Question

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

• A. $\frac{x^2}{9}+\frac{y^2}{16}=1$
• B. $\frac{9x^2}{25}+\frac{12y^2}{25}=1$
• C. $\frac{9x^2}{25}+\frac{y^2}{25}=1$
• D. $\frac{x^2}{25}+\frac{12y^2}{25}=1$

Answer 1

Answer: (B)

Let ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Latus rectum: $\frac{2b^2}{a}=\frac{5}{2}$
$\frac{2\left(a^2\left(1-e^2\right)\right)}{a}=\frac{5}{4}\Rightarrow a\left(1-\frac{1}{4}\right)=\frac{5}{4}\Rightarrow a=\frac{5}{3}$
But $\frac{2b^2}{a}=\frac{5}{2}\Rightarrow \frac{b^2}{a}=\frac{5}{4}$
$b^2=\frac{5}{4}\times \frac{5}{3}=\frac{25}{12}$
$\therefore \frac{x^2}{\frac{25}{9}}+\frac{y^2}{\frac{25}{2}}=1\Rightarrow \frac{9x^2}{25}+\frac{12y^2}{25}=1$