由原点引抛物线y=x^2+2x+4的两条切线,设切点分别为a,b求两切点oa,ob与此抛物线所围城的

Question

由原点引抛物线y=x^2+2x+4的两条切线,设切点分别为a,b求两切点oa,ob与此抛物线所围城的

Answer 1

y' = 2x + 2
切点A(a,a^2 + 2a + 4)
切线斜率k = 2a + 2
切线:y - (a^2 + 2a + 4) = (2a + 2)(x - a)
过原点:-(a^2 + 2a + 4) = (2a + 2)(- a)
a = 2,A(2,12),切线y = 6x
a = -2,B(-2,4),切线y = -2x
面积 = ∫(x^2 + 2x + 4 + 2x)dx (-2 -> 0)
+ ∫(x^2 + 2x + 4 - 6x)dx (0 -> 2)
= (x^3/3 + 2x^2 + 4x) (-2 -> 0)
+ (x^3/3 - 2x^2 + 4x) (0 -> 2)
= 0 - (-8/3 + 8 - 8) + (8/3 - 8 + 8) - 0
= 16/3