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AP EAPCET 04-Jul-2022 Shift 1 Paper
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© examsnet.com
Question : 55
Total: 160
The locus of a variable point whose chord of contact w.r.t. the hyperbola
x
2
a
2
−
y
2
b
2
=
1
subtends a right angle at the origin is
x
2
4
a
2
−
y
2
4
b
2
=
1
(
x
2
a
2
−
y
2
b
2
)
=
x
2
a
4
+
y
2
b
4
x
a
−
y
b
=
1
a
2
+
1
b
2
x
2
a
4
+
y
2
b
4
=
1
a
2
−
1
b
2
Validate
Solution:
👈: Video Solution
Equation of hyperbola
x
2
a
2
−
y
2
b
2
=
1
......(i)
Let
(
h
,
k
)
be the point whose chord of contact subtends a right angle at the centre of hyperbola. Equation of chord of contact
h
x
a
2
−
k
y
b
2
=
1
Squaring both the sides, we get
(
h
x
a
2
)
2
+
(
k
y
b
2
)
2
−
2
h
x
a
2
⋅
k
y
b
2
=
(
1
)
2
⇒
h
2
x
2
a
4
+
k
2
y
2
b
4
−
2
k
h
a
2
b
2
⋅
x
y
=
1
.....(ii)
Now, from Eqs. (i) and (ii), we have
h
2
x
2
a
4
+
k
2
y
2
b
4
−
2
h
k
a
2
b
2
⋅
x
y
=
x
2
a
2
−
y
2
b
2
⇒
(
h
2
a
4
−
1
a
2
)
x
2
+
(
k
2
b
4
+
1
b
2
)
y
2
−
2
h
k
a
2
b
2
⋅
(
x
y
)
=
0
The above equation represents a pair of perpendicular lines if
coefficient of
x
2
+
coefficient of
y
2
=
0
h
2
a
4
−
1
a
2
+
k
2
b
4
+
1
b
2
=
0
⇒
h
2
a
4
+
k
2
b
4
=
1
a
2
−
1
b
2
Replacing
h
→
x
and
k
→
y
,
x
2
a
4
+
y
2
b
4
=
1
a
2
−
1
b
2
© examsnet.com
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