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.

Relevance
• 3 years ago

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)