请问这一道解析几何怎么做已知椭圆C：x²/a²+y²/b²=1(a＞b＞0)上一

椭圆C：x²/a²+y²/b²=1(a＞b＞0)有一点P（m,n）,
∴m^2/a^2+n^2/b^2=1,①
曲线Q：（x-m）²+（y-n）²=m²+n²经过椭圆C的长轴端点(土a,0),
∴(土a-m)^2=m^2,a^2干2am=0,a=2m,②
Q与坐标轴的相交弦长相等,m>0,n>0,
∴m=n,
OP=√2m=2,∴m=√2,
代入②,a=2√2,
都代入①,2/b^2=3/4,b^2=8/3,
∴椭圆方程是x^2/8+3y^2/8=1.③
(2) (2)当过G的直线l与曲线Q的相交弦长最大时,l过圆心(√2,√2),
l:x=m（y-√2)+√2,④
把④代入③,整理得(m^2+3)y^2+2√2m(1-m)y+2m^2-4m-6=0,
△=8m^2(1-m)^2-8(m^2+3)(m^2-2m-3)
=8[m^4-2m^3+m^2-(m^4-2m^3-6m-9)]
=8(m+3)^2,
∴|AB|=2|m+3|√[2(1+m^2)]/(m^2+3),
过点G(√2-√2m,0)且于l垂直的直线：x=-y/m+√2-√2m⑤交椭圆于C,D,把⑤代入③,（1/m^2+3）y^2-2(√2-√2m)y/m+2m^2-4m-6=0,
(△1)/4=(√2-√2m)^2/m^2-(1+3m^2)(2m^2-4m-6)/m^2
=(2/m^2)[(1-m)^2-(3m^2+1)(m^2-2m-3)]
=(-2/m^2)[3m^4-6m^3-9m^2-4],
|CD|=√△1/（1/m^2+3）*√(1+1/m^2)
=2√[(-2)( 3m^4-6m^3-9m^2-4)(m^2+1)]/(1+3m^2),
∴四边形ACBD的面积=(1/2)|AB||CD|=4,
∴(1+m^2)|m+3|√[-( 3m^4-6m^3-9m^2-4)]=(m^2+3)(1+3m^2)
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