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(2012•蓝山县模拟)设双曲线x2a2-y2b2=1(a>0,b>0)的离心率为54,抛物线y2=20x的准线过双曲线的左焦点,则此双曲线的方程为()A.x24-y23=1B.x23-y24=1C.x216-y29=1D.x29-y216=1

题目详情
(2012•蓝山县模拟)设双曲线
x2
a2
-
y2
b2
=1(a>0,b>0)的离心率为
5
4
,抛物线y2=20x的准线过双曲线的左焦点,则此双曲线的方程为(  )

A.
x2
4
-
y2
3
=1
B.
x2
3
-
y2
4
=1
C.
x2
16
-
y2
9
=1
D.
x2
9
-
y2
16
=1
x2
a2
-
y2
b2
=1(a>0,b>0)的离心率为
5
4
,抛物线y2=20x的准线过双曲线的左焦点,则此双曲线的方程为(  )

A.
x2
4
-
y2
3
=1
B.
x2
3
-
y2
4
=1
C.
x2
16
-
y2
9
=1
D.
x2
9
-
y2
16
=1
x2
a2
x2x2x2x22a2a2a2a22
y2
b2
=1(a>0,b>0)的离心率为
5
4
,抛物线y2=20x的准线过双曲线的左焦点,则此双曲线的方程为(  )

A.
x2
4
-
y2
3
=1
B.
x2
3
-
y2
4
=1
C.
x2
16
-
y2
9
=1
D.
x2
9
-
y2
16
=1
y2
b2
y2y2y2y22b2b2b2b22
5
4
,抛物线y2=20x的准线过双曲线的左焦点,则此双曲线的方程为(  )

A.
x2
4
-
y2
3
=1
B.
x2
3
-
y2
4
=1
C.
x2
16
-
y2
9
=1
D.
x2
9
-
y2
16
=1
5
4
55442

x2
4
-
y2
3
=1
B.
x2
3
-
y2
4
=1
C.
x2
16
-
y2
9
=1
D.
x2
9
-
y2
16
=1
x2
4
x2x2x2x2244
y2
3
=1
B.
x2
3
-
y2
4
=1
C.
x2
16
-
y2
9
=1
D.
x2
9
-
y2
16
=1
y2
3
y2y2y2y2233
x2
3
-
y2
4
=1
C.
x2
16
-
y2
9
=1
D.
x2
9
-
y2
16
=1
x2
3
x2x2x2x2233
y2
4
=1
C.
x2
16
-
y2
9
=1
D.
x2
9
-
y2
16
=1
y2
4
y2y2y2y2244
x2
16
-
y2
9
=1
D.
x2
9
-
y2
16
=1
x2
16
x2x2x2x221616
y2
9
=1
D.
x2
9
-
y2
16
=1
y2
9
y2y2y2y2299
x2
9
-
y2
16
=1
x2
9
x2x2x2x2299
y2
16
=1
y2
16
y2y2y2y221616
▼优质解答
答案和解析
∵双曲线
x2
a2
y2
b2
=1(a>0,b>0)的离心率为
5
4

c
a
5
4
即c=
5
4
a
∵抛物线y2=20x的准线:x=-5过双曲线的左焦点(-c,0),
∴c=5,
∴a=4
而c2=a2+b2=16+b2=25,
∴b2=9,
∴双曲线的方程是
x2
16
y2
9
=1,
故选C.
x2
a2
x2x2x22a2a2a22−
y2
b2
y2y2y22b2b2b22=1(a>0,b>0)的离心率为
5
4

c
a
5
4
即c=
5
4
a
∵抛物线y2=20x的准线:x=-5过双曲线的左焦点(-c,0),
∴c=5,
∴a=4
而c2=a2+b2=16+b2=25,
∴b2=9,
∴双曲线的方程是
x2
16
y2
9
=1,
故选C.
5
4
555444,
c
a
5
4
即c=
5
4
a
∵抛物线y2=20x的准线:x=-5过双曲线的左焦点(-c,0),
∴c=5,
∴a=4
而c2=a2+b2=16+b2=25,
∴b2=9,
∴双曲线的方程是
x2
16
y2
9
=1,
故选C.
c
a
cccaaa=
5
4
555444即c=
5
4
a
∵抛物线y2=20x的准线:x=-5过双曲线的左焦点(-c,0),
∴c=5,
∴a=4
而c2=a2+b2=16+b2=25,
∴b2=9,
∴双曲线的方程是
x2
16
y2
9
=1,
故选C.
5
4
555444a
∵抛物线y22=20x的准线:x=-5过双曲线的左焦点(-c,0),
∴c=5,
∴a=4
而c22=a22+b22=16+b22=25,
∴b22=9,
∴双曲线的方程是
x2
16
y2
9
=1,
故选C.
x2
16
x2x2x22161616−
y2
9
y2y2y22999=1,
故选C.