Question

高中数学圆锥曲线第二问 详细解答过程谢谢

Answer 1

联立y=x+n与x^2/6+y^2/3=1消去x并整理得
3y^2-2ny+n^2-6=0
因为方程有两个不等根，所以（-2n)^2-12(n^2-6)>0
解得-3<n<3
从而由根与系数粗键的关系得y1+y2=2n/3,y1*y2=(n^2-6)/3
又直线y=x+n的搭世交知凳肢点为（-n,0)
所以三角形面积S=|n||y1-y2|/2
=|n|√(y1+y2）^2-4y1y2/2
<=3√2/2

Answer 2

椭圆方程x^2/6+y^2/3=1与直线春缓乱扒档y=x+n联立得到x^ 2+2(x+n)^2=6
即有3x^2+4nx+2n ^2-6=0
x1+x2=-4n/3, x1x2=(2n^2-6)/3
(x1-x2)^2=(x1+x2) ^2-4x1x2=16n^2/9-(8n^2-24)/3=(72-8n^2)/9
S(AOB)=1/2|n|*|x1-x2|=1/2|n|*[根号(72-8n^2)]/3=1/6根号(72n^2-8n^4)=1/6根号[-8(n^4-9n^2)]=1/6根号[-8(n^2-9/2)^2+81*2]
故当n^2=9/2,即有n=3/2根号2时,S有最大值是1/6*9根号2=3/2根哪敬号2