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)
即
(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)