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 $\frac{x^{2}}{49}+\frac{y^{2}}{36}=1$
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 $\frac{x^{2}}{49}+\frac{y^{2}}{36}=1$
The given equation is $\frac{x^2}{49}+\frac{y^{2}}{36}=1$ or $\frac{x^2} {7^{2}}+\frac{y^{2}}{6^{2}}=1$
Here, the denominator of $\frac{x^{2}}{49}$ is greater than the denominator 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 $b=6$
$\therefore c=\sqrt{a^{2}-b^{2}}=\sqrt{49-36}=\sqrt{13}$
Therefore,
The coordinates of the foci are $(\pm \,\sqrt{13}, 0)$
The coordinates of the vertices are $(±7,\,0)$
Length of major axis $=2 a =14$
Length of minor axis $=2 b =12$
Eccentricity, $e=\frac{c}{a}=\frac{\sqrt{13}}{7}$
Length of latus rectum $=\frac{2 b^{2}}{a}=\frac{2 \times 36}{7}=\frac{72}{7}$
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- [JEE MAIN 2019]