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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 $16 x^{2}-9 y^{2}=576$
Solution
The given equation is $16 x^{2}-9 y^{2}=576$
It can be written as
$16 x^{2}-9 y^{2}=576$
$\Rightarrow \frac{x^{2}}{36}-\frac{y^{2}}{64}=1$
$\Rightarrow \frac{x^{2}}{6^{2}}-\frac{y^{2}}{8^{2}}=1$ ………… $(1)$
On comparing equation $(1)$ with the standard equation of hyperbola i.e., $\frac{ x ^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1,$
we obtain $a=6$ and $b=8$
We know that $a^{2}+b^{2}=c^{2}$
$\therefore c^{2}=36+64=100$
$\Rightarrow c=10$
Therefore,
The coordinates of the foci are $(±10,\,0)$.
The coordinates of the vertices are $(±6,\,0)$
Eccentricity, $e=\frac{c}{a}=\frac{10}{6}=\frac{5}{3}$
Length of latus rectum $=\frac{2 b^{2}}{a}=\frac{2 \times 64}{6}=\frac{64}{3}$
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:
