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Mathematics
Mathematics
Problem
If
e
e
e
and
e
′
e'
e
′
are the eccentricities of the hyperbola
x
2
a
2
−
y
2
b
2
=
1
\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1
a
2
x
2
−
b
2
y
2
=
1
and
x
2
b
2
−
y
2
a
2
=
1
\frac{x^2}{b^2} - \frac{y^2}{a^2} = 1
b
2
x
2
−
a
2
y
2
=
1
, then the point
(
1
e
,
1
e
′
)
\left(\frac{1}{e}, \frac{1}{e'}\right)
(
e
1
,
e
′
1
)
lies on the circle a.
x
2
+
y
2
=
1
x^2 + y^2 = 1
x
2
+
y
2
=
1
b.
x
2
+
y
2
=
2
x^2 + y^2 = 2
x
2
+
y
2
=
2
c.
x
2
+
y
2
=
3
x^2 + y^2 = 3
x
2
+
y
2
=
3
d.
x
2
+
y
2
=
4
x^2 + y^2 = 4
x
2
+
y
2
=
4
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