Precalc

Find the center and vertices of the ellipse.

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

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asked by pat
  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:

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    posted by Henry
  2. CORRECTION:

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

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

    V3(8 , -1)

    V4(8 , 3).

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    posted by Henry

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