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 Some important formulae to be kept in mind before solving the question:

The Standard equation of ellipse is given by :  $\frac{x^{2}}{a^{2}}$  $+$   $\frac{y^{2}}{b^{2}}$   = 1

The distance of the focus of the ellipse from the centre   $\left ( 0,0 \right )$  can be found using $c=\pm \sqrt{a^{2}-b^{2}}$

The length of the latus rectum(the chord passing through the focus perpendicularly) of the ellipse is :  $2\times \frac{b^{2}}{a}$

Now, coming back to the question, let the intesection points of the tangent with the $x$-$axis$ and $y$-$axis$ in the first quadrant

be $\left ( d,0 \right )$  and $\left ( 0,c \right )$  respectively. 

Also the co-ordinates of the focus in the first quadrant would be  $c=\pm \sqrt{9-5}$ = $\pm 2$  i.e. $\left ( 2,0 \right )$

The ends of latus rectum in the first quadrant will be  $\left ( 2,\frac{5}{3} \right )$ . So, now the equation of tangent to the ellipse at point  $\left ( x_{1},y_{1} \right )$ is given by  :

$\left ( b^{2}x_{1} \right )x$  +   $\left ( a^{2}y_{1} \right )y$  $=$  $a^{2}b^{2}$  (Source: Tangent to ellipse)

Using the above equation, the equation of tangent would be:  $\left ( 5\times 2 \right )x$  $+$   $\left ( 9\times \frac{5}{3} \right ) y$  = $45$  i.e.   $2x+3y=9$

The points   $\left ( d,0 \right )$  and $\left ( 0,c \right )$ satisfy the above equation.

So,  $2\times d=9 \Rightarrow d=\frac{9}{2}$   and  $3\times c=9\Rightarrow c=3$.

Now that we have got the end points of the tangents in the first quadrant, we know that area of the whole quadrilateral would be :

$4\times$  area of the triangle in the first quadrant (by symmetry)

Area of the triangle formed by the tangent with $x$-$axis$ and $y$-$axis$ in the first quadrant is: $\frac{1}{2}\times \frac{9}{2}\times3$ = $\frac{27}{4}$.

$\therefore$  Area of the quadrilateral is : $4\times \frac{27}{4}=27$  $sq.$  $units$

Option (A) is the correct answer.

 

 

 

 

 

 

 

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