如图,直线AC过原点且与双曲线 y=6/x交于AC两点,A在第二象限,直线AC绕原点旋转,以Ac为对角线作正方形ABCD
1个回答
题目显然有问题.双曲线应当为y = -6/x,否则y = 6/x在第一和第三象限.
(1)
A(a, b), C(-a, -b), a < 0, b > 0
AC⊥BD, AC的斜率为 = (-b - b)/(-a – a)= b/a
BD的斜率为-a/b,BD的方程:y = -ax/b,D(m,-am/b),m >0
OD² = OA² = a² + b² = m² + a²m²/b² =(a² + b²)m²/b²
m² = b², m = b (因m >0, b > 0)
-am/b = -a
D(b, -a)
AD的方程: (y + a)/(b + a) = (x - b)/(a - b)
令x= 0, y = (a² + b²)/(b - a)
OE = (a² + b²)/(b - a)
CD的方程: (y + b)/(-a + b) = (x + a)/(b + a)
令y = 0, x = (a² + b²)/(b - a)
OF = (a² + b²)/(b - a)
OE = OF
(2)
A(a, b), ab = -6
显然原点为BD的中点, B(-b, a)
(-b)a = -ab = 6
b(-a) = -ab = 6
B,D始终在同一函数y = 6/x的图像上
相关问题
