优秀教案高中数学第二册上:8.6 抛物线习题课
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- 上传日期:2011-08-31
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抛 物 线 习 题 课
教学目标:熟练掌握抛物线的性质及其求法。重点:抛物线的求法难点:抛物线的证明教学过程:
1 复习回顾
简单回顾抛物线的四种方程及其性质
2. 练习:
⑴选择题:
1,以F(0,1)为焦点,以L:y ( ( 1为准线的拋物线的方程式为何?
(A) y2 ( 4x (B) y2 ( ( 4x (C) x2 ( 4y (D) x2 ( ( 4y (E) y ( x2
答案:C
2.下列何者为拋物线y ( ax2 ( bx ( c的顶点在第四象限的充分条件?
(A) a ( 0,b ( 0,c ( 0 (B) a ( 0,b ( 0,c ( 0
(C) a ( 0,b ( 0,c ( 0 (D) a ( 0,b ( 0,b2 ( 4ac ( 0
答案:C3.设y ( y ( ax2 ( bx ( c的图形如右,下列何者正确?
A) a ( 0 (B) b ( 0 (C) c ( 0
(D) a ( b ( c ( 0 (E) b2 ( 4ac ( 0
答案:B,D,E ⑵填空题:
1.与直线2x (
...点击查看全部>>
3y ( 2 ( 0及点(1,( 1)等距离的点的轨迹方程式为
9x2 ( 12xy ( 4y2 ( 34x ( 14y ( 22 ( 0
2.与y2 ( 4x ( 6y ( 5 ( 0共轴、共焦点且过(3,1)之拋物线方程为
(y ( 3)2 ( ( 16(x ( 4)或(y ( 3)2 ( 4(x ( 1)
3.拋物线C1:y2 ( 4x,椭圆C2:bx2 ( 9y2 (
9b有共同之焦点F1,P为C1,C2位于x轴上方之交点,F2为C2之另一焦点,
且(PF2F1 ( (,(PF1F2 ( (,求cos(.cos( ( ⑶证明题
1.设线段PQ为拋物线C
的焦点弦(过焦点的弦),L为C的准线,F为焦点,如图所示,过P,Q分别作L的垂线
,令垂足依序为A,B,且M为AB的中点,试证:
(1)MP⊥MQ
(2)MF⊥PQ
证明:(1) F为拋物线C 的焦点,且弦PQ过焦点F,L为准线,M为AB中点,
PA⊥L,QB⊥L,所以PA(PF,QB(QF,因此AP(BQ(PF(QF(PQ。过M作MN//AP,交PQ于N,则
MN(1/2 (AP(BQ)(PQ,又N为PQ中点,
所以MN(1/2 PQ(PN(QN,
因此∠MPN (∠PMN,∠QMN (∠MQN
所以∠PMQ (∠PMN (∠QMN ( 90(,即MP⊥MQ
(2)如图,∠1 (∠2,∠3 (∠4,又
∠1 (∠2 (∠APF (∠3 (∠4 (∠BQF ( 360(
又∠APF (∠BQF ( 180(,所以,
∠1 (∠2 (∠3 (∠4 ( 2(∠2 (∠4) ( 180(
因此,∠2 (∠4 ( 90(,可得∠AFB ( 90(,M为直角三角形AFB的斜边中点
所以AM(BM(MF,又AP(FP,MP(MP,所以
△AMP △FMP
因此∠PFM (∠PAM ( 90(,即PF⊥MF,亦即MF⊥PQ
⑷解答题:
1.两拋物线y ( x2 ( 3x ( 2,y ( 2x2 ( 5x ( a相交于两相异点P,Q,
(1)求a的范围
(2)求PQ的方程式
(3)若PQ( 4,则a (?
(1) 2x2 ( 5x ( a ( x2 ( 3x ( 2有两解 (x2 ( 2x ( (a ( 2) ( 0有相异两实根
(判别式 ( ( ( 2)2 ( 4(a ( 2) ( 0 (a ( 3
3.设动圆C与圆x2 ( y2 ( 8x ( 12 ( 0及直线x ( 2 (
0相切,则动圆C之圆心轨迹方程式为何?
解:x2(y2(8x(12(0 ((x(4)2(y2( 4,圆心A(4,0),半径2,
令动圆C之圆心P(x,y)
-----------------------
3.设y ( y ( ax2 ( bx ( c的图形如右,下列何者正确? * a ( 0 (B) b ( 0 (C) c ( 0
(D) a ( b ( c ( 0 (E) b2 ( 4ac ( 0( ( 2 ( ( ( y ( ( x ( 4 ( a ( PQ:x ( y ( 4 ( a ( 0
(2)(3)设x2 ( 2x ( (a ( 2) ( 0两根为(,(,则可令P((,(2 ( 3( ( 2),Q((,(2 ( 3(
( 2)
( PQ2 ( { }2
( (( ( ()2[1 ( (( ( ( ( 3)2]
因故PQ2( (( ( ()2[1 ( (2 ( 3)2] ( 2(( ( ()2
( 2[(( ( ()2 ( 4((]( 2[4 ( 4(a ( 2)]
( 24 ( 8a,24 ( 8a ( 16
( a ( 1外切时,d(P,L) (PA(2
(|x(2|( (2
(y2(16x内切时,d(P,L) (PA(2
(|x(2|( (2
(y2(8x(16
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