Question

The eccentricy of an ellipse is $\frac{2}{3}$ , latus-rectum is $5$ and center is $(0,0)$ . The equation of the ellipse is

• A. $\frac{3x^2}{81}+\frac{3y^2}{45}=1$
• B. $\frac{4x^2}{81}+\frac{4y^2}{45}=1$
• C. $\frac{5x^2}{9}+\frac{5y^2}{5}=1$
• D. $\frac{6x^2}{81}+\frac{6y^2}{45}-1$

Answer 1

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

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

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

Eccentricity$=\frac{2}{3}$

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

$b^2=a^2(1-\frac{4}{9})=a^2\left(\frac{5}{9}\right)$

$\frac{b^2}{a^2}=\frac{5}{9}$

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

$\Rightarrow 2a=9$ (or) $a=\frac{9}{2}$

$b^2=\frac{5a}{2}=\frac{5\times 9}{2\times 2}=\frac{45}{4}$

$a^2=\frac{81}{4}$

Since the centre is at origin, the equation of ellipse is

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

(or) $\frac{4x^2}{81}+\frac{4y^2}{45}=1$.