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BITSAT 2017 Slot 2 Solved Paper
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© examsnet.com
Question : 108
Total: 150
If
p
and
p
′
denote the lengths of the perpendicular from a focus and the centre of an ellipse with semi-major axis of length
a
, respectively, on a tangent to the ellipse and
r
denotes the focal distance of the point, then
a
p
=
r
p
′
r
p
=
a
p
′
a
p
=
r
p
′
+
1
a
p
′
+
r
p
=
1
Validate
Solution:
Let the ellipse be
x
2
a
2
+
y
2
b
2
=
1
and let
x
cos
θ
a
+
y
sin
θ
b
=
1
......(i)
be a tangent to it at point
(
a
cos
θ
,
b
sin
θ
)
. Then,
p
=
length of the perpendicular from
S
(
a
e
,
0
)
on Eq. (i)
⇒
p
=
|
e
cos
θ
−
1
√
cos
2
θ
a
2
+
sin
2
θ
b
2
|
p
′
=
length of the perpendicular from
O
(
0
,
0
)
on Eq. (i)
⇒
p
′
=
|
1
√
cos
2
θ
a
2
+
sin
2
θ
b
2
|
and
r
=
a
e
cos
θ
−
a
Clearly,
r
p
′
=
a
p
© examsnet.com
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