Question

The area (in sq.units) of the quadrilateral formed by the tangents at the end point of the latus rectum to the ellipse $\frac{x^2}{9}+\frac{y^2}{5}=1$, is

• A. $\frac{27}{2}$
• B. $27$
• C. $\frac{27}{4}$
• D. $18$

Answer 1

The correct option is B $27$

We have to find the area of quadrilateral formed by the tangents at the end points of the latus rectum of the ellipse $\frac{x^2}{9}+\frac{y^2}{5}=1$

$\therefore$ $a^2=9$ and $b^2=5$

$\Rightarrow$ Ecentricity $\left(e\right)=\sqrt{1-\frac{b^2}{a^2}}$

$=\sqrt{1-\frac{5}{9}}$

$=\sqrt{\frac{4}{9}}$

$=\frac{2}{3}$

$\Rightarrow$ Focii $=(\pm ae,0)-(\pm 3\times \frac{2}{3},0)=(\pm 2,0)$

$\Rightarrow$ Ends of latus rectum $=(\pm ae,\pm \frac{b^2}{a})=(\pm 2,\pm \frac{5}{3})$

Let the tangent at point $(2,\frac{5}{3})$ $=\frac{2\times x}{9}+\frac{5}{3}\times \frac{y}{5}=\frac{2x}{9}+\frac{y}{3}$

It cuts the coo-ordinates axes at $P(\frac{9}{2},0)$ and $Q(0,3)$

Now, by symmetry area of quadrilateral $ABC=4\times$ Area of $△POQ$

$=4\times \frac{1}{2}\times \frac{9}{2}\times 3$

$=27sq.units$