在平面直角坐标系xoy中,直线l1同时与椭圆c1：2分之x2加y2=1和抛物线y2=4x相切,求直线l的方程

设直线l的方程为 y = kx + b
代入椭圆方程:x²/2 + (kx + b)² = 1
(2k² + 1)x² + 4kbx + 2(b² - 1) = 0
∆ = (4kb)² - 4(2k²+1)*2(b² -1) = 8(2k² - b² + 1) = 0
2k² - b² + 1 = 0 (1)
直线l的方程代入抛物线方程:(kx + b)² = 4x
k²x² + (2kb - 4)x + b² = 0
∆ - (2kb-4)² - 4k²b² = 16(1 - kb) = 0
kb = 1,b = 1/k (2)
(2)代入(1):2k² - 1/k² + 1 = 0
2k⁴ + k² - 1 = 0
k² = [-1 + √(1 + 8)]/4 = 1/2
k² = [-1 - √(1 + 8)]/4 < 0,舍去
k = ±1/√2,b = ±√2
y = x/√2 + √2或y = -x/√2 - √2