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 $\frac{x^{2}}{16}+\frac {y^2} {9}=1$.
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 $\frac{x^{2}}{16}+\frac {y^2} {9}=1$.
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 $=\frac{2 b^{2}}{a}=\frac{2 \times 9}{4}=\frac{9}{2}$
Similar Questions
Let the tangents at the points $P$ and $Q$ on the ellipse $\frac{x^{2}}{2}+\frac{y^{2}}{4}=1$ meet at the point $R(\sqrt{2}, 2 \sqrt{2}-2)$. If $S$ is the focus of the ellipse on its negative major axis, then $SP ^{2}+ SQ ^{2}$ is equal to.
Let the tangents at the points $P$ and $Q$ on the ellipse $\frac{x^{2}}{2}+\frac{y^{2}}{4}=1$ meet at the point $R(\sqrt{2}, 2 \sqrt{2}-2)$. If $S$ is the focus of the ellipse on its negative major axis, then $SP ^{2}+ SQ ^{2}$ is equal to.
- [JEE MAIN 2022]
The equation of an ellipse whose eccentricity is $1/2$ and the vertices are $(4, 0)$ and $(10, 0)$ is
The equation of an ellipse whose eccentricity is $1/2$ and the vertices are $(4, 0)$ and $(10, 0)$ is
The normal at $\left( {2,\frac{3}{2}} \right)$ to the ellipse, $\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{3} = 1$ touches a parabola, whose equation is
The normal at $\left( {2,\frac{3}{2}} \right)$ to the ellipse, $\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{3} = 1$ touches a parabola, whose equation is
- [AIEEE 2012]
An ellipse having foci at $(3, 3) $ and $(- 4, 4)$ and passing through the origin has eccentricity equal to
An ellipse having foci at $(3, 3) $ and $(- 4, 4)$ and passing through the origin has eccentricity equal to
In an ellipse, the distance between its foci is $6$ and minor axis is $8.$ Then its eccentricity is :
In an ellipse, the distance between its foci is $6$ and minor axis is $8.$ Then its eccentricity is :