Question

已知椭圆x^2/2+y^2=1,过点A(2,0)的直线与椭圆交与P、Q两点

Answer 1

2^2/2 + 0 = 2 > 1

点A(2,0)在椭圆x^2/2 + y^2 = 1外,

可设过点A且与椭圆相交的直线的直线方程为 y = k(x-2),

1 = x^2/2 + [k(x-2)]^2 = x^2/2 + k^2x^2 - 4k^2x + 4k^2,

x^2(1 + 2k^2) - 8k^2x + 8k^2 - 2 = 0,

0 < 64k^4 - 4(1 + 2k^2)(8k^2 - 2) = 64k^4 - 32k^2 - 64k^4 + 8 + 16k^2

= 8 - 16k^2 = 16(1/2 - k^2)

-2^(-1/2) < k < 2^(-1/2).0 -1/2,1/2 >= 1/2 - k^2 > 0

所以,0 < D = (1/2 - k^2)^(1/2)