Pope
Lv 7
Pope asked in Science & MathematicsMathematics · 3 years ago

Nearest point on a hyperbola?

Hyperbola: 2x² - 7y² + 20 = 0

Point: (5, 0)

Find the point(s) on the hyperbola nearest the given point.

2 Answers

Relevance
  • Favourite answer

    d^2 = (x[2] - x[1])^2 + (y[2] - y[1])^2

    2x^2 - 7y^2 + 20 = 0

    2x^2 + 20 = 7y^2

    (1/7) * (2x^2 + 20) = y^2

    d^2 = (5 - x)^2 + (0 - y)^2

    d^2 = 25 - 10x + x^2 + y^2

    d^2 = 25 - 10x + x^2 + (1/7) * (2x^2 + 20)

    d^2 = (1/7) * (7 * 25 - 7 * 10x + 7 * x^2 + 2x^2 + 20)

    d^2 = (1/7) * (175 + 20 - 70x + 9x^2)

    d^2 = (1/7) * (9x^2 - 70x + 195)

    2d * dd/dx = (1/7) * (18x - 70)

    dd/dx = 0

    2d * 0 = (1/7) * (18x - 70)

    0 = 18x - 70

    0 = 9x - 35

    35 = 9x

    35/9 = x

    2x^2 - 7y^2 + 20 = 0

    7y^2 = 2x^2 + 20

    7y^2 = 2 * (35/9)^2 + 20

    7y^2 = 2 * (1225/81) + 20 * 81/81

    7y^2 = (2 * 1225 + 20 * 81) / 81

    7y^2 = 2 * (1225 + 810) / 81

    7y^2 = 2 * 2035 / 81

    y^2 = 2 * 5 * 407 / (81 * 7)

    y^2 = 2 * 5 * 7 * 407 / (81 * 7^2)

    y = +/- sqrt(407 * 7 * 10) / (9 * 7)

    y = +/- sqrt(28490) / 63

    (35/9 , sqrt(28490) / 63)

    (35/9 , -sqrt(28490) / 63)

  • 3 years ago

    Let d be the distance from (5, 0) to the hyperbola 7y^2-2x^2=20, then

    d^2=(x-5)^2+y^2

    Let

    F=(x-5)^2+y^2+a(7y^2-2x^2), where a=the multiplier

    Let

    Fx=(x-5)-2ax=0-------(1)

    Fy=y+7ay=0---------(2)

    From (2), get

    y(1+7a)=0=>

    a= -1/7---------(3)

    putting (3) into (1) and eliminating a, get

    x-5-2(-1/7)x=0=>

    x=35/9=3.8889

    7y^2-2(3.8889)^2=20=>

    y=+/-2.6792

    =>

    the points required are

    (3.8889, 2.6792) & (3.8889, -2.6792)

Still have questions? Get answers by asking now.