# Find the Equation of an Ellipse Whose Eccentricity is 2/3, the Latus-rectum is 5 and the Centre is at the Origin. - Mathematics

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

#### Solution

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


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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 26 Ellipse
Exercise 26.1 | Q 12 | Page 23