过椭圆x^2/4+y^2/3=1的右焦点F2作一倾斜角为派/4的直线交与该椭圆与A,B两点.求:1)弦AB的长

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  • 解决方法:①题意,C =√^ 2-B ^ 2 = 1,∴F2(1,0)的k =tanπ/ 4 = 1

    ∴直线方程为y-0 = 1(x-1的),它是为y = x-1的

    为y = x-1的生成到椭圆x 2/4 + Y 2/3 = 1以简化的精加工7倍^ 2-8倍速8 = 0

    | AB | =√1 +1 ^ 2 *√(8/7)^ 2-4 * -8 / 7 = 24/7

    ( 2)易问F1的直线距离D = | -1-1 | /√1 ^ 2 +(-1)^ 2 =√2

    ∴S△AF1B = 1 / 2 * 24/7 *√2 = 12√2/7

    ③7倍^ 2-8倍速-8 = 0有X1 + X2 = 8/7横坐标弦AB的中点(1次+×2)/ 2 = 4/7

    Y1 =的x1-1 Y2 = x2的-1 Y1 + Y2 =×1 +×2-2 = 10 -6 / 7

    所以弦AB的中点的垂直坐标(Y1 + Y2)/ 2 = -3 / 7

    AB中点的坐标为(4/7,-3 / 7)