Question

The area (in s units) of the quadrilateral formed by the tangents at the end points of the latera recta to the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{5}=1$, is:

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

Answer 1

Answer: (D)

27

Answer 2

$\frac{x^{2}}{9}+\frac{y^{2}}{5}=1$
$a=3$
$b=\sqrt{5}$
$e^{2}=1-\frac{b^{2}}{a^{2}}$
$=1-\frac{5}{9}=\frac{4}{9}$
$e=\frac{2}{3}$
now the quadrilateral formed will be a rhombu with area $=\frac{2 a^{2}}{e}$
$=\frac{2.9}{2} \times 3$
$=27$