Question

Find the eccentricity, coordinates of foci, length of the latus-rectum of the ellipse:

 5x² + 4y² = 1

Answer 1

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