(d) Let $P$ $(a\sec \theta ,\,b\tan \theta );\,Q(a\sec \theta ,\, – b\tan \theta )$ be end points of double ordinates

and $C(0,\,0)$, is the centre of the hyperbola.

Now $PQ = 2b\,\tan \theta $

$CQ = CP = \sqrt {{a^2}{{\sec }^2}\theta + {b^2}{{\tan }^2}\theta } $

Since $CQ = CP = PQ$,

$\therefore \,4{b^2}{\tan ^2}\theta = {a^2}{\sec ^2}\theta + {b^2}{\tan ^2}\theta $

==> $3{b^2}{\tan ^2}\theta = {a^2}{\sec ^2}\theta $

==> $3{b^2}{\sin ^2}\theta = {a^2}$

==> $3{a^2}({e^2} – 1){\sin ^2}\theta = {a^2}$

==> $3({e^2} – 1){\sin ^2}\theta = 1$

==> $\frac{1}{{3({e^2} – 1)}} = {\sin ^2}\theta < 1$,

$ \Rightarrow $ $\frac{1}{{{e^2} – 1}} < 3$

$\Rightarrow \,\,{e^2} – 1 > \frac{1}{3}$

$ \Rightarrow \,\,{e^2} > \frac{4}{3}$

$ \Rightarrow \,e > \frac{2}{{\sqrt 3 }}$.