The area (in sq. units) of the quadrilateral formed by the tangents at the end points of the latera recta to the ellipse `` `(x^2)/9+(y^2)/5=1` , is: (1) `(27)/4` (2) 18 (3) `(27)/2` (4) 27

Options

A. `27/4`
B. 9
C. `27/2`
D. 27

Answer

D

Explanation

We have, `x^(2)/9 + y^(2)/5 = 1`
Let e be the eccentricity of this ellipse. Then,
`e^(2) = 1 - 5/9 rArr e = 2/3`
The coordinates of the end - points of latusrecta are
`L(2,5//3), M(-2, 5//3), M'(-2, -5//3) and L'(2, -5//3)`
The equations of tangents at these points are
`2x + 3y - 9 = 0" "...(i)`
`-2x + 3y - 9 = 0 " "...(ii)`
2x + 3y + 9 = 0 " "...(iii)`
-2x + 3y + 9 = 0" "...(iv)`
Clearly, these tagents form a parallelogram whose area is given by
`A=|([9-(-9)]xx{9(-9)})/(|{:(2,3),(-2",",3):}|)|=(18xx18)/(12)` Sq. units
`rArr A = 27` sq. units.