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JEE Main 5 Apr 2024 Shift 1 Solved Paper
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© examsnet.com
Question : 3
Total: 90
Let two straight lines drawn from the origin
O
intersect the line
3
x
+
4
y
=
12
at the points
P
and
Q
such that
△
OPQ
is an isosceles triangle and
∠
POQ
=
90
∘
. If
l
=
OP
2
+
PQ
2
+
QO
2
, then the greatest integer less than or equal to
l
is :
[5 Apr 2024 Shift 1]
44
48
46
42
Validate
Solution:
👈: Video Solution
3
x
+
4
y
=
12
3
(
r
cos
θ
)
+
4
(
r
s
i
n
θ
)
=
12
r
(
3
cos
θ
+
4
s
i
n
θ
)
=
12
.
.
.
(
1
)
3
(
−
r
s
i
n
θ
)
+
4
(
r
cos
θ
)
=
12
r
(
−
3
s
i
n
θ
+
4
cos
θ
)
=
12
.
.
.
(
2
)
(
12
r
)
2
+
(
12
r
)
2
=
(
3
cos
θ
+
4
s
i
n
θ
)
2
+
(
−
3
s
i
n
θ
+
4
cos
θ
)
2
2
(
12
r
)
2
=
9
+
16
2
×
144
r
2
=
25
⇒
288
=
25
r
2
⇒
288
25
=
r
2
⇒
√
2
(
12
5
)
=
r
ℓ
=
OP
2
+
PQ
2
+
QO
2
ℓ
=
r
2
+
r
2
+
r
2
(
cos
θ
+
s
i
n
θ
)
2
+
r
2
(
s
i
n
θ
+
cos
θ
)
2
=
2
r
2
+
r
2
(
1
+
s
i
n
2
θ
+
1
−
2
s
i
n
2
θ
)
=
2
r
2
+
2
r
2
=
4
r
2
=
4
(
288
25
)
=
1152
25
=
46.08
[
ℓ
]
=
46
© examsnet.com
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