Answer :

The given equation is $\dfrac{x^{2}}{16}-\dfrac{y^{2}}{9}=1$ or $\dfrac{x^{2}}{4^{2}}-\dfrac{y^{2}}{3^{2}}=1$

On comparing this equation with the standard equation of hyperbola i.e., $\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$,

we obtain $a=4$ and $b=3$

We know that $a^{2}+b^{2}=c^{2}$

$\therefore \ \ c^{2}=4^{2}+3^{2}=25$

$\Rightarrow c=5$

Therefore,

The coordinates of the foci are $\left( \pm 5,0\right)$

The coordinates of the vertices are $\left( \pm 4,0\right)$

Eccentricity, $e=\dfrac{c}{a}=\dfrac{5}{4}$

Length of latus rectum $=\dfrac{2 b^{2}}{a}=\dfrac{2 \times 9}{4}=\dfrac{9}{2}$

Answer :

The given equation is

$ \dfrac{y^{2}}{9}-\dfrac{x^{2}}{27}=1 \text{ or } \dfrac{y^{2}}{3^{2}}-\dfrac{x^{2}}{\left(\sqrt{27}\right)^{2}}=1 $

On comparing this equation with the standard equation of hyperbola i.e., $\dfrac{y^{2}}{a^{2}}-\dfrac{x^{2}}{b^{2}}=1$,

we obtain $a=3$ and $b=\sqrt{27}$

We know that $a^{2}+b^{2}=c^{2}$

$ \begin{aligned} & \therefore \ \ c^{2}=3^{2}+\left(\sqrt{27}\right)^{2}=9+27=36 \\ \\ & \Rightarrow c=6 \end{aligned} $

Therefore,

The coordinates of the foci are $\left(0, \pm 6\right)$

The coordinates of the vertices are $\left(0, \pm 3\right)$

Eccentricity, $e=\dfrac{c}{a}=\dfrac{6}{3}=2$

Length of latus rectum $=\dfrac{2 b^{2}}{a}=\dfrac{2 \times 27}{3}=18$

Answer :

The given equation is $9 y^{2} - 4 x^{2}=36$

It can be written as $9 y^{2} - 4 x^{2}=36$

or, $\dfrac{y^{2}}{4}-\dfrac{x^{2}}{9}=1$

or, $\dfrac{y^{2}}{2^{2}}-\dfrac{x^{2}}{3^{2}}=1\qquad\ldots\mathrm{(1)}$

On comparing equation $\left(1\right)$ with the standard equation of hyperbola i.e., $\dfrac{y^{2}}{a^{2}}-\dfrac{x^{2}}{b^{2}}=1$,

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 $\left(0, \pm \sqrt{13}\right)$

The coordinates of the vertices are $\left(0, \pm 2\right)$

Eccentricity, $e=\dfrac{c}{a}=\dfrac{\sqrt{13}}{2}$

Length of latus rectum $ =\dfrac{2 b^{2}}{a}=\dfrac{2 \times 9}{2}=9 $

Answer :

The given equation is $16 x^{2} - 9 y^{2}=576$

It can be written as

$16 x^{2}-9 y^{2}=576$

$\Rightarrow \dfrac{x^{2}}{36}-\dfrac{y^{2}}{64}=1$

$\Rightarrow \dfrac{x^{2}}{6^{2}}-\dfrac{y^{2}}{8^{2}}=1\qquad\ldots\mathrm{(1)}$

On comparing equation $\left(1\right)$ with the standard equation of hyperbola i.e., $\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$,

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 $\left( \pm 10,0\right)$

The coordinates of the vertices are $\left( \pm 6,0\right)$

Eccentricity, $e=\dfrac{c}{a}=\dfrac{10}{6}=\dfrac{5}{3}$

Length of latus rectum $=\dfrac{2 b^{2}}{a}=\dfrac{2 \times 64}{6}=\dfrac{64}{3}$

Answer :

The given equation is $5 y^{2} - 9 x^{2}=36$

$\Rightarrow \dfrac{y^{2}}{\left(\dfrac{36}{5}\right)}-\dfrac{x^{2}}{4}=1$

$\Rightarrow \dfrac{y^{2}}{\left(\dfrac{6}{\sqrt{5}}\right)^{2}}-\dfrac{x^{2}}{2^{2}}=1\qquad\ldots\mathrm{(1)}$

On comparing equation $\left(1\right)$ with the standard equation of hyperbola i.e., $\dfrac{y^{2}}{a^{2}}-\dfrac{x^{2}}{b^{2}}=1$,

we obtain $a=\dfrac{6}{\sqrt{5}}$ and $b= 2.$ We know that $a^{2}+b^{2}=c^{2}$

$\therefore \ \ c^{2}=\dfrac{36}{5}+4=\dfrac{56}{5}$

$\Rightarrow c=\sqrt{\dfrac{56}{5}}=\dfrac{2 \sqrt{14}}{\sqrt{5}}$

Therefore, the coordinates of the foci are $\left(0, \pm \dfrac{2 \sqrt{14}}{\sqrt{5}}\right)$

The coordinates of the vertices are $\left(0, \pm \dfrac{6}{\sqrt{5}}\right)$

Eccentricity, $e=\dfrac{c}{a}=\dfrac{\left(\dfrac{2 \sqrt{14}}{\sqrt{5}}\right)}{\left(\dfrac{6}{\sqrt{5}}\right)}=\dfrac{\sqrt{14}}{3}$ $=\dfrac{2 b^{2}}{a}=\dfrac{2 \times 4}{\left(\dfrac{6}{\sqrt{5}}\right)}=\dfrac{4 \sqrt{5}}{3}$

Length of latus rectum $=\dfrac{2 b^{2}}{a}=\dfrac{2 \times 4}{\dfrac{6}{\sqrt{5}}}=\dfrac{64}{3}$

Answer :

The given equation is $49 y^{2} - 16 x^{2}=784$

It can be written as

$49 y^{2} - 16 x^{2}=784$

or, $\dfrac{y^{2}}{16}-\dfrac{x^{2}}{49}=1$

or, $\dfrac{y^{2}}{4^{2}}-\dfrac{x^{2}}{7^{2}}=1\qquad\ldots\mathrm{(1)}$

On comparing equation $\left(1\right)$ with the standard equation of hyperbola i.e., $\dfrac{y^{2}}{a^{2}}-\dfrac{x^{2}}{b^{2}}=1$,

we obtain $a=4$ and $b=7$ We know that $a^{2}+b^{2}=c^{2}$

$\therefore \ \ c^{2}=16+49=65$

$\Rightarrow c=\sqrt{65}$

Therefore,

The coordinates of the foci are $\left(0, \pm \sqrt{65}\right)$

The coordinates of the vertices are $\left(0, \pm 4\right)$

Eccentricity, $e=\dfrac{c}{a}=\dfrac{\sqrt{65}}{4}$

Length of latus rectum $=\dfrac{2 b^{2}}{a}=\dfrac{2 \times 49}{4}=\dfrac{49}{2}$

Answer :

Vertices $\left( \pm 2,0\right)$, foci $\left( \pm 3,0\right)$

Here, the vertices are on the $x$-axis.

Therefore, the equation of the hyperbola is of the form $\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$

Since the vertices are $\left( \pm 2,0\right), a=2$

Since the foci are $\left( \pm 3,0\right), c=3$

We know that $a^{2}+b^{2}=c^{2}$

$\therefore \ \ 2^{2}+b^{2}=3^{2}$

$\quad b^{2}=9-4=5$

Thus, the equation of the hyperbola is $\dfrac{x^{2}}{4}-\dfrac{y^{2}}{5}=1$

Answer :

Vertices $\left(0, \pm 5\right)$, foci $\left(0, \pm 8\right)$

Here, the vertices are on the $y$-axis.

Therefore, the equation of the hyperbola is of the form $\dfrac{y^{2}}{a^{2}}-\dfrac{x^{2}}{b^{2}}=1$

Since the vertices are $\left(0, \pm 5\right), a=5$

Since the foci are $\left(0, \pm 8\right), c=8$

We know that $a^{2}+b^{2}=c^{2}$

$\therefore \ \ 5^{2}+b^{2}=8^{2}$

$\quad b^{2}=64-25=39$

Thus, the equation of the hyperbola is $\dfrac{y^{2}}{25}-\dfrac{x^{2}}{39}=1$

Answer :

Vertices $\left(0, \pm 3\right)$, foci $\left(0, \pm 5\right)$

Here, the vertices are on the $y$-axis.

Therefore, the equation of the hyperbola is of the form $\dfrac{y^{2}}{a^{2}}-\dfrac{x^{2}}{b^{2}}=1$

Since the vertices are $\left(0, \pm 3\right), a=3$

Since the foci are $\left(0, \pm 5\right), c=5$

We know that $a^{2}+b^{2}=c^{2}$

$\therefore \ \ 3^{2}+b^{2}=5^{2}$

$\Rightarrow b^{2}=25 - 9=16$

Thus, the equation of the hyperbola is $\dfrac{y^{2}}{9}-\dfrac{x^{2}}{16}=1$

Answer :

Foci $\left( \pm 5,0\right),$ the transverse axis is of length $8 .$

Here, the foci are on the $x$-axis.

Therefore, the equation of the hyperbola is of the form $\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$

Since the foci are $\left( \pm 5,0\right), c=5$

Since the length of the transverse axis is $8,$

$2 a=8 \Rightarrow a=4$

We know that $a^{2}+b^{2}=c^{2}$

$\therefore \ \ 4^{2}+b^{2}=5^{2}$

$\Rightarrow b^{2}=25 - 16=9$

Thus, the equation of the hyperbola is $\dfrac{x^{2}}{16}-\dfrac{y^{2}}{9}=1$

Answer :

Foci $\left(0, \pm 13\right)$, the conjugate axis is of length $24$ .

Here, the foci are on the $y$-axis.

Therefore, the equation of the hyperbola is of the form $\dfrac{y^{2}}{a^{2}}-\dfrac{x^{2}}{b^{2}}=1$

Since the foci are $\left(0, \pm 13\right), c=13$

Since the length of the conjugate axis is $24,$

$2 b=24 \Rightarrow b=12$

We know that $a^{2}+b^{2}=c^{2}$

$\therefore \ \ a^{2}+12^{2}=13^{2}$

$\Rightarrow a^{2}=169 - 144=25$

Thus, the equation of the hyperbola is $\dfrac{y^{2}}{25}-\dfrac{x^{2}}{144}=1$

Answer :

Foci $\left( \pm 3 \sqrt{5}, 0\right)$, the latus rectum is of length $8 .$

Here, the foci are on the $x$-axis.

Therefore, the equation of the hyperbola is of the form $\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$

Since the foci are $\left( \pm 3 \sqrt{5}, 0\right), c= \pm 3 \sqrt{5}$

Length of latus rectum $=8$

$\Rightarrow \dfrac{2 b^{2}}{a}=8$

$\Rightarrow b^{2}=4 a$

We know that $a^{2}+b^{2}=c^{2}$

$\therefore \ \ a^{2}+4 a=45$

$\Rightarrow a^{2}+4 a - 45=0$

$\Rightarrow a^{2}+9 a - 5 a - 45=0$

$\Rightarrow\left(a+9\right)\left(a - 5\right)=0$

$\Rightarrow a=- 9,5$

Since $a$ is non-negative, $a=5$

$\therefore \ \ b^{2}=4 a=4 \times 5=20$

Thus, the equation of the hyperbola is $\dfrac{x^{2}}{25}-\dfrac{y^{2}}{20}=1$

Answer :

Foci $\left( \pm 4,0\right)$, the latus rectum is of length $12 .$

Here, the foci are on the $x$-axis.

Therefore, the equation of the hyperbola is of the form $\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$

Since the foci are $\left( \pm 4,0\right), c=4$

Length of latus rectum $=12$

$\Rightarrow \dfrac{2 b^{2}}{a}=12$

$\Rightarrow b^{2}=6 a$

We know that $a^{2}+b^{2}=c^{2}$

$\therefore \ \ a^{2}+6 a=16$

$\Rightarrow a^{2}+6 a - 16=0$

$\Rightarrow a^{2}+8 a - 2 a - 16=0$

$\Rightarrow\left(a+8\right)\left(a - 2\right)=0$

$\Rightarrow a=- 8,2$

Since $a$ is non-negative, $a=2$

$\therefore \ \ b^{2}=6 a=6 \times 2=12$

Thus, the equation of the hyperbola is $\dfrac{x^{2}}{4}-\dfrac{y^{2}}{12}=1$

Answer :

Given,

Vertices $\left( \pm 7,0\right)$,

$ e=\dfrac{4}{3} $

Here, the vertices are on the $x$-axis.

Therefore, the equation of the hyperbola is of the form $\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$

Since the vertices are $\left( \pm 7,0\right), a=7$

It is given that $e=\dfrac{4}{3}$

$\therefore \ \ \dfrac{c}{a}=\dfrac{4}{3} \quad[e=\dfrac{c}{a}]$

$\Rightarrow \dfrac{c}{7}=\dfrac{4}{3}$

$\Rightarrow c=\dfrac{28}{3}$

We know that $a^{2}+b^{2}=c^{2}$

$\therefore \ \ 7^{2}+b^{2}=\left(\dfrac{28}{3}\right)^{2}$

$\Rightarrow b^{2}=\dfrac{784}{9}-49$

$\Rightarrow b^{2}=\dfrac{784-441}{9}=\dfrac{343}{9}$

Thus, the equation of the hyperbola is $\dfrac{x^{2}}{49}-\dfrac{9 y^{2}}{343}=1$

Answer :

Foci $\left(0, \pm \sqrt{10}\right)$, passing through $\left(2,3\right)$

Here, the foci are on the $y$-axis.

Therefore, the equation of the hyperbola is of the form $\dfrac{y^{2}}{a^{2}}-\dfrac{x^{2}}{b^{2}}=1$

Since the foci are $\left(0, \pm \sqrt{10}\right), c=\sqrt{10}$

We know that $a^{2}+b^{2}=c^{2}$

$\therefore \ \ a^{2}+b^{2}=10$

$\Rightarrow b^{2}=10 - a^{2}\qquad \ldots(1)$

Since the hyperbola passes through point $\left(2,3\right)$,

$\dfrac{9}{a^{2}}-\dfrac{4}{b^{2}}=1\qquad \ldots(2)$

From equations $\left(1\right)$ and $\left(2\right),$

we obtain, $\dfrac{9}{a^{2}}-\dfrac{4}{\left(10-a^{2}\right)}=1$

$\Rightarrow 9\left(10-a^{2}\right)-4 a^{2}=a^{2}\left(10-a^{2}\right)$

$\Rightarrow 90-9 a^{2}-4 a^{2}=10 a^{2}-a^{4}$

$\Rightarrow a^{4}-23 a^{2}+90=0$

$\Rightarrow a^{4}-18 a^{2}-5 a^{2}+90=0$

$\Rightarrow a^{2}\left(a^{2}-18\right)-5\left(a^{2}-18\right)=0$

$\Rightarrow\left(a^{2}-18\right)\left(a^{2}-5\right)=0$

$\Rightarrow a^{2}=18$ or $5$

In hyperbola, $c>a$, i.e., $c^{2}>a^{2}$

$\therefore \ \ a^{2}=5$

$\Rightarrow b^{2}=10 - a^{2}=10 - 5=5$

Thus, the equation of the hyperbola is $\dfrac{y^{2}}{5}-\dfrac{x^{2}}{5}=1$