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Byju's Answer
Standard XII
Mathematics
Distance Formula
Two points P ...
Question
Two points P and Q are given.R is a variable point on one side of the line PQ such that
∠
R
P
Q
-
∠
R
Q
P
is a positive constant
2
α
. Find the locus of the point R.
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Solution
The diagrammatic representation of the problem is shown below.
In
Δ
R
M
P
,
t
a
n
θ
=
R
M
M
P
=
y
1
a
−
x
1
.
Also, in
Δ
R
Q
M
,
t
a
n
ϕ
=
R
M
Q
M
=
y
1
a
+
x
1
Given
∠
R
P
Q
−
∠
R
Q
P
=
2
α
⇒
θ
−
α
=
2
α
⇒
t
a
n
(
θ
−
α
)
=
t
a
n
(
2
α
)
⇒
t
a
n
θ
−
t
a
n
α
1
+
t
a
n
θ
t
a
n
α
=
t
a
n
(
2
α
)
⇒
t
a
n
(
2
α
)
=
y
1
a
−
x
1
−
y
1
a
+
x
1
1
+
y
1
a
−
x
1
×
y
1
a
+
x
1
⇒
t
a
n
(
2
α
)
=
2
x
1
y
1
a
2
−
x
2
1
+
y
2
1
⇒
a
2
−
x
2
1
+
y
2
1
=
2
x
1
y
1
c
o
t
2
α
⇒
x
2
1
−
y
2
1
+
2
x
1
y
1
c
o
t
2
α
=
a
2
Thus the locus of the point
R
(
x
,
y
)
is given by
x
2
−
y
2
+
2
x
y
c
o
t
2
α
=
a
2
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0
Similar questions
Q.
Given P= (a,0) and Q= (-a,0) and R is a variable point on one side of the line PQ such that
∠
R
P
Q
−
∠
R
Q
P
=
2
α
.
the locus of the point R is
Q.
P
is the point
(
−
1
,
2
)
, a variable line through
P
cuts the co-ordinate axes in
A
and
B
respectively.
Q
is a point on
A
B
such that
P
A
,
P
Q
and
P
B
are in
H
.
P
.
Show that the locus of
Q
is the line
y
=
2
x
.
Q.
P
(
→
p
)
and
Q
(
→
q
)
are the position vectors of two fixed points and
R
(
→
r
)
is the position vector of a variable point. If
R
moves such that
(
→
r
−
→
p
)
×
(
→
r
−
→
q
)
=
→
0
, then the locus of
R
is
Q.
P
and
Q
are two variable points on the axes of
x
and
y
respectively such that
|
O
P
|
+
|
O
Q
|
=
a
, then the locus of foot of perpendicular from origin on
P
Q
is
Q.
P and Q are any two points on the X and Y axes: respectively such that PQ =7. If the point R divides PQ internally in the ratio
4
:
3
, find the equation of locus of R.
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