Find the coordinates of the foci, the vertices, the lengths of major and minor axes and the eccentricity of the ellipse $9 x^{2}+4 y^{2}=36$.
Find the coordinates of the foci, the vertices, the lengths of major and minor axes and the eccentricity of the ellipse $9 x^{2}+4 y^{2}=36$.
The given equation of the ellipse can be written in standard form as
$\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$
since the denominator of $\frac{y^{2}}{9}$ is larger than the denominator of $\frac{x^{2}}{4},$ the major axis is along the $y-$ axis. Comparing the given equation with the standard equation
$\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1$, we have $b=2$ and $a=3$
Also $c=\sqrt{a^{2}-b^{2}}$ $=\sqrt{9-4}=\sqrt{5}$
and $e=\frac{c}{a}=\frac{\sqrt{5}}{3}$
Hence the foci are $(0,\, \sqrt{5})$ and $(0,\,-\sqrt{5}),$ vertices are $(0,\,3)$ and $(0,\,-3),$ length of the major axis is $6$ units, the length of the minor axis is $4$ units and the eccentricity of the cllipse is $\frac{\sqrt{5}}{3}$.
Similar Questions
Let $S=\left\{(x, y) \in N \times N : 9(x-3)^{2}+16(y-4)^{2} \leq 144\right\}$ and $ T=\left\{(x, y) \in R \times R :(x-7)^{2}+(y-4)^{2} \leq 36\right\}$ Then $n ( S \cap T )$ is equal to $......$
Let $S=\left\{(x, y) \in N \times N : 9(x-3)^{2}+16(y-4)^{2} \leq 144\right\}$ and $ T=\left\{(x, y) \in R \times R :(x-7)^{2}+(y-4)^{2} \leq 36\right\}$ Then $n ( S \cap T )$ is equal to $......$
- [JEE MAIN 2022]
The equation of the tangent at the point $(1/4, 1/4)$ of the ellipse $\frac{{{x^2}}}{4} + \frac{{{y^2}}}{{12}} = 1$ is
The equation of the tangent at the point $(1/4, 1/4)$ of the ellipse $\frac{{{x^2}}}{4} + \frac{{{y^2}}}{{12}} = 1$ is
Let $E_1$ and $E_2$ be two ellipses whose centers are at the origin. The major axes of $E_1$ and $E_2$ lie along the $x$-axis and the $y$-axis, respectively. Let $S$ be the circle $x^2+(y-1)^2=2$. The straight line $x+y=3$ touches the curves $S, E_1$ ad $E_2$ at $P, Q$ and $R$, respectively. Suppose that $P Q=P R=\frac{2 \sqrt{2}}{3}$. If $e_1$ and $e_2$ are the eccentricities of $E_1$ and $E_2$, respectively, then the correct expression$(s)$ is(are)
$(A)$ $e_1^2+e_2^2=\frac{43}{40}$
$(B)$ $e_1 e_2=\frac{\sqrt{7}}{2 \sqrt{10}}$
$(C)$ $\left|e_1^2-e_2^2\right|=\frac{5}{8}$
$(D)$ $e_1 e_2=\frac{\sqrt{3}}{4}$
Let $E_1$ and $E_2$ be two ellipses whose centers are at the origin. The major axes of $E_1$ and $E_2$ lie along the $x$-axis and the $y$-axis, respectively. Let $S$ be the circle $x^2+(y-1)^2=2$. The straight line $x+y=3$ touches the curves $S, E_1$ ad $E_2$ at $P, Q$ and $R$, respectively. Suppose that $P Q=P R=\frac{2 \sqrt{2}}{3}$. If $e_1$ and $e_2$ are the eccentricities of $E_1$ and $E_2$, respectively, then the correct expression$(s)$ is(are)
$(A)$ $e_1^2+e_2^2=\frac{43}{40}$
$(B)$ $e_1 e_2=\frac{\sqrt{7}}{2 \sqrt{10}}$
$(C)$ $\left|e_1^2-e_2^2\right|=\frac{5}{8}$
$(D)$ $e_1 e_2=\frac{\sqrt{3}}{4}$
- [IIT 2015]
The smallest possible positive slope of a line whose $y$-intercept is $5$ and which has a common point with the ellipse $9 x^2+16 y^2=144$ is
The smallest possible positive slope of a line whose $y$-intercept is $5$ and which has a common point with the ellipse $9 x^2+16 y^2=144$ is
- [KVPY 2011]
The eccentricity of the ellipse $9{x^2} + 5{y^2} - 30y = 0$, is
The eccentricity of the ellipse $9{x^2} + 5{y^2} - 30y = 0$, is