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Q.
The area of the quadrilateral formed by the tangents at the end points of latus rectum to ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{5}=1$, is
Application of Integrals
Solution:
Given equation of ellipse is $\frac{x^{2}}{9}+\frac{y^{2}}{5}=1$.
To find tangents at the end points of latus rectum we find
ae, i.e., ae $=\sqrt{a^{2}-b^{2}}=\sqrt{4}=2$
By symmetry the quadrilateral is rhombus.
So, area of rhombus is four times the area of the right angled triangle formed by the tangent and axes in the first quadrant.
$\Rightarrow \quad$ Equation of tangent at $\left[ae, \sqrt{b^{2}\left(1-e^{2}\right)}\right]=\left(2,\frac{5}{3}\right)$ is
$\frac{2}{9}x+\frac{5}{3}\cdot\frac{y}{5}=1 \Rightarrow \quad2x+3y=9$
$\therefore \quad$ Area of quadrilateral $ABCD = 4$(area of $\Delta AOB)$
$=4\int\limits_{0}^{9 /2}\frac{9-2x}{3} dx=\left|\frac{4}{3}\left[9x-x^{2}\right]_{0}^{9/ 2}\right|$
$=\frac{4}{3}\left[\frac{81}{2}-\frac{81}{4}\right]=27$ sq. units
