VITEEE 2013 Paper

Let P (a sec θ, b tan θ), Q (a sec θ, - b tan θ) be end points of double ordinates and (0, 0) is the centre of the hyperbola.
So, PQ = 2 b tan θ

OQ = OP = √a2sec2θ+b2tan2θ
Since, OQ = OP = PQ
∴ 4b2tan2θ = a2sec2θ+b2tan2θ
⇒ 3b2tan2θ = a2sec2θ
⇒ 3b2sin2θ = a2
⇒ 3a2(e2−1)sin2θ = a2
⇒ 3 (e2 - 1) sin2θ = 1
⇒

1

3(e2−1)

= sin2θ < 1 , (Since sin2θ < 1)
⇒

1

e2−1

< 3 ⇒ e2 - 1 >

1

3

⇒ e2 >

4

3

∴ e >

2

√3