
已知点a(2a-b,5+a),b(2b-1,-a+b)
1个回答
关注

展开全部
设直线l方程为y+3=kx则y=kx-3①x^2=-8y②把①代入②得x^2=-8(kx-3)x^2+8kx-24=0x1+x2=-8kx1*x2=-24y1+y2=k(x1+x2)-6=-8k^2-6y1*y2=(kx1-3)*(kx2-3)=kx1x2-3k(x1+x2)+9=-24k+24k^2+9AB^2=(x1-x2)^2+(y1-y2)^2=(x1+x2)^2-4x1*x2+(y1+y2)^2-4y1*y2=64k^2+96+64k^4+96k^2+36
咨询记录 · 回答于2023-01-26
已知点a(2a-b,5+a),b(2b-1,-a+b)
请把题完整发过来
设直线l方程为y+3=kx则y=kx-3①x^2=-8y②把①代入②得x^2=-8(kx-3)x^2+8kx-24=0x1+x2=-8kx1*x2=-24y1+y2=k(x1+x2)-6=-8k^2-6y1*y2=(kx1-3)*(kx2-3)=kx1x2-3k(x1+x2)+9=-24k+24k^2+9AB^2=(x1-x2)^2+(y1-y2)^2=(x1+x2)^2-4x1*x2+(y1+y2)^2-4y1*y2=64k^2+96+64k^4+96k^2+36