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高中双曲线题如图,直线l:y=根号3(x-2)和双曲线C:/a?-y?/b?=1(a>0,b>0)交于A,B两点,且|AB|=根号3,又l关于直线l1:y=b/ax对称的直线l2与x轴平行(1)求双曲线C的离心率(2)求双曲线C的方程求
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高中双曲线题
如图,直线l:y=根号3(x-2)和双曲线C:/a?-y?/b?=1(a>0,b>0)交于A,B两点,且|AB|=根号3,又l关于直线l1:y=b/ax对称的直线l2与x轴平行(1)求双曲线C的离心率(2)求双曲线C的方程求详细过程....
如图,直线l:y=根号3(x-2)和双曲线C:/a?-y?/b?=1(a>0,b>0)交于A,B两点,且|AB|=根号3,又l关于直线l1:y=b/ax对称的直线l2与x轴平行(1)求双曲线C的离心率(2)求双曲线C的方程求详细过程....
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答案和解析
(1)直线l:y = √3(x – 2),倾斜角为arctan(√3) = π/3,l关于直线l1:y=bx/a对称的直线l2与x轴平行,因为直线l2的倾斜角为0,所以直线l1的倾斜角为(π/3 + 0)/2 = π/6,所以kl1 = tan(π/6) = b/a,即b/a = √3/3,设b = k(k > 0),则a = √3k,c2 = a2 + b2 = 3k2 + k2 = 4k2 => c = 2k => 双曲线C的离心率e = c/a = 2k/√3k = 2√3/3 ;
(2)设法同(1),所以双曲线C:x2 /a2 – y2 /b2 = 1 => x2 /3k2 – y2 /k2 = 1 => x2 – 3y2 = 3k2,与y = √3(x – 2)联立可得x2 – 3*3(x – 2)2 = 3k2 => 8x2 – 36x + 3k2 + 36 = 0,Δ= 362 – 32(3k2 + 36) = 144 – 96k2 > 0 => 96k2 < 144 => k2 < 3/2 => k∈(0,√6/2),设A(x1,y1),B(x2,y2),由弦长公式可得AB = √(1 + 3)*√(144 – 96k2)/8 = √3*√(3 – 2k2) = √3 => √(3 – 2k2) = 1 => 3 – 2k2 = 1 => 2 = 2k2 => k2 = 1 => k = 1 => 双曲线的方程为x2 /3 – y2 = 1 .
(2)设法同(1),所以双曲线C:x2 /a2 – y2 /b2 = 1 => x2 /3k2 – y2 /k2 = 1 => x2 – 3y2 = 3k2,与y = √3(x – 2)联立可得x2 – 3*3(x – 2)2 = 3k2 => 8x2 – 36x + 3k2 + 36 = 0,Δ= 362 – 32(3k2 + 36) = 144 – 96k2 > 0 => 96k2 < 144 => k2 < 3/2 => k∈(0,√6/2),设A(x1,y1),B(x2,y2),由弦长公式可得AB = √(1 + 3)*√(144 – 96k2)/8 = √3*√(3 – 2k2) = √3 => √(3 – 2k2) = 1 => 3 – 2k2 = 1 => 2 = 2k2 => k2 = 1 => k = 1 => 双曲线的方程为x2 /3 – y2 = 1 .
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