Question

高中椭圆的求方程的题

椭圆ax^2+by^2=1与直线x+y-1=0相交与A，B两点，C是A,B中点，若AB=2√2，OC的斜率为√2/2,求椭圆的方程。

Answer 1

解：直线x+y-1=0
y=-x+1
代入ax²+by²=1
ax²+b(-x+1)²=1
ax²+bx²-2bx+b²-1=0
(a+b)x²-2bx+b²-1=0
设A，B的坐标分别为（xA，yA）（xB，yB）
xA+xB=2b/(a+b)
xA×xB=(b²-1)/(a+b)
点C的横坐标b/(a+b)
纵坐标(yA+yB)/2=(-xA+1-xB+1)/2=-(xA+xB-2)/2=-b/(a+b)+1=a/(a+b)
根据题意
[a/(a+b)]/[b(a+b)]=√2/2
a/b=√2/2
b=√2a(1)
AB=√(xA-xB)²+(yA-yB)²=√(xA-xB)²+(xA-xB)²=√2[(xA+xB)²-4xA×xB]
2√2=√2[4b²/(a+b)²-4(b²-1)/(a+b)]
1=b²/(a+b)²-(b²-1)/(a+b)
a²+2ab+b²=b²-(b²-1)(a+b)
a²+2ab+ab²+b²×b-a-b=0
b=√2a
a²+2ab+ab²+b²×√2a-a-√2a=0
a不为0
a+2√2a+2a²+2√2a²-1-√2=0
(2+2√2)a²+(2√2+1)a-(√2+1)=0
2a²+(3-√2)a-1=0
a=√2-1或a=-(√2+1)/2（舍去）
a=√2-1
b=2-√2
(√2-1)x²+(2-√2)y²=1
x²/(√2+1)+y²/(2+√2)/2=1

Answer 2

将y=1-x代入椭圆方程,有ax^2+b(1-x)^2-1=0,(a+b)x^2-2bx+b-1=0,设A（Xa,Ya),B(Xb,Yb),Xb>Xa,
Xa+Xb=b/(a+b),XaXb=(b-1)/(a+b);同理可得到Ya+Yb=a/(a+b),YaYb=(a-1)/(a+b),
Xc=(Xa+Xb)/2,Yc=(Ya+Yb)/2,OC斜率为k=Yc/Xc=(Ya+Yb)/(Xa+Xb)=a/b=√2/2,b=√2a,
IABI^2=(Xb-Xa)^2+(Yb-Ya)^2=(Xa+Xb)^2+(Ya+Yb)^2-4XaXb-4YaYb
=(b-1)^2/(a+b)^2+(a-1)^2/(a+b)^2-4(b-1)/(a+b)-4(a-1)/(a+b)=(2√2)^2
将b=√2a代入上面各式,得到
(b-1)^2+(a-1)^2=(√2a-1)^2+(a-1)^2=2a^2-2√2a+1+a^2-2a+1=3a^2-(2+2√2)a+2,
[4(b-1)+4(a-1)](a+b)=[4(√2a-1)+4(a-1)](1+√2)a=[(4+4√2)a-8](1+√2)a
=(12+8√2)a^2-(8√2+8)a
3a^2-(2+2√2)a+2-(12+8√2)a^2+(8√2+8)a=8[(√2+1)a]^2
(33+24√2)a^2-(6√2+6)a-2=0
21a^2-(30-18√2)a-(16√2-22)=0
a=5/7-3√2/7+√(66√2-75)/21;
b=-6/7+5√2/7+√(132√2-150)/21
(5/7-3√2/7+√(66√2-75)/21)x^2+(-6/7+5√2/7+√(132√2-150)/21)y^2=1