# Find the Equation of the Ellipse in the Case: (Ii) Eccentricity E = 2 3 and Length of Latus Rectum = 5 - Mathematics

Find the equation of the ellipse in the case:

eccentricity e = $\frac{2}{3}$ and length of latus rectum = 5

#### Solution

$e = \frac{2}{3} \text{ and length of the latus rectum } = 5$
$\text{ We have } \frac{2 b^2}{a} = 5$
$\Rightarrow 2 b^2 = 5a$
$\Rightarrow b^2 = \frac{5a}{2}$
$\text{ Now, } e = \sqrt{1 - \frac{b^2}{a^2}}$
$\Rightarrow \frac{2}{3} = \sqrt{1 - \frac{\frac{5a}{2}}{a^2}}$
$\text{ On squaring both sides, we get:}$
$\frac{4}{9} = \frac{2a - 5}{2a}$
$\Rightarrow 8a = 18a - 45$
$\Rightarrow a = \frac{9}{2}$
$\therefore b^2 = \frac{45}{4}$
$\text{ Substituting the values of } a^2 \text{ and } b^2 ,\text{ we get:}$
$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$
$\Rightarrow \frac{4 x^2}{81} + \frac{4 y^2}{45} = 1$
$\Rightarrow \frac{20 x^2 + 36 y^2}{405} = 1$
$\Rightarrow 20 x^2 + 36 y^2 = 405$
$\text{ This is the required equation of the ellipse.}$

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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 26 Ellipse
Exercise 26.1 | Q 5.02 | Page 22