - k = b,
Y4 = k + b,
Y4 = 1 + 4 = 5.

Center and Vertices:

C(8 , 1)

V1(4 , 1)

V2(12 , 1)

V3(8 , -3)

V4(8 , 5).











Complete the square:

Answer 2

CORRECTION:

b = 2,
Y3 = k - b,
Y3 = 1 - 2 = -1.

Y4 = k + b,
Y4 = 1 + 2 = 3.

V3(8 , -1)

V4(8 , 3).