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Q.
If the length of the latus rectum of ellipse is $\frac{5}{2}$ and eccentricity is $\frac{1}{2}$. Then the equation of the ellipse in standard form is
Conic Sections
Solution:
Let ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Latus rectum: $\frac{2 b ^2}{ a }=\frac{5}{2}$
$\frac{2\left( a ^2\left(1- e ^2\right)\right)}{ a }=\frac{5}{4} \Rightarrow a \left(1-\frac{1}{4}\right)=\frac{5}{4} \Rightarrow a =\frac{5}{3}$
But $\frac{2 b ^2}{ a }=\frac{5}{2} \Rightarrow \frac{ b ^2}{ a }=\frac{5}{4}$
$b ^2=\frac{5}{4} \times \frac{5}{3}=\frac{25}{12} $
$\therefore \frac{ x ^2}{\frac{25}{9}}+\frac{ y ^2}{\frac{25}{2}}=1 \Rightarrow \frac{9 x ^2}{25}+\frac{12 y ^2}{25}=1$