Anonymous
Anonymous asked in Science & MathematicsMathematics · 6 years ago

Find equations of the parabola?

Hi, please help me with this.

Find the equations of the hyperbola, assuming form (x^2/a^2) - (y^2/b^2) = 1:

1. Distances between directrices unity, passing

through (2,3).

2. Latus rectum 6, distance between foci twice the

distance between directrices.

3. Foci (4,0)(-4,0), slope of asymptotes +3 and -3.

I only need solutions to these. The answers given

are:

1. 3x^2 - y^2 =3; 12x^2 - y^2 =39

2. x^2 -y^2 = 9

3. 45x^2 - 5y^2 = 72

Thanks in adv to those who'll answer.

1 Answer

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  • Pope
    Lv 7
    6 years ago
    Favorite Answer

    Your opening condition has the center at the origin and the transverse axis is horizontal.

    [1]

    Distance between the directrices is unity.

    Let the directices be x = -1/2 and x = 1/2.

    The corresponding foci are F(-u, 0) and G(u, 0), where u > 0.

    Let the given point be P(2, 3).

    Let P be projected horizontally, meeting the directrices at Q(-1/2, 3) and R(1/2, 3).

    PF/PQ = PG/PR

    √[(2 + u)² + 3²] / (5/2) = √[(2 - u)² + 3²] / (3/2)

    3√[(2 + u)² + 3²] = 5√[(2 - u)² + 3²]

    9[(2 + u)² + 3²] = 25[(2 - u)² + 3²]

    9(u² + 4u + 13) = 25(u² - 4u + 13)

    16u² - 136u + 208 = 0

    2u² - 17u + 26 = 0

    (u - 2)(2u - 13) = 0

    u = 2 or u = 13/2

    That gives you the foci and the directrices. Can you finish that?

    [2]

    Let the directrices be x = -v and x = v, for v > 0.

    Foci: (-2v, 0), (2v, 0)

    Latus rectum is 6, so we have four points on the section.

    Here is one of them: (2v, 3)

    Use a proportion similar to the previous one.

    √[(4v)² + 3²] / (3v) = 3 / v

    √[(4v)² + 3²] = 9

    16v² + 9 = 81

    v² = 9/2

    v = 3/√(2)

    Directrices: x = -3/√(2), x = 3/√(2)

    Foci: (-3√(2), 0), (3√(2), 0)

    [3]

    Draw vertical lines through the vertices. They intersect the asymptotes at four points, each of them a distance of 4 from the origin.

    Let w be the transverse radius, That makes 3w the conjugate radius.

    w² + (3w)² = 4²

    10w² = 16

    w = √(8/5)

    x²/(8/5) - y²/(72/5) = 1

    45x² - 5y² - 72 = 0

    I felt like I had to finish one of them.

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