Question

The distance between the foci of an ellipse is 10 and its latus rectum is 15. Its equation referred to its axes as axes of coordinates is

• A. $3x^2+4y^2=300$
• B. $2x^2+y^2=50$
• C. $10x^2+15y^2=300$
• D. None of these

Answer 1

Answer: (A)

Let the equation of the ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$
If S and S' be the foci, then $S{S}^{\prime }=10$.
But $S{S}^{\prime }=CS+C{S}^{\prime }=2ae,C$ being the centre
$\therefore 2ae=10$, or $ae=5...\left(1\right)$
Also $2\frac{b^2}{a}=15...\left(2\right)$
Also $b^2=a^2\left(1-e^2\right)=a^2-a^2{e}^2=a^2-25$ [using $\left(1\right)$]
By $\left(2\right),2b^2=15a$; or $=2\left(a^2-25\right)=15a$
$\therefore a=-\frac{5}{2}$ or $a=10$.
But a cannot be negative,
$\therefore a=10;\therefore b^2=\frac{15\times 10}{2}=75$.
The equation to the ellipse is therefore
$\frac{x^2}{100}+\frac{y^2}{75}=1$; or $3x^2+4y^2=300.$