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

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  • 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