求抛物线y=x∧2+ax与抛物线y=x∧2+bx(b≠a)的公切线方程

Question

求抛物线y=x∧2+ax与抛物线y=x∧2+bx(b ≠a)的公切线方程

Answer 1

令切线为y = kx + c
与前者联立:x² + (a - k)x - c = 0,(a - k)² + 4c = 0
类似地,(b - k)² + 4c = 0
(a - k)² = (b - k)²
a ≠ b:a - k = k - b,k = (a + b)/2
c = -(a - b)²/16
y = (a + b)x/2 - (a - b)²/16