If the latus rectum subtends a right angle at the centre of the hyperbola x2a2−y2b2=1, then eccentricity e is
A
√3+12
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B
√5+12
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C
√23
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D
√32
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Solution
The correct option is B√5+12
The coordinates of any end of latus rectum is (ae,b2a)
From figure, we have ae=b2a⇒a2e=a2(e2−1)⇒e=e2−1 ⇒e=√5+12
Alternate Solution:
For the given hyperbola equation of latus rectum will be x=±ae
Homogenizing with equation of hyperbola, we get x2a2−y2b2=x2a2e2 ∵ Pair of lines are perpendicular to each other. ∴ Coeff. of x2+ coeff. of y2=0 ⇒1a2−1a2e2−1b2=0⇒1a2−1a2e2−1a2(e2−1)=0⇒e=±(e2−1)∴e=√5+12(∵e>1)