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Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola $9 y^{2}-4 x^{2}=36$
Solution
The given equation is $9 y^{2}-4 x^{2}=36$
It can be written as
$9 y^{2}-4 x^{2}=36$
Or, $\frac{y^{2}}{4}-\frac{x^{2}}{9}=1$
Or, $\frac{y^{2}}{2^{2}}-\frac{x^{2}}{3^{2}}=1$ ……….. $(1)$
On comparing equation $(1)$ with the standard equation of hyperbola i.e., $\frac{y^{2}}{a^{2}}-\frac{ x ^{2}}{b^{2}},$ we obtain $a=2$ and $b=3$
We know that $a^{2}+b^{2}=c^{2}$
$\therefore c^{2}=4+9=13$
$\Rightarrow c=\sqrt{13}$
Therefore,
The coordinates of the foci are $(0, \,\pm \sqrt{13})$
The coordinates of the vertices are $(0,\,±2)$
Eccentricity, $e=\frac{c}{a}=\frac{\sqrt{13}}{2}$
Length of latus rectum $=\frac{2 b^{2}}{a}=\frac{2 \times 9}{2}=9$
Similar Questions
Let $H : \frac{ x ^2}{ a ^2}-\frac{ y ^2}{ b ^2}=1$, where $a > b >0$, be $a$ hyperbola in the $xy$-plane whose conjugate axis $LM$ subtends an angle of $60^{\circ}$ at one of its vertices $N$. Let the area of the triangle $LMN$ be $4 \sqrt{3}$..
List $I$ | List $II$ |
$P$ The length of the conjugate axis of $H$ is | $1$ $8$ |
$Q$ The eccentricity of $H$ is | $2$ ${\frac{4}{\sqrt{3}}}$ |
$R$ The distance between the foci of $H$ is | $3$ ${\frac{2}{\sqrt{3}}}$ |
$S$ The length of the latus rectum of $H$ is | $4$ $4$ |
The correct option is: