已知曲线(2 x2 3 y2) 3 z2=6 和(x2−y2) z2=0 ,则
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设P是双曲线右支上的一点,设|PF1|=m,|PF2|=n.则m-n=23m+n=26,解得mn=3.|F1F2|=4.∴cos∠F1PF2=m2+n2-422mn=(m+n)2-2mn-422mn=
⑴x^2+y^2+2kx+(4k+10)y+10k+20=0x^2+2kx+k^2+y^2+2(2k+5)y+(2k+5)^2=-10k-20+k^2+(2k+5)^2(x+k)^2+(y+2k+5)
y=2x-5x²+y²=k,x²+(2x-5)²=k5x²-20x+25-k=0∆=(-20)²-4*5*(25-k)=20(k
(1)由D2+E2-4F=4+16-4m=20-4m>0,解得m<5; (4分)(2)设M(x1,y1),N(x2,y2),联立直线x+2y-
(1)设直线的方程为y=k(x+2),代入椭圆x23+y2=1,消去y,可得(1+3k2)x2+12k2x+12k2-3=0由△=0,可得k2-1=0设l1,l2的斜率分别为k1,k2,∴k1=-1,
已知2x=3y,求xy/(x^2+y^2)-y^2/(x^2-y^2)的值2x=3y-->x=(3/2)yx^2=(9/4)y^2xy/(x^2+y^2)-y^2/(x^2-y^2)==(3/2)y*
(x²+y²)²+(x²+y²)-6-6=0(x²+y²)²+(x²+y²)-12=0(x²
(1)设A(x1,y1)、B(x2,y2),由已知|m|1+k2=32,得m2=34(k2+1),把y=kx+m代入椭圆方程,整理得(3k2+1)x2+6kmx+3m2-3=0,∴x1+x2=−6km
由于双曲线x23−y2=1可得a=3,b=1,故可得c=2由双曲线方程的形式知,其右焦点坐标是(2,0)又抛物线y2=2px的焦点与双曲线x23−y2=1的右焦点重合∴p2=2,得p=4故选D
x^2+y^2-k=0,y^=k-x^,①代入3x^+2y^=6x得3x^+2(k-x^)=6x,x^-6x+9=9-2k,当9-2k>=0②时x=3土√(9-2k),代入①得k-[3土√(9-2k)
由2a^2+b^2≤2,可知点P在曲线C围成的区域内部.直线I方程为y-b=k(x-a),中点(m,n),两交点(x1,y1),(x2,y2)m=(x1+x2)/2,n=(y1+y2)/2;k=(y1
如图:可知A(-3,0),设C(0,m),OP⊥AC,由四边形ABCD的面积S=4S△AOC=23m=23,解得m=1,由等面积可知12×OA×OC=12×AC×OP,代入数据可得3m=3+m2×b,
(Ⅰ)设y=kx+t(k>0),由题意,t>0,由方程组y=kx+tx23+y2=1,得(3k2+1)x2+6ktx+3t2-3=0,由题意△>0,所以3k2+1>t2,设A(x1,y1),B(x2,
(x-1)^2+(y-1)^2=1圆心(1,1),半径=1直线x/a+y/b=1bx+ay-ab=0圆心到切线距离=半径所以|b+a-ab|/√(a^2+b^2)=1(a+b-ab)^2=a^2b^2
手机不行,希望你能看明白吧,主要考察我们对于圆锥曲线跟圆的知识
x²-2xy+y²/x²-y²=(x-y)²/(x-y)(x+y)=(x-y)/(x+y)因为x=3,y=-5,所以(3-(-5))/(3+(-5))
由题意可得,y2=3x−3x22由y2≥0可得3x−3x22≥0解可得,0≤x≤2设t=x2+y2=x2+3x−3x22=−12x2+3x=−12(x2−6x)=−12(x−3)2+92∵0≤x≤2又
1.y1=y2:-x+2=3x+4=>4x=-2=>x=-1/2y1-1/2y1>y2:-x+2>3x+4=>xx=1y=1带入y=ax+7=>a=-6
3x2+2y2-6x=0x2+y2=1/2(6x-x2)=9/2-1/2(x2-6x+9)=9/2-2-1/2(x-3)2当x=3时,Z最大=4.5