• # question_answer The area of the quadrilateral formed by the tangents at the end points of latus rectum to the ellipse $\frac{{{x}^{2}}}{9}+\frac{{{y}^{2}}}{5}=1$, is [IIT Screening 2003] A)            27/4 sq. unit                              B)            9 sq. unit C)            27/2 sq. unit                              D)            27 sq. unit

By symmetry the quadrilateral is a rhombus. So area is four times the area of the right angled triangle formed by the tangent and axes in the Ist quadrant. Now, $ae=\sqrt{{{a}^{2}}-{{b}^{2}}}\Rightarrow ae=2$            Þ Tangent (in first quadrant) at end of latus rectum $\left( 2,\frac{5}{3} \right)$ is $\frac{2}{9}x+\frac{5}{3}\frac{y}{5}=1$ i.e., $\frac{x}{9/2}+\frac{y}{3}=1$                    Area $=4.\,\frac{1}{2}.\,\frac{9}{2}.3=27$ sq. unit.