Area Ques 12
- The area of the quadrilateral formed by the tangents at the end points of latusrectum to the ellipse $\frac{x^2}{9}+\frac{y^2}{5}=1$, is
(2003, 1M)
(a) $27 / 4$ sq units
(b) $9$ sq units
(c) $27 / 2$ sq units
(d) $27$ sq units
Show Answer
Answer:
Correct Answer: 12.(d)
Solution: (d) Given, $\frac{x^2}{9}+\frac{y^2}{5}=1$
To find tangents at the end points of latusrectum, we find ae.
i.e. $\quad a e=\sqrt{a^2-b^2}=\sqrt{4}=2$
and $\sqrt{b^2\left(1-e^2\right)}=\sqrt{5\left(1-\frac{4}{9}\right)}=\frac{5}{3}$
By symmetry, the quadrilateral is a rhombus.
So, area is four times the area of the right angled triangle formed by the tangent and axes in the $Ist$ quadrant.
$\therefore \quad $ Equation of tangent at $\left(2, \frac{5}{3}\right)$ is
$ \frac{2}{9} x+\frac{5}{3} \cdot \frac{y}{5}=1 \Rightarrow \frac{x}{9 / 2}+\frac{y}{3}=1 $
$\therefore \quad $ Area of quadrilateral $A B C D$
$=4 \quad $ [area of $\triangle A O B$]
$ =4\left(\frac{1}{2} \cdot \frac{9}{2} \cdot 3\right)=27 $ $\mathrm{sq} \text { units }$
BETA
