求与椭圆x^2/49+y^2/24=1有公共点,且离心率e=5/4的双曲线

就是有交点
若双曲线x^2/a^2-y^2/b^2=1
c/a=5/4
(a^2+b^2)/a^2=25/16
a^2=16b^2/9
9x^2/16b^2-y^2/b^2=1
x^2=16(b^2+y^2)/9
代入椭圆
16(b^2+y^2)/(9*49)+y^2/24=1
整理825y^2=10584-384b^2≥0
所以b^2≤441/16
a^2=16b^2/9≤49
若取等号
则x^2/49-16y^2/441=1
或者a^2≤49，且b^2=9a^2/16亦可

若双曲线是y^2/a^2-x^2/b^2=1
c/a=5/4
(a^2+b^2)/a^2=25/16
a^2=16b^2/9
9y^2/16b^2-x^2/b^2=1
y^2=16(x^2+b^2)/9
代入椭圆
x^2/49+2(x^2+b^2)/27=1
x^2=(1323-98b^2)/125≥0
b^2≤27/2
a^2=16b^2/9≤24
若取等号
y^2/24-2x^2/27=1
或者a^2≤24，且b^2=9a^2/16亦可