Find the center and vertices of the ellipse.
4x^2 + 16y^2 - 64x - 32y + 208 = 0
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:
CORRECTION:
b = 2,
Y3 = k - b,
Y3 = 1 - 2 = -1.
Y4 = k + b,
Y4 = 1 + 2 = 3.
V3(8 , -1)
V4(8 , 3).
To determine the center and vertices of an ellipse, we need to put the equation in the standard form of an ellipse:
((x - h)^2 / a^2) + ((y - k)^2 / b^2) = 1,
where (h, k) represents the center of the ellipse, a is the length of the semi-major axis, and b is the length of the semi-minor axis.
To rearrange your equation into the standard form, complete the square separately for the x and y terms.
First, let's work on the x terms:
4x^2 - 64x = 0
4(x^2 - 16x) = 0
4(x^2 - 16x + 64) = 4(64)
4(x - 8)^2 = 256
Next, let's work on the y terms:
16y^2 - 32y = -208
16(y^2 - 2y) = -208
16(y^2 - 2y + 1) = -208 + 16
16(y - 1)^2 = -192
Combining the x and y terms:
4(x - 8)^2 + 16(y - 1)^2 = 256 - 192
4(x - 8)^2 + 16(y - 1)^2 = 64
Dividing through by 64, we get:
(x - 8)^2 / 16 + (y - 1)^2 / 4 = 1
Now, we can compare this equation to the standard form:
((x - h)^2 / a^2) + ((y - k)^2 / b^2) = 1
From this comparison, we can identify the center of the ellipse as (h, k) = (8, 1).
The length of the semi-major axis (a) is √16 = 4, and the length of the semi-minor axis (b) is √4 = 2.
Therefore, the center of the ellipse is (8, 1), and the vertices are (8 ± 4, 1), which gives us (12, 1) and (4, 1).