1. If the latus rectum subtends a right angle at the centre of the hyperbola then its
eccentricity
Answers
Answer:Given LR of hyperbola subtends 90
0
at its centre
Let there be a hyperbola :
a
2
x
2
−
b
2
y
2
=1----------------(1)
So, eccentricity , e =
1+
a
2
b
2
Ends of LR, L: (ae,
a
b
2
) &L'(ae,
a
−b
2
)
C : centre(0,0)
∠LCL
′
=90
0
⇒△LCL
′
=right angled triangle
⇒ It must follow pythogores theorem (H
2
=P
2
+B
2
)
ForLC and L
′
Cand LL
′
LL'=length of
a
2b
2
LC
=
(ae−0)
2
+(
a
b
2
−0)
=
a
2
e
2
+
a
2
b
4
=
a
2
a
4
e
2
+b
4
--------------------(2)
LC
=
L
′
C
due to symmetry along x axis--------------(3)
⇒ Applying Pythagoras theorem , (
LL
′
)
2
=(
LC
)
2
(
L
′
C
)
2
⇒
a
2
4b
4
=2(LC)
2
from equation (3)
⇒
a
2
4b
4
=2(a
2
e
2
+
a
2
b
4
) from equation (2)
⇒
a
2
4b
4
−
a
2
2b
4
=2a
2
e
2
⇒
a
2
2b
4
=2a
2
e
2
⇒b
4
=a
4
e
2
⇒(a
2
(e
2
−1))
2
=a
4
e
2
(a
2
(e
2
−1))
⇒a
4
(e
2
−1)
2
=a
4
e
2
⇒(e
2
−1)
2
=e
2
⇒e
2
−1=±e⇒e
2
±e−1=0
e=
2
±1±
1+4
=
2
±1±
5
e
=
2
−1−
5
(negative),
2
1−
5
e=
2
5
−1
,
2
1+
5
.
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