已知点是F抛物线C&1：x2=4y与椭圆C&2：y2a2+x2b2=1(a＞b＞0)的公共焦点，且椭圆的离心率为12（1）求椭圆的方程；（2）过抛物线上一点P，作抛物线的切线l，切点P在第一象限，如图，设切线l与椭圆相交于不同的两点A、B，记直线OP，FA，FB的斜率分别为k，k1，k2（其中O为坐标原点），若k&1+k2=203k，求点P的坐标．
分析：（1）利用抛物线与椭圆有公共焦点，且椭圆的离心率为12，建立方程组，求出几何量，即可求椭圆的方程；（2）设出切线方程与椭圆方程联立，利用韦达定理及斜率公式，即可求得结论．解答：解：（1）∵点F的坐标为（0，1），则有a2-b2=1a2-b2a=12∴a=2，b=3∴椭圆方程为y24+x23=1；（2）设P（2t，t2），由y2=x2，得切线的斜率为t，从而切线l的方程为y=tx-t2，直线l与椭圆方程联立，得（3t2+4）x2-6t3x+3t4-12=0设A（x1，y1），B（x2，y2），则x1+x2=6t33t2+4，x1x2=3t4-123t2+4∴k1+k2=y1-1x1+y2-1x2=2t-2t3(t2+1)t4-4∵k=t22t=t2∴2t-2t3(t2+1)t4-4=10t3∵t＞0，∴5t4+3t2-8=0∴t2=1∴t=1∴P的坐标为（2，1）．点评：本题考查椭圆的标准方程，考查直线与椭圆的位置关系，考查学生的计算能力，属于中档题．
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科目：高中数学
（;嘉兴一模）已知F为抛物线C：y2=4x焦点，其准线交x轴于点M，点N是抛物线C上一点（Ⅰ）如图1，若MN的中垂线恰好过焦点F，求点N的y轴的距离（Ⅱ）如图2，已知直线l交抛物线C于点P，Q，若在抛物线C上存在点R，使FPRQ为平行四边形，试探究直线l是否过定点？并说明理由．
科目：高中数学
设抛物线C：x2=2py（p＞0）的焦点为F，A（x0，y0）（x0≠0）是抛物线C上的一定点．（1）已知直线l过抛物线C的焦点F，且与C的对称轴垂直，l与C交于Q，R两点，S为C的准线上一点，若△QRS的面积为4，求p的值；（2）过点A作倾斜角互补的两条直线AM，AN，与抛物线C的交点分别为M（x1，y1），N（x2，y2）．若直线AM，AN的斜率都存在，证明：直线MN的斜率等于抛物线C在点A关于对称轴的对称点A1处的切线的斜率．
科目：高中数学
来源：学年浙江省台州中学高二（下）期中数学试卷（理科）（解析版）
题型：解答题
已知点是F抛物线C与椭圆C的公共焦点，且椭圆的离心率为（1）求椭圆的方程；（2）过抛物线上一点P，作抛物线的切线l，切点P在第一象限，如图，设切线l与椭圆相交于不同的两点A、B，记直线OP，FA，FB的斜率分别为k，k1，k2（其中O为坐标原点），若k，求点P的坐标．
科目：高中数学
来源：2013年浙江省嘉兴市高考数学一模试卷（文科）（解析版）
题型：解答题
已知F为抛物线C：y2=4x焦点，其准线交x轴于点M，点N是抛物线C上一点（Ⅰ）如图1，若MN的中垂线恰好过焦点F，求点N的y轴的距离（Ⅱ）如图2，已知直线l交抛物线C于点P，Q，若在抛物线C上存在点R，使FPRQ为平行四边形，试探究直线l是否过定点？并说明理由．ASCII Chart:
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STX start of text
ETX end of text
EOT end of
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ENQ enquiry
ACK acknowledege
BS backspace [\b]
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SO shift out
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DC4 device control
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SYN synchronize
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CAN cancel
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SUB substitute
ESC escape
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US unit separator
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dollar sign
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upper case A
upper case B
upper case C
upper case D
upper case E
upper case F
upper case G
upper case H
upper case I
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upper case K
upper case L
upper case M
upper case N
upper case O
upper case P
upper case Q
upper case R
upper case S
upper case T
upper case U
upper case V
upper case W
upper case X
upper case Y
upper case Z
left square bracket
backslash, not slash!
right square bracket
hat, circumflex
underscore
grave, rhymes with have
lower case a
lower case b
lower case c
lower case d
lower case e
lower case f
lower case g
lower case h
lower case i
lower case j
lower case k
lower case l
lower case m
lower case n
lower case o
lower case p
lower case q
lower case r
lower case s
lower case t
lower case u
lower case v
lower case w
lower case x
lower case y
lower case z
left curly brace
vertical bar
right curly brace
DEL delete
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PostScript (') quotesingle
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PostScript (,) cedilla
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USA - WMJ Marine Corporation Copy Right: WMJ Marine Corporation1.对任意实数k,圆C：x2+y2-6x-8y+12=0与直线l：kx-y-4k+3=0的位置关系是2.若PQ是圆x2+y2=9的铉,PQ的中点为M（1.2）,则实现PQ的方程是3.已知满足a2+b2=4,则√（a-3）2+（b-4）2的最小值与最大值分别是4.已知_百度作业帮
1.对任意实数k,圆C：x2+y2-6x-8y+12=0与直线l：kx-y-4k+3=0的位置关系是2.若PQ是圆x2+y2=9的铉,PQ的中点为M（1.2）,则实现PQ的方程是3.已知满足a2+b2=4,则√（a-3）2+（b-4）2的最小值与最大值分别是4.已知
1.对任意实数k,圆C：x2+y2-6x-8y+12=0与直线l：kx-y-4k+3=0的位置关系是2.若PQ是圆x2+y2=9的铉,PQ的中点为M（1.2）,则实现PQ的方程是3.已知满足a2+b2=4,则√（a-3）2+（b-4）2的最小值与最大值分别是4.已知点（-a.3）在远x2+y2+2ax+4y+a2-4a+7=0外则实数a的取值范围是5.已知园C：（x-1）2+y2=4内衣店P（2.1）,则国电P最短铉所在的直线方程是 ,最长选所在的直线方程是6.点P 在园C1：x2+y2-8x-4y+11=0上,点Q在园C2：x2+y2+4x+2y+1=0上求PQ的最小值
1,圆C：（x-3）?+(y-4)?=13,圆心（3,4）,半径√13,直线l：y=k（x--4）+3,恒过定点（4,3）,其中（4,3）在圆内,故相交.2,圆心为原点,弦中点与圆心连续垂直弦,故PQ斜率-1/2,则方程为y-2= -1/2（x-1）,即y= -x/2-3/2.3,就是问圆心为原点,半径为2的圆上一点到（3,4）距离,显然最长最短的点都在圆心与点连线上,故可得最大值7,最小值3.4,将点（-a.3）代入圆方程,在圆外,则应（-a）?+3?+2a·（-a）+4·3+a?-4a+7＞0,解得a≤7,5,P（2.1）,圆心C（1,0）,最长弦是CP直线,即斜率1,方程y=x-1,最短弦是过P的CP垂直线,则斜率-1,则方程y= -x+1,6,两个圆心连线上,得3√5 -5.
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