Question

Find the equation of an ellipse whose eccentricity is 2/3, the latus-rectum is 5 and the centre is at the origin.

Answer 1

\[\text{ Let the ellipse be }\frac{x^2}{a^2}+\frac{y^2}{b^2}=1. ...\left( 1 \right)\]
\[e = \frac{2}{3} \text{ and latus rectum }=5 (\text{ Given })\]
\[\text{ Now }, \frac{2 b^2}{a} = 5\]
\[ \Rightarrow 2 b^2 = 5a . . . (2)\]
\[ \Rightarrow 2 a^2 (1 - e^2 ) = 5a [ \because b^2 = a^2 (1 - e^2 )]\]
\[ \Rightarrow 2 a^2 \left[ 1 - \frac{4}{9} \right] = 5a\]
\[ \Rightarrow 2 a^2 \times \frac{5}{9} = 5a\]
\[ \Rightarrow 10 a^2 = 45a\]
\[ \Rightarrow a = \frac{9}{2}\]
\[\text{ Substituting the value ofain eq. (2), we get }:\]
\[2 b^2 = 5 \times \frac{9}{2}\]
\[ \Rightarrow b^2 = \frac{45}{4}\]
\[\text{ Substituting the values of } a^2\text{ and } b^2 \text{ in eq }. (1), \text{ we get }:\]
\[\frac{x^2}{\frac{81}{4}}+\frac{y^2}{\frac{45}{4}}=1\]
\[ \Rightarrow \frac{4 x^2}{81} + \frac{4 y^2}{45} = 1\]
\[\text{ This is the required equation of the ellipse }.\]
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