$x^{2}+y^{2}=\frac{9}{4} \quad y=4 x$

Equation tangent in slope form

$y=m x \pm \frac{3}{2} \sqrt{\left(1+m^{2}\right)}$      …..$(1)$

$y=m x+\frac{1}{m}$      …..$(2)$

compare $(1)$ and $(2)$

$\pm \frac{3}{2} \sqrt{\left(1+m^{2}\right)}=\frac{1}{m^{2}}$

$9 m^{2}\left(1+m^{2}\right)=4$

$9 m^{4}+9 m^{2}-4=0$

$9 m^{4}+12 m^{2}-3 m^{2}-4=0$

$3 m^{2}\left(3 m^{2}+4\right)-\left(3 m^{2}+4\right)=0$

$m^{2}=-\frac{4}{3}(\text { Rejected })$

$m^{2}=\frac{1}{3} \Rightarrow m=\pm \frac{1}{\sqrt{3}}$

Equation of common tangent

$y =\frac{1}{\sqrt{3}} x +\sqrt{3}$

$\text { on } X \text { axis } y =0$

$OQ =-3$

$b =\mid OQ\mid =3$

$a =6$

$b ^{2}= a ^{2}\left(1- e ^{2}\right) \Rightarrow e ^{2}=1-\frac{9}{36}=\frac{3}{4} ~\\ e =\frac{2 b ^{2}}{ a }=\frac{2 \times 9}{6}=3 ~\\ \frac{ e }{ e ^{2}}=\frac{3}{3 / 4}=4$

$\frac{ e }{ e ^{2}}=\frac{3}{3 / 4}=4$