Question

The eccentricity of an ellipse is $\frac{2}{3}$ latus rectum is $5$ and a centre $(0,0)$ the equation of ellipse is

• A. $\frac{x^2}{81}+\frac{y^2}{45}=1$
• B. $\frac{4x^2}{81}+\frac{4y^2}{45}=1$
• C. $\frac{x^2}{9}+\frac{y^2}{5}=1$
• D. None of these

Answer 1

The correct option is B $\frac{4x^2}{81}+\frac{4y^2}{45}=1$

$e=\frac{2}{3}$, latusrectum$=5$

$\frac{2b^2}{a}=5$

$b^2=a^2(1-e^2)$

$\frac{2a^2(1-e^2)}{a}=5$

$2a(1-\frac{4}{9})=5$

$2a\times \frac{5}{9}=5$

$2a=9\Rightarrow a=\frac{9}{2}$

$b^2=\frac{9}{2}\times \frac{9}{2}\times (1-\frac{4}{9})=\frac{81}{4}\times \frac{5}{9}=b=\frac{9}{6}\sqrt{5}$

$b=\frac{3}{2}\sqrt{5}$

Equation of ellipse$=\frac{{(x-0)}^2}{a^2}+\frac{{(y-0)}^2}{b^2}=1$

$=\frac{4x^2}{81}+\frac{y^2}{45}\times 4=1$

$=\frac{4x^2}{81}+\frac{4y^2}{45}=1$