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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 $4 x ^{2}+9 y ^{2}=36$
Solution
The given equation is $4 x ^{2}+9 y ^{2}=36$
It can be written as
$4 x^{2}+9 y^{2}=36$
Or , $\frac{ x ^{2}}{9}+\frac{y^{2}}{4}=1$
Or, $\frac{x^{2}}{3^{2}}+\frac{y^{2}}{2^{2}}=1$ ……… $(1)$
Here, the denominator of $\frac{ x ^{2}}{3^{2}}$ is greater than the denominator of $\frac{y^{2}}{2^{2}}$
Therefore, the major axis is along the $x-$ axis, while the minor axis is along the $y-$ axis.
On comparing the given equation with $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1,$ we obtain $a=3$ and $b=2$
$\therefore c=\sqrt{a^{2}-b^{2}}=\sqrt{9-4}=\sqrt{5}$
Therefore,
The coordinates of the foci are $(\pm \sqrt{5}, \,0)$
The coordinates of the vertices are $(±3,\,0)$
Length of major axis $=2 a=6$
Length of minor axis $=2 b=4$
Eccentricity, $e=\frac{c}{a}=\frac{\sqrt{5}}{3}$
Length of latus rectum $=\frac{2 b^{2}}{a}=\frac{2 \times 4}{3}=\frac{8}{3}$
Similar Questions
Consider the ellipse
$\frac{x^2}{4}+\frac{y^2}{3}=1$
Let $H (\alpha, 0), 0<\alpha<2$, be a point. A straight line drawn through $H$ parallel to the $y$-axis crosses the ellipse and its auxiliary circle at points $E$ and $F$ respectively, in the first quadrant. The tangent to the ellipse at the point $E$ intersects the positive $x$-axis at a point $G$. Suppose the straight line joining $F$ and the origin makes an angle $\phi$ with the positive $x$-axis.
$List-I$ | $List-II$ |
If $\phi=\frac{\pi}{4}$, then the area of the triangle $F G H$ is | ($P$) $\frac{(\sqrt{3}-1)^4}{8}$ |
If $\phi=\frac{\pi}{3}$, then the area of the triangle $F G H$ is | ($Q$) $1$ |
If $\phi=\frac{\pi}{6}$, then the area of the triangle $F G H$ is | ($R$) $\frac{3}{4}$ |
If $\phi=\frac{\pi}{12}$, then the area of the triangle $F G H$ is | ($S$) $\frac{1}{2 \sqrt{3}}$ |
($T$) $\frac{3 \sqrt{3}}{2}$ |
The correct option is: