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Mathematics
Mathematics
Problem
Find the point(s) on
x
2
4
+
y
2
36
=
1
\frac{x^{2}}{4}+\frac{y^{2}}{36}=1
4
x
2
+
36
y
2
=
1
that are closest to
(
0
,
1
)
(0,1)
(
0
,
1
)
.
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