The line 2y=3x-6 intersects the curve xy=12 at the point P and Q.
Find the equation of the perpendicular of PQ
要过程,谢谢~
The line 2y=3x-6 intersects the curve xy=12 at the point P and Q.
Find the equation of the perpendicular of PQ
要过程,谢谢~
问题:直线2y = 3x-6 ,与曲线xy = 12 相交于两点P,Q;
求过P、Q两点的垂直线方程?
解答:
y=3/2x-3;则xy = 3/2x^2-3x;又xy = 12;所以3/2^2-3x = 12,即:x^2-2x = 8 =〉x = -2或4 ; =〉y=。。。。
把P、Q两点都求出来了,应该没问题了吧