10-2. Parabola, Ellipse, Hyperbola
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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 $\frac{x^{2}}{16}+\frac {y^2} {9}=1$.

Option A
Option B
Option C
Option D

Solution

The given equation is $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$ or $\frac{x^{2}}{4^{2}}+\frac{y^{2}}{3^{2}}=1$

Here, the denominator of $\frac{x^{2}}{16}$ is greater than the denominator of $\frac{y^{2}}{9}$.

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 $a=4$ and $b=3$

$\therefore c=\sqrt{a^{2}-b^{2}}=\sqrt{16-9}=\sqrt{7}$

Therefore,

The coordinates of the foci are $(\pm \sqrt{7}, \,0)$

The coordinates of the vertices are $(±4,\,0)$

Length of major axis $=2 a=8$

Length of minor axis $=2 b=6$

Eccentricity, $e=\frac{c}{a}=\frac{\sqrt{7}}{4}$

Length of latus rectum $=\frac{2 b^{2}}{a}=\frac{2 \times 9}{4}=\frac{9}{2}$

Standard 11
Mathematics

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