If there are two tangent lines to the curve y=4x−x2 that pass through point P(2,5), how do you find the x coordinates of point of tangency?
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Explanation:
The equation of a generic straight passing for
p
0
=
{
2
,
5
}
is given by
r
→
y
=
5
+
m
(
x
−
2
)
where
m
is the straight declivity.
The intersection points between
r
and the parabola
p
→
y
=
4
x
−
x
2
are found by solving the condition
5
+
m
(
x
−
2
)
=
4
x
−
x
2
for
x
.
We found
x
=
1
2
(
4
−
m
−
√
m
2
−
4
)
,
y
=
1
2
(
10
−
m
2
−
m
√
m
2
−
4
)
and
x
=
1
2
(
4
−
m
+
√
m
2
−
4
)
,
y
=
1
2
(
10
−
m
2
+
m
√
m
2
−
4
)
So each generic straight with declivity
m
intersects the parabola in two points. By conveniently choosing
m
we can make the two points to be coincident.
So, making
m
=
{
−
2
,
2
}
we have two straights
t
1
→
y
=
5
−
2
(
x
−
2
)
and
t
2
→
y
=
5
+
2
(
x
−
2
)
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