Question

Area of the quadrilateral formed by tangents at the ends of latus rectum of an ellipse $\frac{x^2}{9}+\frac{y^2}{5}=1$ is

• A. $18$ sq. units
• B. $27$ sq. units
• C. $\frac{9}{2}$ sq. units
• D. $\frac{21}{4}$ sq. units

Answer 1

Answer: (B)

$27$ sq. units

Here $a^2=9,b^2=5,b^2=a^2(1-e^2)$
$\Rightarrow e^2=\frac{4}{9}\Rightarrow e=\frac{2}{3}$

Foci $=(\pm ae,0)=(\pm 2,0)$

Ends of latus rectum $=(\pm ae,\pm \frac{b^2}{a})=(\pm 2,\pm \frac{5}{3})$
The equation of tangent at one end point of latus rectum $(2,\frac{5}{3})$ is
$2\times \frac{x}{9}+\frac{5}{3}\times \frac{y}{5}=1$

It cuts the co-ordinate axes at $(\frac{9}{2},0)$ and $(0,3)$
$x$ intercept $=\frac{9}{2}$, $y$ intercept $=3$

Area formed by co-ordinate axis and the tangent is same as the area formed by right angled triangle with base $\frac{9}{2}$ and height $3$

$\therefore$ Area formed by co-ordinate axis and the tangent is

$=\frac{1}{2}\times \frac{9}{2}\times 3=\frac{27}{4}$

Now, Area of quadrilateral formed by the tangents drawn at the end points of the latum rectums $=4\times \frac{27}{4}=27$ sq. units.