Let the common tangents to the curves 4( x ^ 2 + y ^ 2 ) = 9 and y...

Given curves are ${x}^{2} + {y}^{2} = \dfrac{9}{4} \text{ and } {y}^{2} = 4x$

Equation of tangent in slope form to the circle and parabola will be

$y = mx \pm \dfrac{3}{2}\sqrt{\left(1 + {m}^{2}\right)}$ \ldots $\left(i\right)$

$y = mx + \dfrac{1}{m} ⋯ \left(ii\right)$

Comparing \left(i\right)& \left(ii\right), we get

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

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

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

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

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

here ${m}^{2} = - \dfrac{4}{3}$ (Rejected)

so ${m}^{2} = \dfrac{1}{3} ⇒ m = \pm \dfrac{1}{\sqrt{3}}$

So, the equation of common tangent will be

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

When $y = 0$

$b = \left|OQ\right| = 3$

$a = 6$

i.e. ${b}^{2} = {a}^{2}\left(1 - {e}^{2}\right) ⇒ {e}^{2} = 1 - \dfrac{9}{36} = \dfrac{3}{4}$

Also $l = \dfrac{2{b}^{2}}{a} = \dfrac{2 \times 9}{6} = 3$

$\dfrac{l}{{e}^{2}} = \dfrac{3}{\dfrac{3}{4}} = 4$