Question

Answer it with explanation.
Please solve it .

Answer 1

HELLO DEAR



WE HAVE TO FIND THE AREA OF THE QUADRILATERAL FORMED BY THE TANGENT AT THE END POINTS OF THE LATUS RACTA OF THE ELLIPSE
$\frac{ {x}^{2} }{9} + \frac{ {y}^{2} }{5} = 1 \\ = > \frac{ {x}^{2} }{ {3}^{2} } + \frac{ {y}^{2} }{ { (\sqrt{5} })^{2} } \\ = > so \: \: {a}^{2} = 9 \\ = > {b}^{2} = 5 \\ = > ecentricity \: \: e = \sqrt{1 - \frac{5}{9} } = \sqrt{ \frac{9 - 5}{9} } = \frac{2}{3} \\ = > focii = ( |ae| \: \: \: 0) = ( |3 \times \frac{2}{3} | \: \: \: 0) \\ = > = ( |2| \: 0) \\ = >$
ENDS OF LATUS RECTUM =(+-ae,+-b²/a)
=>(+-2,+-5/3)



LET THE TANGENT AT THE POINT (2,5/3)


$\frac{2 \times x}{9} + \frac{5}{3} \times \frac{y}{5} = \frac{2x}{9} + \frac{y}{3}$
IT CUTS THE COORDINATE AXES AT
p(9/2,0) &Q(0,3)


NOW BY SYMMETRY AREA OF QUADRILATERAL ABCD=4

AREA IF TRIANGLE( POQ)

$4( \frac{1}{2} \times \frac{9}{2} \times 3) \\ = > 27 {unit}^{2}$
I HOPE ITS HELP YOU DEAR,
THANKS