Question

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $\frac{x^2}{49}+\frac{y^2}{36}=1.$

Answer 1

The given equation is $\frac{x^2}{49}+\frac{y^2}{36}=1$ or $\frac{x^2}{7^2}+\frac{y^2}{6^2}=1$
Her the donominator of $\frac{x^2}{49}$is greater than the donominator of $\frac{y^2}{36}$
Therefore the major axis is along the $x$-axis while the minor axis is along the $y$-axis
On comparing the given equation with $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ we obtain $a=7$ and $b=6$
$\therefore ae=c=\sqrt{a^2-b^2}=\sqrt{49-36}=\sqrt{13}$
Therefore, the coordinates of the foci are $(\pm \sqrt{13},0)$
The coordinates of the vertices are $(\pm 7,0)$
Length of major axis $=2a=14$
Length of minor axis $=2b=12$
Eccentricity $e=$ $\frac{c}{a}=$ $\frac{\sqrt{13}}{7}$
Length of latus rectum $=$ $\frac{2b^2}{a}=\frac{2\times 36}{7}=\frac{72}{7}$