【解析】抛物线x2=2py的焦点为(0"," p/2),所以可得b=p/2,因为2a=4√2⇒a=2√2,所以双曲线方程为x^2/8-(4y^2)/p^2 =1,可求得其渐近线方程为y=±p/(4√2)x,不妨设y=kx-1与y=p/(4√2)x平行,则有k=p/(4√2).

　　联立方程{■(y=p/(4√2) x"-" 1"," @x^2=2py"," )┤得x2-p^2/(2√2)x+2p=0,所以Δ=("-" p^2/(2√2))^2-8p=0,解得p=±4,又p>0,故p=4.

　　【答案】A

10.已知抛物线y2=2px(p>0)的焦点为F,△ABC的顶点都在抛物线上,且满足(FA) ⃗+(FB) ⃗=-(FC) ⃗,则1/k_AB +1/k_BC +1/k_CA =.　

　　【解析】设A,B,C三点的坐标分别为(x1,y1),(x2,y2),(x3,y3).

　　∵(FA) ⃗+(FB) ⃗=-(FC) ⃗,∴△ABC的重心是F.

　　又∵抛物线y2=2px的焦点F的坐标为(p/2 "," 0),

　　∴y1+y2+y3=0.

　　又∵点A,B在抛物线上,∴y_1^2=2px1,y_2^2=2px2,两式相减,得y_1^2-y_2^2=2p(x1-x2),

　　∴kAB=2p/(y_1+y_2 ),同理kBC=2p/(y_2+y_3 ),kCA=2p/(y_1+y_3 ),

　　∴1/k_AB +1/k_BC +1/k_CA =(y_1+y_2)/2p+(y_2+y_3)/2p+(y_3+y_1)/2p=(y_1+y_2+y_3)/p=0.

　　【答案】0

11.已知椭圆C1的中心在坐标原点,两焦点分别为双曲线C2:x^2/2-y2=1的顶点,直线x+√2y=0与椭圆C1交于A,B两点,且点A的坐标为(-√2,1),点P是椭圆C1上异于A,B的任意一点,点Q满足(AQ) ⃗·(AP) ⃗=0,(BQ) ⃗·(BP) ⃗=0,且A,B,Q三点不共线.

(1)求椭圆C1的方程;

(2)求点Q的轨迹方程;

(3)求△ABQ面积的最大值及此时点Q的坐标.

　　【解析】(1)∵双曲线C2:x^2/2-y2=1的顶点为F1(-√2,0),F2(√2,0),

　　∴椭圆C1两焦点分别为F1(-√2,0),F2(√2,0).

　　设椭圆C1方程为x^2/a^2 +y^2/b^2 =1(a>b>0),

　　∵椭圆C1过点A(-√2,1),

　　∴2/a^2 +1/b^2 =1.　①

　　∵a2=b2+2,　②

　　　　由①②解得a2=4,b2=2.

　　∴椭圆C1的方程为x^2/4+y^2/2=1.

　　(2)设点Q(x,y),点P(x1,y1),

　　由点A(-√2,1)及椭圆C1关于原点对称可得B(√2,-1),

　　∴(AQ) ⃗=(x+√2,y-1),(AP) ⃗=(x1+√2,y1-1),

　　(BQ) ⃗=(x-√2,y+1),(BP) ⃗=(x1-√2,y1+1).

　　由(AQ) ⃗·(AP) ⃗=0,得(x+√2)(x1+√2)+(y-1)(y1-1)=0,

　　即(x+√2)(x1+√2)=-(y-1)(y1-1).　①

　　同理,由(BQ) ⃗·(BP) ⃗=0,得(x-√2)(x1-√2)=-(y+1)(y1+1).　②

　　①×②得(x2-2)(x_1^2-2)=(y2-1)(y_1^2-1).　③

　　由于点P在椭圆C1上,则(x_1^2)/4+(y_1^2)/2=1,得x_1^2=4-2y_1^2,

　　代入③式得-2(y_1^2-1)(x_^2-2)=(y2-1)(y_1^2-1).

当y_1^2-1≠0时,有2x2+y2=5;