Question

The equation of the ellipse whose centre is at origin, major axis is along x-axis with eccentricity $\frac{3}{4}$ and latus rectum $4$ units is

• A. $\frac{x^2}{1024}+\frac{7y^2}{64}=1$
• B. $\frac{49x^2}{1024}+\frac{7y^2}{64}=1$
• C. $\frac{7x^2}{1024}+\frac{49y^2}{64}=1$
• D. $\frac{x^2}{1024}+\frac{y^2}{64}=1$

Answer 1

The correct option is B $\frac{49x^2}{1024}+\frac{7y^2}{64}=1$

Let equation be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Major axis is along x axis $\therefore b^2=a^2(1-e^2)$

So, equation of ellipse is $\frac{x^2}{a^2}+\frac{y^2}{a^2(1-e^2)}=1$ ..... equation (1)

Given, $e=\frac{3}{4}\therefore (1-e^2)=\frac{7}{16}$

Also, latus rectum $=4=\frac{2b^2}{a}=\frac{2a^2(1-e^2)}{a}$

$a=\frac{2}{1-e^2}=\frac{32}{7}$

Put values of $a$ and $e$ in equation (1):

$\frac{49x^2}{1024}+\frac{7y^2}{64}=1$

Option (B)