#2
將\(\displaystyle y^2=mx \)代入
\(\displaystyle x^2-4x+4+mx=3,x^2-(4-m)x+1=0 \)
\(\displaystyle x_1+x_2=4-m \)
另一方面
\(\displaystyle \frac{y^4}{m^2}-\frac{4y^2}{m}+4+y^2=3,y^4-(4m-m^2)y^2+m^2=0 \)
\(\displaystyle y_1^2+y_2^2=4m-m^2,y_1^2y_2^2=m^2 \)
因為m是正數,以及交點在第一象限,所以
\(\displaystyle y_1y_2=m \)
\(\displaystyle (y_1+y_2)^2=6m-m^2 \)
\(\displaystyle y_1+y_2=\sqrt{6m-m^2} \)
中點在y=x上
\(\displaystyle 4-m=\sqrt{6m-m^2} \)
\(\displaystyle 16-8m+m^2=6m-m^2 \)
\(\displaystyle 2m^2-14m+16=0,m^2-7m+8=0 \)
\(\displaystyle m=\frac{7+\sqrt{17}}{2}, or m=\frac{7-\sqrt{17}}{2} \)
但是m要小於4
所以
\(\displaystyle m=\frac{7-\sqrt{17}}{2} \)