Question

一道关于椭圆的题。。急求

Answer 1

解：椭圆x­­
2
/a
2
+
y
2
/b
2
=
1的右焦点F(c，0)，设直线PF的方程为y
=
k(x
–
c)，（k是非零实数），即kx
–
y
–
ck
=
0，圆(x
–
c/3)
2
+
y
2
=
b
2
/9的圆心M坐标为(c/3，0)，半径为b/3，由点到直线距离公式可得：点M到直线PF的距离等于圆M的半径，即d
=
|ck/3
–
ck|/√[k
2
+
(-1)
2
]
=
(2c/3)|k|/√(k
2
+
1)
=
b/3
=>
(2c)|k|/√(k
2
+
1)
=
b
=>
4c
2
k
2
/(k
2
+
1)
=
b
2
=>
4c
2
k
2
=
b
2
(k
2
+
1)
=>
4c
2
k
2
=
b
2
k
2
+
b
2
=>
(4c
2
–
b
2
)k
2
=
b
2
=>
k
=
±√[b
2
/(4c
2
–
b
2
)]，如图所示，当点P，Q在x轴上方，取正值k
=
√[b
2
/(4c
2
–
b
2
)]
=
b/√(4c
2
–
b
2
)
；
因为点Q是切点，所以QM⊥PF，在RtΔQMF中，因为FM
=
c
–
c/3
=
2c/3，而且QM
=
r
M
=
b/3，所以QF
=
√(FM
2
–
QM
2
)
=
√(4c
2
/9
–
b
2
/9)
=
√(4c
2
–
b
2
)/3，所以PQ
=
2QF
=
2√(4c
2
–
b
2
)/3，所以PF
=
PQ
+
QF
=
√(4c
2
–
b
2
)
=
√(k
2
+
1)*|x
F
–
x
P
|
=
√[b
2
/(4c
2
–
b
2
)
+
1]*(c
–
x
P
)
=
2c(c
–
x
P
)/√(4c
2
–
b
2
)，所以4c
2
–
b
2
=
2c
2
–
2cx
P
=>
2cx
P
=
b
2
–
2c
2
=>
x
P
=
(b
2
–
2c
2
)/2c，而k
=
k
PF
=
(y
P
–
y
F
)/(x
P
–
x
F
)
=
y
P
/(x
P
–
c)，所以y
P
=
k(x
P
–
c)
=
[b/√(4c
2
–
b
2
)]*[(b
2
–
4c
2
)/2c]
=
-b√(4c
2
–
b
2
)/2c，把点P的坐标(x
P
，y
P
)代入椭圆方程可得(b
2
–
2c
2
)
2
/(4a
2
c
2
)
+
b
2
(4c
2
–
b
2
)/(4b
2
c
2
)
=
1
=>
b
2
(b
4
–
4b
2
c
2
+
4c
4
)
+
a
2
b
2
(4c
2
–
b
2
)
=
4a
2
b
2
c
2
=>
(b
6
–
4b
4
c
2
+
4b
2
c
4
)
+
(4a
2
b
2
c
2
–
a
2
b
4
)
=
4a
2
b
2
c
2
=>
b
6
–
4b
4
c
2
+
4b
2
c
4
–
a
2
b
4
=
0
=>
b
4
–
4b
2
c
2
+
4c
4
–
(b
2
+
c
2
)b
2
=
0
=>
4c
4
–
5b
2
c
2
=
0
=>
4c
2
–
5b
2
=
0
=>
b
2
=
(4/5)c
2
=>
a
2
=
b
2
+
c
2
=
(9/5)c
2
=>
a
=
√[(9/5)c
2
]
=
(3√5/5)c，所以椭圆C的离心率e
=
c/a
=
1/(3√5/5)
=
√5/3，答案选
A
。