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Coordinates of the vertices B and C of a triangle ABC are ( 2,0) and (8,0) respectively .The vertex A is varying in such a way that 4tan B/2 tan C/2 =1 .Then find the Locus of A

Coordinates of the vertices B and C of a triangle ABC are ( 2,0) and (8,0) respectively .The vertex A is varying in such a way that 4tan B/2 tan C/2 =1 .Then find the Locus of A

Grade:12

1 Answers

Arun
25750 Points
3 years ago
Given that A,B and C are vertices of a 
△
ABC
The coordinate of B and C are (2,0) and (8,0)
∴
 
a
=
8
−
α
=
6
 units
we can assume the coordinate of A to be 
(
x
,
y
)
⇒
b
=
√
(
x
−
2
)
2
+
y
2
,
c
=
√
(
x
−
8
)
2
+
y
2
From half angle formulas 
tan
B
2
=
√
(
s
−
a
)
(
s
−
c
)
s
(
s
−
b
)
 and 
tan
c
2
=
√
(
s
−
a
)
(
s
−
b
)
s
(
s
−
c
)
s
=
a
+
b
+
c
2
 i.e, perimeter.
Given that 
4
tan
B
2
tan
C
2
=
1
⇒
4
√
(
s
−
a
)
(
s
−
c
)
s
(
s
−
b
)
√
(
s
−
a
)
(
s
−
b
)
s
(
s
−
c
)
=
1
Squaring on both side,
16
[
(
s
−
a
)
(
s
−
c
)
s
(
s
−
b
)
]
[
(
s
−
a
)
(
s
−
b
)
s
(
s
−
c
)
]
=
1
=
16
(
s
−
b
)
2
s
2
[
(
s
−
c
)
(
s
−
a
)
(
s
−
a
)
(
s
−
c
)
]
=
1
⇒
16
(
s
−
b
)
2
=
s
2
taking root,
4
(
s
−
b
)
=
s
⇒
 3s-4b=0
Substituting value of
s
=
s
=
a
+
b
+
c
2
3
[
a
+
b
+
c
2
]
−
4
b
=
0
⇒
3
a
−
5
b
+
3
c
=
0
substituting the values of a,b and c
3
×
6
−
5
√
(
x
−
8
)
2
+
y
2
+
3
√
(
x
−
2
)
2
+
y
2
=
0
∴
18
−
5
√
(
x
−
8
)
2
+
y
2
+
3
√
(
x
−
2
)
2
+
y
2
=
0
,
The Locus of A

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