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VITEEE 2010 Solved Paper
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© examsnet.com
Question : 96
Total: 120
A rod of length
l
slides with its ends on two perpendicular lines. Then, the locus of its mid point is
x
2
+
y
2
=
l
2
4
x
2
+
y
2
=
l
2
2
x
2
−
y
2
=
l
2
4
None of these
Validate
Solution:
Let both of the ends of the rod are on x - axis and y - axis. Let AB be rod of length
l
and coordinates of A and B be (a,0) and (0,b) respectively.
Let P(h,k) be the mid point of the rod AB.
Then,
[
h
=
0
+
a
2
=
a
2
k
=
b
+
0
2
=
b
2
]
.
.
.
(
i
)
In Δ OAB,
OA
2
+
OB
2
=
AB
2
⇒
a
2
+
b
2
=
l
2
⇒
(
2
h
)
2
+
(
2
k
)
2
=
l
2
[
using
eq
.
(
i
)
]
⇒
h
2
+
k
2
=
l
2
4
∴ The equation of locus is
x
2
+
y
2
=
l
2
4
© examsnet.com
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