Question

For the ellipse x² + 4y² = 9
(a) the eccentricity is 1/2
(b) the latus-rectum is 3/2
(c) a focus is $\left(3\sqrt{3},0\right)$
(d) a directrix is x = $-2\sqrt{3}$

Answer 1

$$\begin{gathered} \left(\mathrm{b}\right)\mathrm{the}\mathrm{latus}\mathrm{rectum}\mathrm{is}\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \\ {x}^2+4y^2=9 \\ \Rightarrow \frac{x^2}{9}+\frac{y^2}{{\displaystyle \frac{9}{4}}}=1 \\ \mathrm{Here},a>b \\ \therefore e=\sqrt{1-\frac{b^2}{a^2}} \\ \Rightarrow e=\sqrt{1-\frac{9}{{\displaystyle \frac{9}{4}}}} \\ \Rightarrow e=\sqrt{1-\frac{1}{4}} \\ \Rightarrow e=\frac{\sqrt{3}}{2} \\ \text{Latus rectum=}\frac{2b^2}{a} \\ =\frac{2\times {\displaystyle \frac{9}{4}}}{3} \\ =\frac{3}{2} \\ \text{Focus =}\left(\pm ae,0\right) \\ \text{=}\left(\frac{3\sqrt{3}}{2},0\right) \\ \text{Directrix x=±}\frac{a}{e} \\ =\pm \frac{3}{{\displaystyle \frac{\sqrt{3}}{2}}} \\ =2\sqrt{3} \end{gathered}$$