Question

Find the eccentricity, coordinates of foci, length of the latus-rectum of the ellipse:
 4x² + 9y² = 1

Answer 1

\[4 x^2 + 9 y^2 = 1\]
\[ \Rightarrow \frac{x^2}{\frac{1}{4}} + \frac{y^2}{\frac{1}{9}} = 1\]
\[\text{ This is of the form }  \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \text{ where a} ^2 = \frac{1}{4} \text{ and } b^2 = \frac{1}{9}, i . e . a = \frac{1}{2}\text{ and }  b = \frac{1}{3} . \]
\[\text{ Clearly a } > b\]
\[\text{ Now, } e = \sqrt{1 - \frac{b^2}{a^2}}\]
\[ \Rightarrow e = \sqrt{1 - \frac{\frac{1}{9}}{\frac{1}{4}}}\]
\[ \Rightarrow e = \sqrt{1 - \frac{4}{9}}\]
\[ \Rightarrow e = \frac{\sqrt{5}}{3}\]
\[\text{ Coordinates of the foci } = \left( \pm ae, 0 \right) = \left( \pm \frac{\sqrt{5}}{6}, o \right)\]
\[\text{ Length of the latus rectum} =\frac{2 b^2}{a}\]
\[ = \frac{2 \times \frac{1}{9}}{\frac{1}{2}}\]
\[ = \frac{4}{9}\]