F1(-c,0),F2(c,0)分别是椭圆C:x^2/a^2+y^2/b^2=1(a>b>0)的左,右焦点F1(-c,0),F2(c,0)分别是椭圆C:x^2/a^2+y^2/b^2=1(a>b>0)的左,右焦点,过F1作x轴的垂线交椭圆的上半部分于P,过F2作直线PF2的垂线交直线x=a^2/c于Q(1
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![F1(-c,0),F2(c,0)分别是椭圆C:x^2/a^2+y^2/b^2=1(a>b>0)的左,右焦点F1(-c,0),F2(c,0)分别是椭圆C:x^2/a^2+y^2/b^2=1(a>b>0)的左,右焦点,过F1作x轴的垂线交椭圆的上半部分于P,过F2作直线PF2的垂线交直线x=a^2/c于Q(1](/uploads/image/z/1756161-9-1.jpg?t=F1%28-c%2C0%29%2CF2%28c%2C0%29%E5%88%86%E5%88%AB%E6%98%AF%E6%A4%AD%E5%9C%86C%3Ax%5E2%2Fa%5E2%2By%5E2%2Fb%5E2%3D1%28a%3Eb%3E0%29%E7%9A%84%E5%B7%A6%2C%E5%8F%B3%E7%84%A6%E7%82%B9F1%28-c%2C0%29%2CF2%28c%2C0%29%E5%88%86%E5%88%AB%E6%98%AF%E6%A4%AD%E5%9C%86C%EF%BC%9Ax%5E2%2Fa%5E2%2By%5E2%2Fb%5E2%3D1%28a%3Eb%3E0%29%E7%9A%84%E5%B7%A6%2C%E5%8F%B3%E7%84%A6%E7%82%B9%2C%E8%BF%87F1%E4%BD%9Cx%E8%BD%B4%E7%9A%84%E5%9E%82%E7%BA%BF%E4%BA%A4%E6%A4%AD%E5%9C%86%E7%9A%84%E4%B8%8A%E5%8D%8A%E9%83%A8%E5%88%86%E4%BA%8EP%2C%E8%BF%87F2%E4%BD%9C%E7%9B%B4%E7%BA%BFPF2%E7%9A%84%E5%9E%82%E7%BA%BF%E4%BA%A4%E7%9B%B4%E7%BA%BFx%3Da%5E2%2Fc%E4%BA%8EQ%EF%BC%881)
F1(-c,0),F2(c,0)分别是椭圆C:x^2/a^2+y^2/b^2=1(a>b>0)的左,右焦点F1(-c,0),F2(c,0)分别是椭圆C:x^2/a^2+y^2/b^2=1(a>b>0)的左,右焦点,过F1作x轴的垂线交椭圆的上半部分于P,过F2作直线PF2的垂线交直线x=a^2/c于Q(1
F1(-c,0),F2(c,0)分别是椭圆C:x^2/a^2+y^2/b^2=1(a>b>0)的左,右焦点
F1(-c,0),F2(c,0)分别是椭圆C:x^2/a^2+y^2/b^2=1(a>b>0)的左,右焦点,过F1作x轴的垂线交椭圆的上半部分于P,过F2作直线PF2的垂线交直线x=a^2/c于Q
(1)若点Q的坐标为(4,4)求椭圆C的方程
(2)证明直线PQ与椭圆C只有一个交点
F1(-c,0),F2(c,0)分别是椭圆C:x^2/a^2+y^2/b^2=1(a>b>0)的左,右焦点F1(-c,0),F2(c,0)分别是椭圆C:x^2/a^2+y^2/b^2=1(a>b>0)的左,右焦点,过F1作x轴的垂线交椭圆的上半部分于P,过F2作直线PF2的垂线交直线x=a^2/c于Q(1
显然,有a^2/c=4,P(-c,b^2/a),KPF2=-b^2/(2ac),KQF2=2ac/b^2,QF2的方程为:y=2ac/b^2(x-c),所以,2ac/b^2(4-c)=4,即有:a^2=4c 及2ac(4-c)=4(a^2-c^2),解得:a=2,c=1,b=sqrt(3).所以椭圆的方程为:x^2/4+y^2/3=1.
(2) PQ的方程为:y=1/2(x-4)+4,显然只证其与过P的切线方程相同即可,因切线方程为:-x/4+3/2*y/3=1,即为x+2y=4显然相同.
终于找到答案了,在链接里。是2012安徽理科高考题目,如果链接打不开。楼主去百度文库里找吧。希望我的答案能帮到你,祝愉快O(∩_∩)O~