[tex]\huge\mathcal{\fcolorbox{pink} {black} {\pink{Question}}}[/tex]
The locus of the mid point of the focal radii of a variable point moving on the parabola, y² = 4ax
is a parabola whose
(A) Latus rectum is half the latus rectum of the original parabola
(B) Vertex is (a/2,0)
(C) Directrix is y-axis
(D) Focus has the co-ordinates (a,0)
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Step-by-step explanation:
The correct options are
A Latus rectum is half the latus rectum of the original parabola
B Vertex is
(
a
2
,
0
)
C Directrix is
y
−
axis
D Focus has the co-ordinates
(
a
,
0
)
Any point on the parabola is
P
(
a
t
2
,
2
a
t
)
Therefore, the midpoint of
S
(
a
,
0
)
and
P
(
a
t
2
,
2
a
t
)
is,
R
(
a
+
a
t
2
2
,
a
t
)
≡
(
h
,
k
)
⇒
h
=
a
+
a
t
2
2
,
k
=
a
t
Eliminating
t
,
2
x
=
a
(
1
+
y
2
a
2
)
=
a
+
y
2
a
⇒
2
a
x
=
a
2
+
y
2
⇒
y
2
=
2
a
(
x
−
a
2
)
It is a parabola with vertex at
(
a
2
,
0
)
and latus rectum 2a.
The directrix is
x
−
a
2
=
−
a
2
⇒
x
=
0
The focus is
x
−
a
2
=
a
2
⇒
x
=
a
Therefore the focus is
(
a
,
0
)
flag
Suggest CorrectionsThe correct options are
A Latus rectum is half the latus rectum of the original parabola
B Vertex is
(
a
2
,
0
)
C Directrix is
y
−
axis
D Focus has the co-ordinates
(
a
,
0
)
Any point on the parabola is
P
(
a
t
2
,
2
a
t
)
Therefore, the midpoint of
S
(
a
,
0
)
and
P
(
a
t
2
,
2
a
t
)
is,
R
(
a
+
a
t
2
2
,
a
t
)
≡
(
h
,
k
)
⇒
h
=
a
+
a
t
2
2
,
k
=
a
t
Eliminating
t
,
2
x
=
a
(
1
+
y
2
a
2
)
=
a
+
y
2
a
⇒
2
a
x
=
a
2
+
y
2
⇒
y
2
=
2
a
(
x
−
a
2
)
It is a parabola with vertex at
(
a
2
,
0
)
and latus rectum 2a.
The directrix is
x
−
a
2
=
−
a
2
⇒
x
=
0
The focus is
x
−
a
2
=
a
2
⇒
x
=
a
Therefore the focus is
(
a
,
0
)
flag