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.谁懂,x^2/a^
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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.谁懂,x^2/a^](/uploads/image/z/8625375-63-5.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+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.%E8%B0%81%E6%87%82%2Cx%5E2%2Fa%5E)
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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.谁懂,x^2/a^
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.
谁懂,
x^2/a^2 -y^2/b^2=1 这个少了个‘-’
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.谁懂,x^2/a^
斜率为2作业刚做了
我也不会 第一问告诉你
设B点(X1,Y1)D点(X2,Y2)
X1+X2=2 Y1+Y2=6 Y2-Y1=X2-X1
x1^2/a^2 -y1^2/b^2=1
x2^2/a^2 -y2^2/b^2=1
两式相减 化简得3a^2=b^2
易求得C关于B的关系 能求出来离心率
第二问就不会了 我也在问~嘿嘿