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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
- 3 years agoFavorite 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)
- PinkgreenLv 73 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)
