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设椭圆x^2/a^2+y^2/b^2=1(a>b>0)的左右顶点分别为A,B点P在椭圆上且异于A,B两点,O为坐标原点(

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设椭圆x^2/a^2+y^2/b^2=1(a>b>0)的左右顶点分别为A,B点P在椭圆上且异于A,B两点,O为坐标原点(1)若直线AP与BP的斜率之积为-1/2,求椭圆的离心率
(2)若AP的 长等于OA的长,证明直线OP的斜率K满足K的绝对值>√3
设椭圆x^2/a^2+y^2/b^2=1(a>b>0)的左右顶点分别为A,B点P在椭圆上且异于A,B两点,O为坐标原点(
(1)
设P(p,q),p²/a² + q²/b² = 1 (i)
A(-a,0),B(a,0)
AP与BP的斜率分别为(q - 0)/(p+ a) = q/(p + a),(q- 0)/p - a) = q/(p - a)
斜率之积为[q/(p+a)][q/(p - a)] = q²/(p² - a²) = -1/2
p² = a² - 2q² (ii) q² = (a² - p²)/2 (iii)
代入(i):a² = 2b²
e = c/a = √[(a² - b²)/a²] = √[(2b² - b²)/(2b²)] = √2/2
(2)
OA = a
OA² = AP² = a² = (p + a)² + (q - 0)² = p² + 2ap + a² + q² = p² + 2ap + a² + (a² - p²)/2
p² + 4ap + a² = 0
p = (-4a ± √(16a² - 4a²)]/2 = (-2 ± √3)a
的横坐标应在(-a,0)上,舍去p = (-2 - √3)a
p = (-2 + √3)a
p² = (7 - 4√3)a²
q² = (a² - p²)/2 = (2√3 - 3)a²
|K|² = |q²/p²| = (2√3 - 3)/(7 - 4√3)
= (2√3 - 3)(7 + 4√3)/[(7 - 4√3)(7 + 4√3)]
= 3 + 2√3 > 3
|K| > √3