Question

已知直线过点(-2,0 ),且与抛物线y2=8x有公共点,求直线斜率的取值范围 我算不

Answer 1

令直线斜率为k, 其方程为y - 0 = k(x + 2),y = k(x + 2)

y² = 8x

k²(x + 2)² = 8x

k²x² + 4(k² - 2)x + 4k² = 0

判别式 = 16(k² - 2)² - 16k^4 = 64(1 - k²) = 0

k = 1,k = -1

-1 <= k <= 1