Ellipse Ques 37
- The area (in sq units) of the quadrilateral formed by the tangents at the end points of the latusrectum to the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{5}=1$ is
(2015 Main)
(a) $\frac{27}{4}$
18
(c) $\frac{27}{2}$
27
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Answer:
Correct Answer: 37.(d)
Solution:
- Given equation of an ellipse is
$ \begin{array}{lc} & \frac{x^{2}}{9}+\frac{y^{2}}{5}=1 \\ \therefore & a^{2}=9, b^{2}=5 \Rightarrow a=3, b=\sqrt{5} \\ Now, & e=\sqrt{1-\frac{b^{2}}{a^{2}}}=\sqrt{1-\frac{5}{9}}=\frac{2}{3} \end{array} $
Foci $=( \pm a e, 0)=( \pm 2,0)$ and $\frac{b^{2}}{a^{2}}=\frac{5}{9}$
$\therefore$ Extremities of one of the latus rectum are
$ (2, \frac{5}{3}) \text { and } (2, \frac{-5}{3}) $
$\therefore$ Equation of tangent at $(2, \frac{5}{3})$ is
$ \frac{x(2)}{9}+\frac{y(5 / 3)}{5}=1 \quad \text { or } \quad 2 x+3 y=9 $
Since, Eq. (ii) intersects $X$ and $Y$-axes at $(\frac{9}{2}, 0)$ and $(0,3)$, respectively.
$\therefore$ Area of quadrilateral $=4 \times$ Area of $\triangle P O Q$
$ =4 \times (\frac{1}{2} \times \frac{9}{2} \times 3)=54 \text { sq units } $
BETA
