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JEE Mains 28-July-2022 Shift 1 Solved Paper
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© examsnet.com
Question : 7
Total: 90
For
t
∈
(
0
,
2
π
)
, if
ABC
is an equilateral triangle with vertices
A
(
sin
t
,
−
cos
t
)
,
B
(
cost
,
sin
t
)
and
C
(
a
,
b
)
such that its orthocentre lies on a circle with centre
(
1
,
1
3
)
, then
(
a
2
−
b
2
)
is equal to:
[28-Jul-2022-Shift-1]
8
3
8
77
9
80
9
Validate
Solution:
Let
P
(
h
,
k
)
be the orthocentre of
△
ABC
Then
h
=
s
i
n
t
+
cos
t
+
a
3
,
k
=
−
cos
t
+
s
i
n
t
+
b
3
(orthocentre coincide with centroid)
∴
(
3
h
−
a
)
2
+
(
3
k
−
b
)
2
=
2
∴
(
h
−
a
3
)
2
+
(
k
−
b
3
)
2
=
2
9
∵
orthocentre lies on circle with centre
(
1
,
1
3
)
∴
a
=
3
,
b
=
1
∴
a
2
−
b
2
=
8
© examsnet.com
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