(x/3)^2 - (y/4)^2 =1
3^2 + 4^2 = 5^2
因此,F_1 (-5,0) ,F_2(5,0)没看懂么？求“点P到x轴的距离”？如果 P(x, y)(x/3)^2 - (y/4)^2 =1 -> 16x^2- 9y^2= 9*16--(1)向量PF1·向量PF2=0-> (x+5, y-0) ( x-5 , y-0 ) =0-> x^2 - 5^2 + y^2 =0---(2)(1), (2)16( 25 - y^2 ) - 9 y^2 = 9*1616*25 - 25y^2 = 16 *916(25-9) = 25 y^216^2= 5^2 y^2y^2 = (16/5 )^2因此 点P到x轴的距离 -> |y 坐标 |【答案是：16/5】