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

Answer in Brief

Find the equation of the ellipse in the case:

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

 
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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
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