中四A.MATHS

設 P ( 0 , b ) 及 Q ( -2a , a )。
√( 0 + 2a )2 + ( b – a )2 = 4
4a2 + b2 – 2ab + a2 = 16
5a2 – 2ab + b2 = 16

設 M ( x , y ) 為 PQ的中點.
x = ( 0 – 2a ) / 2 = -a
a = - x
y = ( b + a ) / 2
2y = b – x
b = x + 2y

M的軌跡方程是
5(-x)2 – 2(-x)(x + 2y) + (x + 2y)2 = 16
5x2 + 2x2 + 4xy + x2 + 4xy + 4y2 = 16
8x2 + 8xy + 4y2 – 16 = 0
2x2 + 2xy + y2 – 4 = 0