希望十二小时内有解答已知双曲线c:x^2/a^2-y^2/b^2=1(a>0,b>0)的离心率为√3,右准线方程x=√3/3(1)求双曲线的方程(2)设直线l是圆O:x^2+y^2=2上动点P(x0,y0)(x0y0≠0)处的切线,l与双曲线C交于不同两
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![希望十二小时内有解答已知双曲线c:x^2/a^2-y^2/b^2=1(a>0,b>0)的离心率为√3,右准线方程x=√3/3(1)求双曲线的方程(2)设直线l是圆O:x^2+y^2=2上动点P(x0,y0)(x0y0≠0)处的切线,l与双曲线C交于不同两](/uploads/image/z/9308525-5-5.jpg?t=%E5%B8%8C%E6%9C%9B%E5%8D%81%E4%BA%8C%E5%B0%8F%E6%97%B6%E5%86%85%E6%9C%89%E8%A7%A3%E7%AD%94%E5%B7%B2%E7%9F%A5%E5%8F%8C%E6%9B%B2%E7%BA%BFc%3Ax%5E2%2Fa%5E2-y%5E2%2Fb%5E2%3D1%28a%3E0%2Cb%3E0%29%E7%9A%84%E7%A6%BB%E5%BF%83%E7%8E%87%E4%B8%BA%E2%88%9A3%2C%E5%8F%B3%E5%87%86%E7%BA%BF%E6%96%B9%E7%A8%8Bx%3D%E2%88%9A3%2F3%EF%BC%881%EF%BC%89%E6%B1%82%E5%8F%8C%E6%9B%B2%E7%BA%BF%E7%9A%84%E6%96%B9%E7%A8%8B%EF%BC%882%EF%BC%89%E8%AE%BE%E7%9B%B4%E7%BA%BFl%E6%98%AF%E5%9C%86O%EF%BC%9Ax%5E2%2By%5E2%3D2%E4%B8%8A%E5%8A%A8%E7%82%B9P%28x0%2Cy0%29%28x0y0%E2%89%A00%29%E5%A4%84%E7%9A%84%E5%88%87%E7%BA%BF%2Cl%E4%B8%8E%E5%8F%8C%E6%9B%B2%E7%BA%BFC%E4%BA%A4%E4%BA%8E%E4%B8%8D%E5%90%8C%E4%B8%A4)
希望十二小时内有解答已知双曲线c:x^2/a^2-y^2/b^2=1(a>0,b>0)的离心率为√3,右准线方程x=√3/3(1)求双曲线的方程(2)设直线l是圆O:x^2+y^2=2上动点P(x0,y0)(x0y0≠0)处的切线,l与双曲线C交于不同两
希望十二小时内有解答
已知双曲线c:x^2/a^2-y^2/b^2=1(a>0,b>0)的离心率为√3,右准线方程x=√3/3(1)求双曲线的方程(2)设直线l是圆O:x^2+y^2=2上动点P(x0,y0)(x0y0≠0)处的切线,l与双曲线C交于不同两点A,B,证明∠AOB的大小为定值.
希望十二小时内有解答已知双曲线c:x^2/a^2-y^2/b^2=1(a>0,b>0)的离心率为√3,右准线方程x=√3/3(1)求双曲线的方程(2)设直线l是圆O:x^2+y^2=2上动点P(x0,y0)(x0y0≠0)处的切线,l与双曲线C交于不同两
这种题 你就不能怕麻烦,就得死算.
(1) e=c/a = √3,a^2/c =√3/3
a=1,c = √3,b =√2,双曲线方程为
2x^2 -y^2 = 2
x^2+y^2=2上动点P(x0,y0)(x0y0≠0)处的切线方程为
x0x+y0y = 2
(2)
设A,B,的坐标为(Xa,Ya),(Xb,Yb),
则(Xa,Ya),(Xb,Yb) 为方程组
x0x+y0y = 2 (1)
2x^2-y^2 = 2 (2)
的解
(1) 代入(2)消去y,得到
(2-x0^2/y0^2)/x^2 +4x0x/y0^2 - (4/y0^2+2) = 0
XaXb = - (4/y0^2+2)/(2-x0^2/y0^2) = -(4+2y0^2)/(2y0^2-x0^2)
(1) 代入(2)消去x,得到
(2y0^2-x0^2)y^2 -8y0y + 8-2x0^2 = 0
YaYb = (8-2x0^2)/(2y0^2-x0^2)
XaXb+YaYb
= -(4+2y0^2)/(2y0^2-x0^2) + (8-2x0^2)/(2y0^2-x0^2)
= [4-2(x0^2+y0^2)]/(2y0^2-x0^2)
(x0,y0) 是圆x^2+y^2=2的点,上式分母为0,
XaXb+YaYb = 0
向量OA和OB垂直,∠AOB = 90度
好好再看看书上双曲线的概念和特征,这里没有用到特别的方法,其实就是最直接的套公式,解方程而已