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 $( \pm 5,0)$.

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

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

Therefore,

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

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

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, $\dfrac{y^{2}}{2^{2}}-\dfrac{x^{2}}{3^{2}}=1$

On comparing equation (1) 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 $(0, \pm \sqrt{13})$.

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

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

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

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

$\Rightarrow \dfrac{x^{2}}{6^{2}}-\dfrac{y^{2}}{8^{2}}=1$

On comparing equation (1) 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 $( \pm 10,0)$.

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

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 $5 y^{2} - 9 x^{2}=36$.

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

$\Rightarrow \dfrac{y^{2}}{(\dfrac{6}{\sqrt{5}})^{2}}-\dfrac{x^{2}}{2^{2}}=1$

On comparing equation (1) 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 $(0, \pm \dfrac{2 \sqrt{14}}{\sqrt{5}})$.

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

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

Length of latus rectum

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, $\dfrac{y^{2}}{4^{2}}-\dfrac{x^{2}}{7^{2}}=1$

On comparing equation (1) 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 $(0, \pm \sqrt{65})$.

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

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 ( $\pm 2,0)$, foci $( \pm 3,0)$

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 $( \pm 2,0), a=2$.

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

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

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

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

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

Answer :

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

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 $(0, \pm 5), a=5$.

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

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

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

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

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

Answer :

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

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 $(0, \pm 3), a=3$.

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

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 ( $\pm 5,0)$, 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 $( \pm 5,0), c=5$.

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 $(0, \pm 13)$, 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 $(0, \pm 13), c=13$.

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 $( \pm 3 \sqrt{5}, 0)$, 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 $( \pm 3 \sqrt{5}, 0), c= \pm 3 \sqrt{5}$.

Length of latus rectum $=8$ $\Rightarrow \dfrac{2 b^{2}}{a}=8$

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

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

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

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

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

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

$\Rightarrow a=-$ “ 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 $( \pm 4,0)$, 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 $( \pm 4,0), c=4$.

Length of latus rectum $=12$

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

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

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

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

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

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

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

$\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 :

Vertices $( \pm 7,0)$,

$ 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 $( \pm 7,0), a=7$.

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 $a^{2}+b^{2}=c^{2}$.

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

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

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

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

Answer :

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

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 $(0, \pm \sqrt{10}), c=\sqrt{10}$.

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

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

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

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

$\dfrac{9}{a^{2}}-\dfrac{4}{b^{2}}=1$

From equations (1) and (2), we obtain

$\dfrac{9}{a^{2}}-\dfrac{4}{(10-a^{2})}=1$

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

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

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

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

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

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

$\Rightarrow a^{2}=18$ 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 $\dfrac{y^{2}}{5}-\dfrac{x^{2}}{5}=1$.