Question

设椭圆ax平方+by平方=1与直线x+y-1=0相交于A,B两点，点c是AB的中点，若绝对值AB=2倍根号2,OC的斜率为

根号2/2求椭圆方程（分母的2没有根号）最好写详细的步骤

Answer 1

设A（x1，y1），B（x2，y2），用点差法得，
a(x1-x2)(x1+x2)+b(y1-y2)(y1+y2)=0，(y1-y2)/(x1-x2)= -1，
所以OC的斜率=(y1+y2)/(x1+x2)=a/b=√2/2，所以a√2=b， ①
所以椭圆ax^2+(√2)ay^2=1，与直线x+y-1=0联立，得
ax^2+b(x-1)^2=1
(a+b)x^2-2bx+b-1=0
x1+x2=2b/(a+b)
x1x2=(b-1)/(a+b)
再由|AB|=2√2=[√(1+kAB^2)]*√[(x1+x2)^2-4x1x2]=√[(x1+x2)^2-4x1x2]√2，
即2=√[(x1+x2)^2-4x1x2]=√{[2b/(a+b)]^2-4[(b-1)/(a+b)]} 两边平方得
b^2=(a+b)(b-1)化简得　a+b=ab　②
将①②联立方程组
最后解得a=(2+√2)/2，b=1+√2
所以椭圆方程为[(2+√2)/2]x^2+(1+√2)y^2=1

Answer 2

ax^2+by^2=1
l:x+y-1=0
ax^2+b(x-1)^2=1
(a+b)x^2-2bx+b-1=0
x1+x2=2b/(a+b)
x1x2=(b-1)/(a+b)
(x1-x2)^2=(x1+x2)^2-4x1x2=[4b^2-(4b-4)(a+b)]/(a+b)^2
(y1-y2)^2=(x1-x2)^2
|AB|^2=8=(x1-x2)^2+(y1-y2)^2
(x1-x2)^2=4
[ 4b^2-(4b-4)(a+b)]/(a+b)^2=4
b^2-(b-1)(a+b)=(a+b)^2
-(b-1)(a+b)=a^2+2ab a+b=a^2+3ab+b^2
Cx=a^2/(a^2+b^2)
Cy=1-x=b/(a+b)
Cy/Cx=√2/2
(a-b)/b=√2/2
a=(1+√2/2)b
(2+√2/2)b= (2+√2/2)^2b^2+(1+√2/2)b^2
(4+√2)=(9+4√2)b+(2+√2)b
b=(4+√2)/(11+5√2)
a=(1+√2/2)(4+√2)/(11+5√2)=(5+3√2)/(11+5√2)
[(5+3√2)/11+5√2)]x^2 + [ (4+√2)/(11+5√2)] y^2 =1