經(jīng)過(guò)P（2,0）作直線L交橢圓C：x^2/2+y^2=1于A,B兩點(diǎn)若三角形AOB的面積為2/3求直線
設(shè)直線方程為y=k(x-2)=kx-2k,代入橢圓方程得：
x²+2(kx-2k)²-2=0,展開化簡(jiǎn)得：
(1+2k²)x²-8k²x+8k²-2=0
設(shè)直線與橢圓的交點(diǎn)A(x₁,y₁)；B(x₂,y₂)；依韋達(dá)定理有：
x₁+x₂=8k²/(1+2k²)；x₁x₂=(8k²-2)/(1+2k²)；
y₁+y₂=kx₁-2k+kx₂-2k=k(x₁+x₂)-4k=8k³/(1+2k²)-4k=-4k/(1+2k²)
y₁y₂=(kx₁-2k)(kx₂-2k)=k²x₁x₂-2k²(x₁+x₂)+4k²=k²[(8k²-2)/(1+2k²)-16k²/(1+2k²)+4]
=k²[2/(1+2k²)]=2k²/(1+2k²)
弦長(zhǎng)︱AB︱=√[(x₁+x₂)²+(y₁+y₂)²-4(x₁x₂+y₁y₂)]
=√[64k⁴/(1+2k²)²+16k²/(1+2k²)²-4(8k²-2)/(1+2k²)-8k²/(1+2k²)]
=√[(-16k⁴-8k²+8)/(1+2k²)²]=[2/(1+2k²)]√(-4k⁴-2k²+2)
橢圓中心(0,0)到直線kx-y-2k=0的距離h=︱-2k︱/√(1+k²)
△AOB的面積S=(1/2)︱AB︱h=[︱2k︱√(-4k⁴-2k²+2)]/[(1+2k²)√(1+k²)]=2/3
3[︱k︱√(-4k⁴-2k²+2)]=(1+2k²)√(1+k²)
9k²(-4k⁴-2k²+2)=(1+2k²)²(1+k²)
展開化簡(jiǎn)得 40k^6+26k⁴-13k²+1=0.(1)
解方程(1),得實(shí)根k=±1/2,故直線方程為y=±(1/2)(x-2).