The area (in sq. units) of the quadrilateral formed by the tangents...

$\dfrac{{x}^{2}}{9} + \dfrac{{y}^{2}}{5} = 1$

$e = \sqrt{1 - \dfrac{5}{9}} = \dfrac{2}{3}$

${a}^{2} = 9,{b}^{2} = 5$

Foci $= \left(\pm ae , 0\right) = \left(\pm 2, 0\right)$

Ends of latus rectum$= \left(\pm ae , \pm \dfrac{{b}^{2}}{a}\right)$

$= \left(\pm 2, \pm \dfrac{5}{3}\right)$

Tangent at '$L$' is $T = 0$

$\dfrac{2 \times x}{9} + \dfrac{5}{3} \times \dfrac{y}{5} = 1$

It cuts coordinate axes at $P \left(\dfrac{9}{2}, 0\right) \text{ and } Q \left(0, 3\right)$

Area of quadrilateral $PQRS = 4$ (Area of triangle $OPQ$)

$= 4\left(\dfrac{1}{2} \times \dfrac{9}{2} \times 3\right) = 27$ sq. units