Question

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

 4x² + 3y² = 1

Answer 1

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