过x^2-y^2=1右焦点的直线交双曲线于A,B,AB为圆的直径,求圆是否过原点,若过原点则斜率K为多少?
过x^2-y^2=1右焦点的直线交双曲线于A,B,AB为圆的直径,求圆是否过原点,若过原点则斜率K为多少?
双曲线x^2-y^2=1中c=√(a²+b²)=√2
∴右焦点F2(√2,0)
直线AB的斜率为k,那么直线
AB:y=k(x-√2)代入x^2-y^2=1中
得x²-k²(x-√2)²=1
即(1-k²)x²+2√2k²x-2k²-1=0
1-k²≠0,Δ=8k⁴+4(1-k²)(2k²+1)>0
设A(x1,y1),(x2,y2)
则x1+x2=2√2k²/(k²-1),x1x2=(2k²+1)/(k²-1)
若以AB为圆的直径过原点
则OA⊥OB
向量OA●OB=x1x2+y1y2=00
∵y1=k(x1-√2),y2=k(x2-√2)
∴y1y2=k²(x1-√2)(x2-√2)
=k²[x1x2-√2(x1+x2)+2]
=k²x1x2-√2k²(x1+x2)+2k²
∴(1+k²)x1x2-√2k²(x1+x2)+2k²=0
∴(1+k²)(2k²+1)/(k²-1)-4k⁴/(k²-1)+2k²=0
∴(2k⁴+3k²+1)-4k⁴+2k⁴-2k²=0
∴k²+1=0无解
∴以AB为直径圆不过原点
一般情况,不经过原点,只有特定条件才经过原点,
当OA垂直OB时,三角形AOB是RT三角形,才经过原点,
a=1,b=1,c=√2,
AB方程为:y=k(x-√2),
设A(x1,y1),B(x2,y2),
向量OA·OB=0,
x1x2+y1y2=0,
y1=k(x1-√2),
y2=k(x2-√2),
(1+k^2)x1x2-√2k^2(x1+x2)+2k^2=0,(1)
x^2-k^2(x-√2)^2=1,
(1-k^2)x^2+2√2k^2x-2k^2-1=0,
根据韦达定理,
x1+x2=-2√2k^2/(1-k^2),
x1*x2=-(2k^2+1)/(1-k^2),
代入(1)式,
-(1+k^2)(2k^2+1)/(1-k^2)-√2k^2*[-2√2k^2/(1-k^2)]+2k^2=0,
-k^2-1=0,
∴k无解,圆不经过原点.