(1)C:x^2/4+y^2=1.①
(2)y'=x/2,
设M的切点为(2t,t^2),则切线l:y-t^2=t(x-2t),即y=tx-t^2,
代入①*4,得x^2+4(t^2x^2-2xt^3+t^4)=4,
整理得(1+4t^2)x^2-8xt^3+4t^4-4=0,
△/4=16t^6-(1+4t^2)(4t^4-4)=-4t^4+16f^2+4,则
|AB|=√[△(1+t^2)]/(1+4t^2),
O到l的距离h=t^2/√(t^2+1),
∴S△OAB=(1/2)|AB|h=t^2*√△/[2(1+4t^2)]=2√[t^4(-t^4+4t^2+1)]/(1+4t^2),
设u=t^2>=0,f(u)=u^2(-u^2+4u+1)/(1+4u)^2,则
f'(u)=(-4u^3+12u^2+2u)/(1+4u)^2-8u^2(-u^2+4u+1)/(1+4u)^3
=2u[(-2u^2+6u+1)(1+4u)-4u(-u^2+4u+1)]/(1+4u)^3
=2u(-4u^3+6u^2+6u+1)/(1+4u)^3,繁!
超出中学数学范围,还需要继续吗?
追问不需要了,我换了其他方法,思路没问题,谢谢你