JEE Main · 2015 · 04 Apr 2015 · Offline · Ellipse

For $\frac{x^2}{9}+\frac{y^2}{5}=1$ we have $a=3$ and $b^2=5$, so $e=\sqrt{1-\frac{5}{9}}=\frac{2}{3}$ and the latera recta stand at $x=\pm ae=\pm 2$ with half-length $\frac{b^2}{a}=\frac{5}{3}$. The tangent at the end $(2,\frac{5}{3})$ is $\frac{2x}{9}+\frac{y}{3}=1$, cutting the axes at $(\frac{9}{2},0)$ and $(0,3)$. By symmetry the four such tangents frame a rhombus with diagonals $2\times \frac{9}{2}=9$ and $2\times 3=6$, whose area is $\frac{1}{2}\times 9\times 6=27$.