求过P(0,1)到双曲线x^2-y^2=1最小距离的直线方程.

Question

求过P(0,1)到双曲线x^2-y^2=1最小距离的直线方程.

Answer 1

以P为圆心,r为半径的圆与双曲线相切时,点P与距离双曲线最短.
圆的方程:x² + (y - 1)² = r²
减去双曲线的方程:2y² - 2y + 2 - r² = 0 (i)
∆ = 4 - 8(2 - r²) = 8r² - 12 = 0
r² = 3/2
(i)变为:4y² - 4y + 1 = (2y - 1)² = 0
y = 1/2,x = ±√5/2
切点(±√5/2,1/2)
用两点式得二直线为:
x + √5y - √5 = 0
x - √5y + √5 = 0