求高手做一道高中数学解析几何题
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1)由图和题意知:A(0,1),B(0,-1);
M(m,1/2),则AM:(y-1)/(1/2-1)=(x-0)/(m-0),y=-x/(2m)+1;BM:(y+1)/(1/2+1)=(x-0)/(m-0),y=3x/(2m)-1;
对于E,有x^2/4+[-x/(2m)+1]^2=1,x^2(m^2+1)/(4m^2)-x/m=0,显然x≠0,则x=4m/(m^2+1),
所以E[4m/(m^2+1),(m^2-1)/(m^2+1)];
对于F,有x^2/4+[3x/(2m)-1]^2=1,x^2(m^2+9)/(4m^2)-3x/m=0,显然x≠0,则x=12m/(m^2+9),
所以E[12m/(m^2+9),(9-m^2)/(m^2+9)];
则EF斜率k=[(9-m^2)/(m^2+9)-(m^2-1)/(m^2+1)]/[12m/(m^2+9)-4m/(m^2+1)]=(9-m^4)/[4m(m^2-3)]
则EF:y=k[x-4m/(m^2+1)]+(m^2-1)/(m^2+1)
EF和y轴的交点设为G(0,g),则
g=-4mk/(m^2+1)+(m^2-1)/(m^2+1)=-(9-m^4)/[(m^2-3)(m^2+1)]+(m^2-1)/(m^2+1)
=(m^2+3)/(m^2+1)+(m^2-1)/(m^2+1)=2
即EF和y轴的交点为G(0,2),和m无关
M(m,1/2),则AM:(y-1)/(1/2-1)=(x-0)/(m-0),y=-x/(2m)+1;BM:(y+1)/(1/2+1)=(x-0)/(m-0),y=3x/(2m)-1;
对于E,有x^2/4+[-x/(2m)+1]^2=1,x^2(m^2+1)/(4m^2)-x/m=0,显然x≠0,则x=4m/(m^2+1),
所以E[4m/(m^2+1),(m^2-1)/(m^2+1)];
对于F,有x^2/4+[3x/(2m)-1]^2=1,x^2(m^2+9)/(4m^2)-3x/m=0,显然x≠0,则x=12m/(m^2+9),
所以E[12m/(m^2+9),(9-m^2)/(m^2+9)];
则EF斜率k=[(9-m^2)/(m^2+9)-(m^2-1)/(m^2+1)]/[12m/(m^2+9)-4m/(m^2+1)]=(9-m^4)/[4m(m^2-3)]
则EF:y=k[x-4m/(m^2+1)]+(m^2-1)/(m^2+1)
EF和y轴的交点设为G(0,g),则
g=-4mk/(m^2+1)+(m^2-1)/(m^2+1)=-(9-m^4)/[(m^2-3)(m^2+1)]+(m^2-1)/(m^2+1)
=(m^2+3)/(m^2+1)+(m^2-1)/(m^2+1)=2
即EF和y轴的交点为G(0,2),和m无关
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