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Byju's Answer
Standard XII
Mathematics
General Equation of Conics
The locus of ...
Question
The locus of the point of intersection of the lines
(
√
3
)
k
x
+
k
y
−
4
√
3
=
0
and
√
3
x
–
y
–
4
(
√
3
)
k
=
0
is a conic, whose eccentricity is
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Solution
√
3
k
x
+
k
y
=
4
√
3
.
.
.
.
.
.
(
1
)
√
3
k
x
−
k
y
=
4
√
3
k
2
.
.
.
.
.
(
2
)
Adding equation
(
1
)
and
(
2
)
2
√
3
k
x
=
4
√
3
(
k
2
+
1
)
x
=
2
(
k
+
1
k
)
.
.
.
.
(
3
)
Substracting equation
(
1
)
and
(
2
)
y
=
2
√
3
(
1
k
−
k
)
.
.
.
.
(
4
)
By
(
3
)
and
(
4
)
∴
x
2
4
−
y
2
12
=
4
x
2
16
−
y
2
48
=
1
Hyperbola
∴
e
2
=
1
+
48
16
e
=
2
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0
Similar questions
Q.
The locus of the point of intersection of the lines
(
√
3
)
k
x
+
k
y
−
4
√
3
=
0
and
√
3
x
–
y
–
4
(
√
3
)
k
=
0
is a conic, whose eccentricity is
Q.
The locus of the point of intersection of the lines
√
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x
−
y
−
4
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3
k
=
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and
√
3
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x
+
k
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−
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√
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=
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for different values of
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is a
Q.
The locus of the point of intersection of the lines
x
√
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−
y
−
4
√
3
k
=
0
and
k
x
√
3
+
k
y
−
4
√
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Q.
The locus of the point of intersection of the lines
√
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x
−
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√
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t
=
0
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√
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t
x
+
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y
−
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√
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Q.
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√
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k
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y
−
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√
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