Question

经过P（2,0）作直线L交椭圆C：x^2/2+y^2=1于A,B两点若三角形AOB的面积为2/3求直线

Answer 1

经过P（2,0）作直线L交椭圆C：x^2/2+y^2=1于A,B两点若三角形AOB的面积为2/3求直线
解：设直线方程为y=k(x-2)=kx-2k，代入椭圆方程得：
x²+2(kx-2k)²-2=0，展开化简得：
(1+2k²)x²-8k²x+8k²-2=0
设直线与椭圆的交点A(x₁，y₁)；B(x₂，y₂)；依韦达定理有：
x₁橘慧+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)
弦长︱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)
椭圆中心(0，0)到直线kx-y-2k=0的距离h=︱-2k︱/镇宽√(1+k²)
△AOB的面积S=(1/2)︱AB︱h=[︱2k︱√(-4k⁴-2k²+2)]/[(1+2k²)√(1+k²)]=2/圆旅答3
3[︱k︱√(-4k⁴-2k²+2)]=(1+2k²)√(1+k²)
9k²(-4k⁴-2k²+2)=(1+2k²)²(1+k²)
展开化简得 40k^6+26k⁴-13k²+1=0............(1)
解方程(1)，得实根k=±1/2，故直线方程为y=±(1/2)(x-2).

Answer 2

经过P（2,0）作直线L交椭圆C：x^2/2+y^2=1于A,B两点若三角形AOB的面积为2/3求直线
解：设直线方程为y=k(x-2)=kx-2k，代入椭圆方程得：
x²+2(kx-2k)²-2=0，展开化简得：
(1+2k²)x²-8k²x+8k²-2=0
设直线与椭圆的交点A(x₁，y₁)；B(x₂，y₂)；依韦达定理有：
x₁橘慧+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)
弦长︱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)
椭圆中心(0，0)到直线kx-y-2k=0的距离h=︱-2k︱/镇宽√(1+k²)
△AOB的面积S=(1/2)︱AB︱h=[︱2k︱√(-4k⁴-2k²+2)]/[(1+2k²)√(1+k²)]=2/圆旅答3
3[︱k︱√(-4k⁴-2k²+2)]=(1+2k²)√(1+k²)
9k²(-4k⁴-2k²+2)=(1+2k²)²(1+k²)
展开化简得 40k^6+26k⁴-13k²+1=0............(1)
解方程(1)，得实根k=±1/2，故直线方程为y=±(1/2)(x-2).