问一道解析几何题曲面 x^2/4+y^2/9-z^/16=1 上过点A(2,0,0)的直母线方程为?
1个回答

设过点A(2,0,0)的直母线方程为,

(x - 2) = at,y = bt,z = ct.t为参变量.a,b,c为常数.

则因直母线上的点都在曲面上,有

(2+at)^2/4 + b^2t^2/9 - (ct)^2/16 = 1

[4 + 4at + (at)^2]/4 + (bt)^2/9 - (ct)^2/16 - 1 = 0,

t^2[a^2/4 + b^2/9 - c^2/16] + at = 0,

t[a^2/4 + b^2/9 - c^2/16] + a = 0,

要使得对于任意的t上面的等式都成立,只有,

a = 0,b^2/9 = c^2/16,

因此,过点A(2,0,0)的直母线方程为,

x = 2,y = 3t,z = 4t.

或者,

x = 2,y = -3t,z = 4t.

其中,t为任意实数.

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