已知斜率为1的直线l与双曲线c:x^2/a^2- y^2/b^2=1(a〉0,b〉0)相交于B、D两点,且BD的中点为M(1,3)⑴求C的离心率⑵设C的右顶点为A,右焦点为F1│DF│*│BF│=17,证明A、B、D三点的圆与X轴相切.
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![已知斜率为1的直线l与双曲线c:x^2/a^2- y^2/b^2=1(a〉0,b〉0)相交于B、D两点,且BD的中点为M(1,3)⑴求C的离心率⑵设C的右顶点为A,右焦点为F1│DF│*│BF│=17,证明A、B、D三点的圆与X轴相切.](/uploads/image/z/6921912-48-2.jpg?t=%E5%B7%B2%E7%9F%A5%E6%96%9C%E7%8E%87%E4%B8%BA1%E7%9A%84%E7%9B%B4%E7%BA%BFl%E4%B8%8E%E5%8F%8C%E6%9B%B2%E7%BA%BFc%3Ax%5E2%2Fa%5E2%EF%BC%8D+y%5E2%2Fb%5E2%3D1%28a%E3%80%890%2Cb%E3%80%890%29%E7%9B%B8%E4%BA%A4%E4%BA%8EB%E3%80%81D%E4%B8%A4%E7%82%B9%2C%E4%B8%94BD%E7%9A%84%E4%B8%AD%E7%82%B9%E4%B8%BAM%EF%BC%881%2C3%EF%BC%89%E2%91%B4%E6%B1%82C%E7%9A%84%E7%A6%BB%E5%BF%83%E7%8E%87%E2%91%B5%E8%AE%BEC%E7%9A%84%E5%8F%B3%E9%A1%B6%E7%82%B9%E4%B8%BAA%2C%E5%8F%B3%E7%84%A6%E7%82%B9%E4%B8%BAF1%E2%94%82DF%E2%94%82%2A%E2%94%82BF%E2%94%82%3D17%2C%E8%AF%81%E6%98%8EA%E3%80%81B%E3%80%81D%E4%B8%89%E7%82%B9%E7%9A%84%E5%9C%86%E4%B8%8EX%E8%BD%B4%E7%9B%B8%E5%88%87.)
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已知斜率为1的直线l与双曲线c:x^2/a^2- y^2/b^2=1(a〉0,b〉0)相交于B、D两点,且BD的中点为M(1,3)⑴求C的离心率⑵设C的右顶点为A,右焦点为F1│DF│*│BF│=17,证明A、B、D三点的圆与X轴相切.
已知斜率为1的直线l与双曲线c:x^2/a^2- y^2/b^2=1(a〉0,b〉0)相交于B、D两点,且BD的中点为M(1,3)
⑴求C的离心率
⑵设C的右顶点为A,右焦点为F1│DF│*│BF│=17,证明A、B、D三点的圆与X轴相切.
已知斜率为1的直线l与双曲线c:x^2/a^2- y^2/b^2=1(a〉0,b〉0)相交于B、D两点,且BD的中点为M(1,3)⑴求C的离心率⑵设C的右顶点为A,右焦点为F1│DF│*│BF│=17,证明A、B、D三点的圆与X轴相切.
B(x1,y1) D(x2,y2)
BD:y=x+m 过M(1,3)
m=2
y=x+2
x^2/a^2-y^2/b^2=1
b^2x^2-a^2(x+2)^2-a^2b^2=0
(b^2-a^2)x^2-4a^2x+a^2(-4-b^2)=0
x1+x2=4a^2/(b^2-a^2)
x1x2=a^2(-4-b^2)/(b^2-a^2)
b^2=c^2-a^2,e=c/a
x1+x2=4a^2/(c^2-2a^2)=4/(e^2-2)
x1x2=a^2(-4+a^2-c^2)/(c^2-2a^2)
Mx=(x1+x2)/2=[4/(e^2-2)]/2=1
2/(e^2-2)=1
e^2-2=2 e^2=4
e=2