设F1.F2分别为椭圆x^2/3+y^2=1的左右的焦点,点A,B在椭圆上,若向量F1A=5F2B,则A点的坐标2011高考题,浙江.
2019-05-02
设F1.F2分别为椭圆x^2/3+y^2=1的左右的焦点,点A,B在椭圆上,若向量F1A=5F2B,则A点的坐标
2011高考题,浙江.
优质解答
椭圆焦点F1(-√2,0) F2(√2,0);
AB:x=ty+√2
与x^2/3+y^2=1联立消去x得:
(ty+√2)^2+3y^2=3
即:(t^+3)y^2+2√2ty-1=0
设A(x1,y1),B(x2,y2)
y1+y2=-2√2t/(t^2+3) (1)
y1y2=-1/(t^2+3) (2)
∵向量F1A=5F2B
∴y1=-5y2 代入(1)(2):
-4y2=-2√2t/(t^2+3) (3)
-5y2^2=-1/(t^2+3) (4)
(3)^2/(4):t^2=2,t=±√2
y2=1/5 或, y2=- 1/5
∴ y1=-1, x1=0
或 y1=1, x1=0
∴A(0,-1)或A(0,1)
椭圆焦点F1(-√2,0) F2(√2,0);
AB:x=ty+√2
与x^2/3+y^2=1联立消去x得:
(ty+√2)^2+3y^2=3
即:(t^+3)y^2+2√2ty-1=0
设A(x1,y1),B(x2,y2)
y1+y2=-2√2t/(t^2+3) (1)
y1y2=-1/(t^2+3) (2)
∵向量F1A=5F2B
∴y1=-5y2 代入(1)(2):
-4y2=-2√2t/(t^2+3) (3)
-5y2^2=-1/(t^2+3) (4)
(3)^2/(4):t^2=2,t=±√2
y2=1/5 或, y2=- 1/5
∴ y1=-1, x1=0
或 y1=1, x1=0
∴A(0,-1)或A(0,1)