Question

导数题以及圆锥曲线

Answer 1

(20)
(I)
f(x)=x^3/3+ax^2+bx+c-ln(x+2)
a=-1,b=-2
f(x)=x^3/3-x^2-2x+c-ln(x+2)
f'(x)=x^2-2x-2 - 1/(x+2)
= (x^3-6x-5)/(x+2)
= (x+1)(x^2-x-5)/(x+2)
= (x+1)(x+2)(x-3)/(x+2)
单调区域
减小[-1,3]
增加(-∞,-2)U(-2,1]U [3,+∞)
(II)
1/2≤a≤1, b=2, x∈[-1,+∞), f(x)≥2/3, Find: min c

f(x)=x^3/3+ax^2+2x+c-ln(x+2)
f'(x) =x^2+2ax +2 - 1/(x+2) >0 for x∈[-1,+∞) 1/2≤a≤1
f(-1)=-1/3+a-2+c-ln(1)≥2/3
c≥3-a
min c =3-1 = 2
待续！！！
(21)
E1: x^2/16+y^2=1

A左顶点
L : x=8/3
L cuts E1 at B,C
(I)Find : inner circle, G, of △ABC
(II) 2 tangents from M(0,-1) to G, cuts E1 at E,F, find: relationship of line EF and G

Solution:
(I)
E1: y=0
x^2/16=1
x=4 or -4
A(-4,0)

E1: x=8/3
(64/9)/16+y^2=1
y^2= 5/9
y = √5/3 or -√5/3
B(8/3,√5/3), C(8/3,-√5/3)

△ABC中，A(x1，y1)，B(x2，y2)，C(x3，y3)，那么△ABC内心I的坐标是：
(ax1/(a+b+c)+bx2/(a+b+c)+cx3/(a+b+c))，ay1/(a+b+c)+by2/(a+b+c)+cy3/(a+b+c))

|AB|=|AC| = 3√5
|BC|= 2√5/3
let C1, centre of G, be (x0, y0)
x0= [1/(20√5/3) ] [ (2√5/3)(-4) +(8/3)3√5+(8/3)3√5]
= [3/(20√5)].[40√5/3]
=2
y0=0
C1(2, 0)
r= 8/3-2 = 2/3
equation of G
(x-2)^2 +y^2 = (2/3)^2

(II)
2 tangents from M(0,-1) to G, cuts E1 at E,F, find: relationship of line EF and G

Solutions:
let 2 tangents from M(0,-1) to G be
y+1=mx (1)
equation of G:
(x-2)^2 +y^2 = (2/3)^2 (2)

sub (1) into (2)
(x-2)^2 +(mx-1)^2 = (2/3)^2
9(x-2)^2 +9(mx-1)^2 = 4
9(m^2+1)x^2-18(m+2)x +41=0
△ =0

[18(m+2)]^2-4(9(m^2+1))41=0
9(m+2)^2 -41(m^2+1) =0
32m^2-36m+5 =0
m= (9+√41)/16 or (9-√41)/16

2 tangents from M(0,-1) to G
y+1=[(9±√41)/16] x