question_answer

The area of the quadrilateral formed by the tangents at the end points of latusrectum to the ellipse \[\frac{{{x}^{2}}}{9}+\frac{{{y}^{2}}}{5}=1\] is

A)  \[\frac{27}{4}\text{sq}\,\text{unit}\]            

B)  \[\text{9}\,\text{sq}\,\text{unit}\]

C)  \[\frac{27}{2}\text{sq}\,\text{unit}\]            

D)  \[\text{27}\,\text{sq}\,\text{unit}\]

Correct Answer: D

Solution :

Given equation of ellipse is \[\frac{{{x}^{2}}}{9}+\frac{{{y}^{2}}}{5}=1\] \[\therefore \]    \[ae=\sqrt{{{a}^{2}}-{{b}^{2}}}\]                         \[=\sqrt{9-5}=2\] By symmetry, the quadrilateral is a rhombus. So, total area is four times the area of the right angled triangle formed by the tangent and axis in the 1st quadrant. \[\Rightarrow \]Equation of tangent at \[\left( 2,\frac{5}{3} \right)\] is\[\frac{2}{9}x+\frac{5}{3}.\frac{y}{5}=1\] \[\Rightarrow \]\[\frac{x}{9/2}+\frac{y}{3}=1\] \[\therefore \]Area of quadrilateral ABCD= 4(area of\[\Delta AOB\]) \[=4\times \frac{1}{2}\times \frac{9}{2}\times 3\]=27squnit