椭圆x^2+4y^2=8中,AB是长为5/2的动弦,o为坐标原点,求三角形AOB的面积的取值范围

Question

椭圆x^2+4y^2=8中,AB是长为5/2的动弦,o为坐标原点,求三角形AOB的面积的取值范围

Answer 1

设A(m+5cosu/4,n+5sinu/4),B(m-5cosu/4,n-5sinu/4)在椭圆x^2+4y^2=8上,
∴(m+5cosu/4)^2+4(n+5sinu/4)^2=8,①
(m-5cosu/4)^2+4(n-5sinu/4)^2=8.②
①+②,2m^2+8n^2+25(cosu)^2/2+50(sinu)^2=16,
4(m^2+4n^2)=7-75(sinu)^2,③
①-②,5mcosu+20nsinu=0,
∴tanu=-m/(4n),
∴（sinu)^2=(tanu)^2/[1+(tanu)^2]=m^2/(m^2+16n^2)
代入③/4,（m^2+4n^2)(m^2+16n^2)=28n^2-17m^2,
m^4+(20n^2+17)m^2+64n^4-28n^2=0,
△=（20n^2+17)^2-4(64n^4-28n^2)=144n^4+452n^2+289,
∴m^2=[-(20n^2+17)+√△]/2,
AB的斜率=tanu=-m/(4n),
∴AB:y-n=-m(x-m)/(4n),
即mx+4ny-m^2-4n^2=0,
∴O到AB的距离d=(m^2+4n^2)/√(m^2+16n^2)
=(√△-12n^2-17)/√[2(√△+12n^2-17)],
设v=n^2,由③,0