Question

数学求助！步骤详细点！3

Answer 1

(1)
e=(根2)/2, a^2=2c^2=b^2+c^2
所以b=c=a*(根2)/2
M(0,b),F(b,0),B(b*根2,0)
向量MF(b,-b),FB(b*(根2-1),0)
MF*MB=(根2-1)*b^2=根2-1
b=1
a=根2
x^2/2+y^2=1
(2)
F为垂心,所以MF垂直于PQ.MF斜率为-1,所以PQ斜率为1
设PQ方程为:y=x+n,P(x1,y1);Q(x2,y2)
代入椭圆得:3x^2+4nx+2n^2-2=0
x1+x2=-4n/3,x1x2=(2n^2-2)/3
向量QF(1-x2,-y2),向量MP(x1,y1-1)
QF*MP=x1-x1x2-y1y2+y2=x1-x1x2-(x1+n)(x2+n)+x2+n=(1-n)(x1+x2)-2x1x2-n^2+n=0
解得n=1或N=-4/3
当n=1时，PQ过点M，不构成三角形，舍去。
y=x-4/3

(2)
F为垂心,所以MF垂直于PQ.MF斜率为-1,所以PQ斜率为1
设PQ方程为:y=x+n,P(x1,y1);Q(x2,y2)
代入椭圆得:3x^2+4nx+2n^2-2=0
x1+x2=-4n/3,x1x2=(2n^2-2)/3
向量QF(1-x2,-y2),向量MP(x1,y1-1)
QF*MP=x1-x1x2-y1y2+y2=x1-x1x2-(x1+n)(x2+n)+x2+n=(1-n)(x1+x2)-2x1x2-n^2+n=0
解得n=1或N=-4/3
当n=1时，PQ过点M，不构成三角形，舍去。
y=x-4/3

Answer 2

设：A(a,0),B(0,-b)
--->x/a-y/b=1被椭圆C截得的弦长|AB|=2√2=√(a²+b²)=c

过椭圆右焦点且斜率为√3的直线L：y=√3(x-2√2)
代入椭圆方程：b²x²+3a²(x-2√2)²=a²b²
--->(a²-8)x²+3a²(x²-4√2x+8)-a²(a²-8) = 0
--->(4a²-8)x²-12√2a²x+32a²-(a²)² = 0
--->x1+x2=3√2a²/(a²-2), 4x1x2=(32a²-a^4)/(a²-2)

L被椭圆C截得的弦长 = √(1+3)•|x1-x2| = (2/5)2a = 4a/5
--->|x1-x2|² = (2a/5)² = (x1+x2)²-4x1x2
　　　　　　 = 18a^4/(a²-2)² - (32a²-a^4)/(a²-2)
---> 4(a²-2)² = 450a² - 25(32-a²)(a²-2)
---> 4a^4-16a²+16 = 450a² + 25(a^4-34a²+64)
---> 21a^4 - 384a² + 1584 = 0
...

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