已知A(1,0),椭圆c:x^2/4+y^2/3=1,过点A做直线交椭圆c于P,Q两点,AP=2QA,则直线PQ的斜率
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![已知A(1,0),椭圆c:x^2/4+y^2/3=1,过点A做直线交椭圆c于P,Q两点,AP=2QA,则直线PQ的斜率](/uploads/image/z/8594550-54-0.jpg?t=%E5%B7%B2%E7%9F%A5A%281%2C0%29%2C%E6%A4%AD%E5%9C%86c%3Ax%5E2%2F4%2By%5E2%2F3%3D1%2C%E8%BF%87%E7%82%B9A%E5%81%9A%E7%9B%B4%E7%BA%BF%E4%BA%A4%E6%A4%AD%E5%9C%86c%E4%BA%8EP%2CQ%E4%B8%A4%E7%82%B9%2CAP%3D2QA%2C%E5%88%99%E7%9B%B4%E7%BA%BFPQ%E7%9A%84%E6%96%9C%E7%8E%87)
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已知A(1,0),椭圆c:x^2/4+y^2/3=1,过点A做直线交椭圆c于P,Q两点,AP=2QA,则直线PQ的斜率
已知A(1,0),椭圆c:x^2/4+y^2/3=1,过点A做直线交椭圆c于P,Q两点,AP=2QA,则直线PQ的斜率
已知A(1,0),椭圆c:x^2/4+y^2/3=1,过点A做直线交椭圆c于P,Q两点,AP=2QA,则直线PQ的斜率
C:x^2/4+y^2/3=1,A(1,0)
设直线PQ:x=ty+1代入x^2/4+y^2/3=1,
(ty+1)²/4+y²/3=1
即(3t²+4)y²+6ty-9=0
设P(x1,y1),Q(x2,y2)
∴y1+y2=-6t/(3t²+4),
y1y2=-9/(3t²+4)
∵AP=2QA
∴y1=-2y2
那么
-y2=-6t/(3t²+4)
- y²2=-9/(3t²+4)
消去y2得:
36t²/(3t²+4)²=9/(3t²+4)
∴4t²=3t²+4
∴t²=4,t=±2
∴直线PQ的斜率为1/t=±1/2