已知方程x² 16m-y² 9m=1表示双曲线,并且焦距为10-搜答案-智慧作业帮

x²/25-m+y²/m+9=1表示焦点在y轴上的椭圆所以m+9>25-m>08

(m+3)^2+(1-4m^2)^2-(16m^4+9)1或者m

解x=3=m24-3（y+3）=7(y+3)+424-3y-9=7y+21+410y=-10y=-1

x2+y2-2（m+3）x+2(1-4m2)y+16m2+9=0所以(x-m-3)^2+(y+1-4m^2)^2=-7m^2+6m+1因为-7m^2+6m+1>0所以m的范围为（-1/7,1)请采纳回

若是x^2+y^2+2(m+3)x+2(1-4m^2)y+16m^4+9=0[x-(m+3)]^2+[y-(1-4m^2)]^2=m^2+6m+1思路R^2=m^2+6m+1>0m^2+6m+1>0m

1)配方：（x-m-3)^2+(y-1+4m^2)^2=(m+3)^2+(1-4m^2)^2-16m^4-9(x-m-3)^2+(y-1+4m^2)^2=-7m^2+6m+1(x-m-3)^2+(y-

X^2+Y^2-2(m+3)X+2(1-4m^2)Y+16m^4+9=0(X-(m+3))^2+(Y+1-4m^2)^2=(m+3)^2+(1-4m^2)^2-9-16m^4等式左边可以化为左=(m+

圆心显然是X=M+3Y=1-4m^2所以M=X-3代入Y=1-4m^2不就得结果了再注意原方程化为原的标准形式时有定义域就出来了

楼主计算可能有误方程可化为[x-(m+3)]^2+[y-(1-4m^2)]^2=-(7m+1)(m-1)当且仅当-1/7

x²+y²-2m-6+2-8m²+（4m²）²+9=0x²+y²-2m-8m²+（4m²）²+5=0

提示：（1）r=√(D²+E²-4F)/2,求r的最大值即可（2）P点代入方程后左端

由X-Y=9M得X=9M+Y又∵X-Y-(X+2Y)=9M-3M∴Y=-2M∴X=7M∵5X+8Y=38∴35M-16M=38∴M=2

[x-(m+3)]^2+[y+(1-4m^2)]^2=(m+3)^2+(1-4m^2)^2-9-16m^4=m^2+2m+1-8m^2=1+2m-7m^2圆心：x=m+3,y=4m^2-1,m=x-3

(1)原式=(2x+y+4)+m(x-2y-3)=0令2x+y+4=0x-2y-3=0得：x=-1y=-2即该直线一定过点（-1,-2)(2)设该直线方程为y+2=k(x+1)（k

X^2+Y^2-2(m+3)X+2(1-4m^2)Y+16m^4+9=0(X-(m+3))^2+(Y+1-4m^2)^2=(m+3)^2+(1-4m^2)^2-9-16m^4等式左边可以化为左=(m+

证明：x²-(m+3)x+(3m-2)=0判别式△=(m+3)²-4(3m-2)=m²+6m+9-12m+8=m²-6m+17=(m-3)²+8>=0

-1/7

(1)x^2+y^2-2(m+3)x+2(1-4m^2)y+16m^4+9=0[x-(m+3)]^2-(m+3)^2+[y+(1-4m^2)]^2-(1-4m^2)^2+16m^4+9=0[x-(m+

p:双曲线,则系数为一正一负,故有(1-2m)(m+2)1/2或m

由3x-5y=2m,2x+7y=3m+34得x=(29m+170)/31,y+(5m+102)/31由x+y