Question

高中数学解析几何题

Answer 1

(1)
设M(x,y)
k(A1M)*K(A2M)=K(A1N1)K(A2N2)
[y/(x+2)]*[y/(x-2)]=[a/2][b/(-2)]=-3/4
4y²+3(x²-4)=0
x²/4+y²/3=1
(2)
联立：
{3x²+4y²-12=0
{y=kx+m
3x²+4(kx+m)²-12=0
3x²+4[k²x²+m²+2mkx]-12=0
(3+4k²)x²+8mkx+4(m²-3)=0
{x1+x2= - (8mk)/(3+4k²)
x1x2=4(m²-3)/(3+4k²)
因为α，β互补，所以k1=-k2
y2/(x2-1)= - y1/(x1-1)即：
（kx2+m)/(x2-1)= - (kx1+m)/(x1-1)
(kx2+m)(x1-1)+(kx1+m)(x2-1)=0
kx1x2+mx1-kx2-m+kx1x2+mx2-kx1-m=0
2k(x1x2)+(m-k)(x1+x2)-2m=0
2k[4(m²-3)/(3+4k²)]+(m-k)[ - (8mk)/(3+4k²)]-2m=0
8km²-24k-8m²k+8mk²=2m(3+4k²)
-24k+8mk²=6m+8mk²
m=-4k
直线方程：
y=kx-4k
y-0=k(x-4)
过定点P0(4,0)

kx1x2+mx1-kx2-m+kx1x2+mx2-kx1-m=0

Answer 2