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The area (in sq. units) of the quadrilateral formed by the tangents at the end points of the latera recta to the ellipse\[\frac{{{x}^{2}}}{9}+\frac{{{y}^{2}}}{5}=1,\]is: [JEE Main Solved Paper-2015 ]

A)  \[\frac{27}{2}\]                               

B)  27

C) \[\frac{27}{4}\]                                

D) 18

Correct Answer: B

Solution :

Ellipse\[\frac{{{x}^{2}}}{9}+\frac{{{y}^{2}}}{5}=1\] \[e=\sqrt{1-\frac{{{b}^{2}}}{{{a}^{2}}}}\]                               \[=\sqrt{1-\frac{5}{9}}\] \[=\frac{2}{3}\]                                 focus : \[\left( \pm ae,o \right)\]              \[=\left( \pm 2,0 \right)\] Latus rectum co-ordinates : \[=\left( \pm ae,\pm \frac{{{b}^{2}}}{a} \right)\] \[\Rightarrow \left( \pm 2,\pm \frac{5}{3} \right)\] Tangent at L : \[\frac{x.2}{9}+\frac{y.\frac{5}{3}}{5}=1\] \[Q:\left( \frac{9}{2},0 \right)\] \[P:(0,3)\] Area of quadrilateral : \[4(\Delta OPQ)\] \[=4.\frac{1}{2}.\frac{9}{2}.3\]                                   = 27