Find the equation of the ellipse in the case:

eccentricity *e* = \[\frac{2}{3}\] and length of latus rectum = 5

#### Solution

\[ e = \frac{2}{3} \text{ and length of the latus rectum } = 5\]

\[\text{ We have } \frac{2 b^2}{a} = 5\]

\[ \Rightarrow 2 b^2 = 5a\]

\[ \Rightarrow b^2 = \frac{5a}{2}\]

\[\text{ Now, } e = \sqrt{1 - \frac{b^2}{a^2}}\]

\[ \Rightarrow \frac{2}{3} = \sqrt{1 - \frac{\frac{5a}{2}}{a^2}}\]

\[\text{ On squaring both sides, we get:} \]

\[\frac{4}{9} = \frac{2a - 5}{2a}\]

\[ \Rightarrow 8a = 18a - 45\]

\[ \Rightarrow a = \frac{9}{2}\]

\[ \therefore b^2 = \frac{45}{4}\]

\[\text{ Substituting the values of } a^2 \text{ and } b^2 ,\text{ we get:} \]

\[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\]

\[ \Rightarrow \frac{4 x^2}{81} + \frac{4 y^2}{45} = 1\]

\[ \Rightarrow \frac{20 x^2 + 36 y^2}{405} = 1\]

\[ \Rightarrow 20 x^2 + 36 y^2 = 405\]

\[\text{ This is the required equation of the ellipse.} \]