Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the latus rectum of the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$
Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the latus rectum of the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$
since denominator of $\frac{x^{2}}{25}$ is larger than the denominator of $\frac{y^{2}}{9},$ the major axis is along the $x-$ axis.
Comparing the given equation with $\frac{x^{2}}{a^{2}}$ $+\frac{y^{2}}{b^{2}}$ $=1,$ we get $a=5$ and $b=3$ . Also $c=\sqrt{a^{2}-b^{2}}=\sqrt{25-9}=4$
Therefore, the coordinates of the foci are $(-4,\,0)$ and $(4,\,0),$ vertices are $(-5,\,0)$ and $(5,\,0).$ Length of the major axis is $10$ units length of the minor axis $2b$ is $6$ units and the eccentricity is $\frac{4}{5}$ and latus rectum is $\frac{2 b^{2}}{a}=\frac{18}{5}$.
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- [JEE MAIN 2022]