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Q.
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
Conic Sections
Solution:
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 $SS' = 10$.
But $SS' = CS + CS' = 2ae,\, C$ being the centre
$\therefore 2ae =10$, or $ae = 5\quad ...\left(1\right)$
Also $2 \frac{b^{2}}{a} = 15\quad ...\left(2\right)$
Also $b^{2} = a^{2}\left(1-e^{2}\right) = a^{2}-a^{2} e^{2} = a^{2} - 25\quad$ [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,
$\therefore a = 10; \,\therefore b^{2} = \frac{15\times10}{2} = 75$.
The equation to the ellipse is therefore
$\frac{x^{2}}{100} + \frac{y^{2}}{75} = 1$; or $3x^{2} + 4y^{2} = 300.$