双曲线方程x^2/a^2-y^2/b^2=1(a>0,b>0)的一条渐近线为x+2y=0

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  • 设其半焦距为 c (>0);则c^2=a^2+b^2.

    则左焦点(-c,0);右准线:x=a^2/c;则左焦点到右准线的距离为a^2/c+c=(9√5)/10;

    渐近线为x+2y=0,说明虚轴长与实轴长之比为(渐近线的斜率的绝对值) |-1/2| =1/2;

    即a=2b.

    则c^2=5b^2;

    代入a^2/c+c=(9√5)/10得:

    (4/√5)b+√5b=(9√5)/10;

    则b=1/2.

    而a=2b=1,

    则双曲线方程为x^2-4y^2=1.

    其参数方程为

    x=sec θ;

    y=(1/2)tan θ;

    设点C坐标为( sec θ1 ,(1/2)tan θ1 );

    点D坐标为( sec θ2 ,(1/2)tan θ2 );

    则由题意知:点A(1/2,0)在线段CD上.

    于是得到

    [(1/2)tan θ2 -0] / (sec θ2 - 1/2) = [(1/2)tan θ1 -0] / (sec θ1 - 1/2)

    (tan θ2 )/(tan θ1)=(sec θ2 - 1/2) / (sec θ1 - 1/2)

    (sin θ2 )/(sin θ1)=[1 - (1/2)cos θ2 ] / [1 - (1/2)cos θ1 ]

    → sin θ2 - (1/2)cos θ1·sin θ2 =sin θ1 - (1/2)cos θ2·sin θ1

    →(sin θ2 - sin θ1)=(1/2)[cos θ1·sin θ2 - cos θ2·sin θ1]

    =(1/2)·sin(θ2 - θ1).

    sin(θ2 - θ1)=2(sin θ2 - sin θ1)=4*cos[(θ2 + θ1)/2]*sin[(θ2 - θ1)/2]

    则cos[(θ2 - θ1)/2]=2*cos[(θ2 + θ1)/2]

    或者cos[(θ2 + θ1)/2]=(1/2)*cos[(θ2 - θ1)/2]

    可知点M坐标为( sec θ2 ,(-1/2)tan θ2 );

    则直线MC方程斜率k=[(-1/2)tan θ2 - (1/2)tan θ1 ] / [sec θ2 - sec θ1]

    =(-1/2)*(cos θ1·sin θ2 + cos θ2·sin θ1)/(cos θ1 - cos θ2)

    =(-1/2)*sin(θ1 + θ2)/(cos θ1 - cos θ2)

    =(-1/2)*sin(θ1 + θ2)/{-2*sin[( θ1 + θ2)/2]*sin[( θ1 - θ2)/2]}

    =(1/4)*{2*sin[( θ1 + θ2)/2]*cos[( θ1 + θ2)/2]}/{sin[( θ1 + θ2)/2]*sin[( θ1 - θ2)/2]}

    =(1/2)*{cos[( θ1 + θ2)/2] / sin[( θ1 - θ2)/2]}

    将cos[(θ2 + θ1)/2]=(1/2)*cos[(θ2 - θ1)/2]代入得:

    k=(1/4)*{cos[( θ1 - θ2)/2] / sin[( θ1 - θ2)/2]}

    =(1/4)*cot[(θ2 - θ1)/2]

    =[1+cos(θ2 - θ1)]/[4*sin(θ2 - θ1)]

    则MC方程为

    y= (1/2)tan θ1 +k*( x - sec θ1)

    =(1/2)tan θ1 + [1+cos(θ2 - θ1)]/[4*sin(θ2 - θ1)]*( x - sec θ1)

    =(1/2)tan θ1 - secθ1 * [1+cos(θ2 - θ1)]/[4*sin(θ2 - θ1)] + [1+cos(θ2 - θ1)]/[4*sin(θ2 - θ1)]* x

    当x=0时,y=

    (1/2)tan θ1 - secθ1 * [1+cos(θ2 - θ1)]/[4*sin(θ2 - θ1)]

    =2sin θ1 /(4cosθ1) - cot[(θ2 - θ1)/2] /(4cosθ1)

    ={2sin θ1 - cot[(θ2 - θ1)/2] }/(4cosθ1)