Question

双曲线的问题过坐标原点的直线PA与双曲线x^2/a^2-y^2/b^2=1(a>0b>0)相交于P,A两点,直线PC垂直

Answer 1

设P(x0,y0):x0^/a^-y0^/b^=1,则A(-x0,-y0),C(x0,0),

AC的斜率=y0/(2x0),

AC:y=[y0/(2x0)](x-x0),①

代入x^2/a^2-y^2/b^2=1得

b^x^-a^y0^/(4x0^)*(x^-2x0x+x0^)=a^b^,

(4b^x0^)x^-a^y0^(x^-2x0x+x0^)=4a^b^x0^,

(4b^x0^-a^y0^)x^+2a^x0y0^x-a^x0^y0^-4a^b^x0^=0,

xA=-x0,xB=a^x0(y0^+4b^)/(4b^x0^-a^y0^),

代入①,yB=y0(a^y0^+2a^b^-2b^x0^)/(4b^x0^-a^y0^),

PA的斜率k1=y0/x0,

PB的斜率k2=y0(a^y0^+a^b^-3b^x0^)/[x0(a^y0^+2a^b^-2b^x0^)],

k1k2=1/2,

∴2y0^(a^y0^+a^b^-3b^x0^)=x0^(a^y0^+2a^b^-2b^x0^),

2a^y0^4+2a^b^y0^-6b^x0^y0^=a^x0^y0^+2a^b^x0^-2b^x0^4,

2a^y0^4+2a^b^y0^=x0^(a^y0^+2a^b^+6b^y0^-2b^x0^),

把x0^=a^(1+y0^/b^)代入上式÷a^得

2y0^4+2b^y0^=(1+y0^/b^)(a^y0^+2a^b^+6b^y0^-2a^b^-2a^y0^),

2b^y0^4+2b^4y0^=(b^+y0^)(6b^y0^-a^y0^),

约去y0^,得2b^y0^+2b^4=6b^4-a^b^+y0^(6b^-a^),

(b^+y0^)(a^-4b^)=0,

∴a^=4b^,c^=5b^,

∴c/a=√5/2,为所求.