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Byju's Answer
Standard XIII
Mathematics
Distinguishing between Conics from General Equation and Eccentricity
Let Ax1,y1 an...
Question
Let
A
(
x
1
,
y
1
)
and
B
(
x
2
,
y
2
)
are two fixed points.Then locus of point
P
such that
A
∠
A
P
B
=
90
o
is a circle
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B
Area of
△
A
P
B
is minimum is a parabola
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C
A
P
+
P
B
=
K
where
K
<
A
B
is an ellipse
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D
A
P
P
B
=
K
where
K
<
1
is a circle
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Solution
The correct option is
D
A
P
P
B
=
K
where
K
<
1
is a circle
(
1
)
∠
A
P
B
=
90
o
⇒
(
slope of
A
P
)
⋅
(
slope of
P
B
)
=
−
1
⇒
(
y
−
y
1
x
−
x
1
)
(
y
−
y
2
x
−
x
2
)
=
−
1
⇒
(
x
−
x
1
)
(
x
−
x
2
)
+
(
y
−
y
1
)
(
y
−
y
2
)
=
0
which is a circle
(
2
)
Area of
△
A
P
B
is minimum
⇒
△
A
P
B
=
0
⇒
A
,
P
,
B
are collinear
(
3
)
A
P
+
P
B
=
K
;
if
k
=
A
B
:
Straight line
k
>
A
B
:
an ellipse
k
<
A
B
:
no solution
(
4
)
A
P
P
B
=
K
If
k
=
1
then locus is straight line and if
k
<
1
or
k
>
1
then locus is circle
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0
Similar questions
Q.
Let
A
(
x
1
,
y
1
)
and
B
(
x
2
,
y
2
)
are two fixed points.Then locus of point
P
such that
Q.
Let
A
(
x
1
,
y
1
)
and
B
(
x
2
,
y
2
)
are two fixed points.Then locus of point
P
such that
Q.
The distance formula between two points
A
(
x
1
,
y
1
)
and
B
(
x
2
,
y
2
)
is given by
Q.
If
A
(
x
1
,
y
1
,
z
1
)
and
B
(
x
2
,
y
2
,
z
2
)
are two points such that the direction cosines of
¯
A
B
are
l
,
m
,
n
then
l
=
x
2
−
x
1
∣
∣
¯
A
B
∣
∣
,
m
=
y
2
−
y
1
∣
∣
¯
A
B
∣
∣
,
n
=
z
2
−
z
1
∣
∣
¯
A
B
∣
∣
.
Q.
Determine the equation of family of circles passing through two given points
A
(
x
1
,
y
1
)
and
B
(
x
2
,
y
2
)
. If fixed circle
S
1
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