Let $(h,k)$ be the point of intersection of the given lines. Then,

Centre of the hyperbola of form $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ has centre at orgin.

For the hyperbola of the form $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$

$\dfrac{y^2}{a^2}-\dfrac{x^2}{b^2}=1$

The given hyperbola $\dfrac{-x^2}{9} + \dfrac{y^2}{16} = 1$  or $\dfrac{x^2}{9} - \dfrac{y^2}{16} = - 1$ is a standard form of a conjugate hyperbola. 

The given hyperbola $\dfrac{x^2}{16} - \dfrac{y^2}{25} = 1$ is a standard form of hyperbola. 

Since  constant term is negative, so its major axis will be on $y$ axis.

The given hyperbola $\dfrac{-x^2}{9} + \dfrac{y^2}{16} = 1$  or $\dfrac{x^2}{9} - \dfrac{y^2}{16} = - 1$ is a standard form of a conjugate hyperbola. 

The given equation of hyperbola can be re-written as,

Hyperbola $\dfrac{y^2}{a^2}-\dfrac{x^2}{b^2}=1$

The given hyperbola is $-\dfrac{(x-1)^2}{3} + \dfrac{(y+2)^2}{16} = 1$ is a conjugate hyperbola.

The given hyperbola is $-\dfrac{(x-1)^2}{3} + \dfrac{(y+2)^2}{16} = 1$ is a conjugate hyperbola.

The equation at the conjugate axis of the hyperbola

We have given  $\cfrac{a}{ae-a}=2\Rightarrow e=\cfrac{3}{2}$
Using $e^2=1+\cfrac{b^2}{a^2}$
$\cfrac{9}{4}=1+\cfrac{b^2}{a^2}\Rightarrow b^2=\cfrac{5}{4}a^2$
Hence required hyperbola is $\cfrac{x^2}{a^2}-\cfrac{y^2}{b^2}=1$
$\cfrac{x^2}{a^2}-\cfrac{y^2}{\frac{5}{4}a^2}=1$
$\Rightarrow 5x^2-4y^2=5a^2$
Hence option 'D' is correct.

The centre of the hyperbola is the mid-point of the line joining the two
foci. So, the coordinates of the centre are $\left( \cfrac { 8+0 }{ 2 },\cfrac { 3+3 }{ 2 }  \right) \quad $ i.e $(4,3)$
Let $2a,2b$ be the length of the transverse and conjugate axes and let $e$ be the
eccentricity. Then, the equation of the hyperbola is
$\cfrac { {\left( x-4 \right)  }^{ 2 } }{ { a }^{ 2 } } -\cfrac { { \left( y-3 \right)  }^{ 2 } }{ { b }^{ 2 } } =1\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (i)$
Distance between the two foci$ = 2ae$
$\Rightarrow \sqrt { { \left( 8-0 \right)  }^{ 2 }+{ \left( 3-3 \right)  }^{ 2 } } =2ae$
$\Rightarrow ae=4\Rightarrow a=3$
$\therefore\quad { b }^{ 2 }={ a }^{ 2 }\left( { e }^{ 2 }-1 \right) \Rightarrow {b }^{ 2 }=9\left( \cfrac { 16 }{ 9 } -1 \right) =7$
Substituting the value of $a$ and $b$ in $(i)$, we find that the equation of the hyperbola is
$\cfrac{ { \left( x-4 \right)  }^{ 2 } }{ 9 } +\cfrac { { \left( y-3 \right)  }^{ 2 } }{ 7 } =1\quad or\quad 7{ \left( x-4 \right)  }^{ 2 }-9{ \left( y-3 \right)  }^{ 2 }=63$
Hence, option 'A' is correct.

Using Definition of hyperbola
$PS^2=e^2\cdot PM^2$
$(x-2)^2+(y-1)^2=2^2\left(\cfrac{x+2y-1}{\sqrt{5}}\right)^2$
$5(x^2+y^2-4x-2y+5)=4(x^2+4y^2+1+4xy-2x-4y)$
$\Rightarrow x^2 - 16 xy - 11y^2 - 12 x + 6y + 21 = 0$
Hence, option 'B' is correct.

We have ${ e } _{ 1 }=\sqrt { 1-\cfrac { 16 }{ 25 }  } =\cfrac { 3 }{ 5 } $
$\because \quad { e } _{ 1 }{ e } _{ 2 }=1\Rightarrow { e } _{ 2 }=\cfrac { 5 }{ 3 } $
Clearly y-axis is transverse axis of the ellipse.
Thus, coordinates of foci of the ellipse are $(0,\pm b e _1)$ or $\left( 0,\pm 3 \right) $.
Let the equation of hyperbola is, $\dfrac{y^2}{b^2}-\cfrac{x^2}{a^2}=1$ ..... $(1)$
Since, hyperbola passes through foci of the ellipse
$\Rightarrow b^2=9$ and also $a^2=b^2(e^2-1)=9(25/9-1)=16$
Therefore, the required hyperbola is, $\cfrac{x^2}{16}-\cfrac{y^2}{9}=-1$
Hence, option 'B' is correct.

Given $a=3$ and $ae=5\Rightarrow e=\cfrac{5}{3}$
Using $e^2=1+\cfrac{b^2}{a^2}$
$\cfrac{25}{9}=1+\cfrac{b^2}{9}\Rightarrow b^2=16$
Therefore required hyperbola is, $\cfrac{x^2}{9}-\cfrac{y^2}{16}=1$
$\Rightarrow 16x^2-9y^2=144$
Hence, option 'A' is correct.

Using Definition of hyperbola
$PS^2=e^2\cdot PM^2$
$(x-1)^2+(y+1)^2=2\left(\cfrac{x-y+1}{\sqrt{2}}\right)^2$
$(x^2+y^2-2x+2y+2)=(x^2+y^2+1-2xy+2x-2y)$
$\Rightarrow 2 xy - 4 x + 4y + 1 = 0$
Hence, option 'C' is correct.

Given hyperbola is,  $\displaystyle \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}}=1$
It passes through $(2,3)$
$\cfrac{4}{a^{2}} - \cfrac{9}{b^{2}}=1 ..(1)$
Also eccentricity is $2$,
$\Rightarrow e^2=1+\cfrac{b^2}{a^2}=4\Rightarrow \cfrac{b^2}{a^2}=3   ..(2)$
Solving (1) and (2) we get $a=1, b=\sqrt{3}$
Hence, length of transverse axis is $=2a=2$

Given eccentricity of the hyperbola $e=\sqrt{3}$
and distance between focii is 9. $\Rightarrow 2ae=9\Rightarrow a=\cfrac{3\sqrt{3}}{2}$
also $b^2=a^2(e^2-1)=\cfrac{27}{4}(3-1)=\cfrac{27}{2}\Rightarrow b=\cfrac{3\sqrt{3}}{\sqrt{2}}$
Hence equation of required hyperbola is,
$\cfrac{x^2}{\left(\cfrac{3\sqrt{3}}{2}\right )^2}-\cfrac{y^2}{\left (\cfrac{3\sqrt{3}}{\sqrt{2}}\right )^2}=1$
Hence, option 'B' is correct.

We have   $x=5\tan\phi, y=4\sec\phi$
Eliminating $\phi$ we get
$\cfrac{y^2}{16}-\cfrac{x^2}{25}=\sec^2\phi-\tan^2\phi=1$
This is a hyperbola. Therefore, its eccentricity $= \sqrt{1+\cfrac{25}{16}}=\cfrac{\sqrt{41}}{4}$

For hyperbola $\displaystyle \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$
eccentricity $\Rightarrow e=\sqrt{1+\cfrac{b^2}{a^2}}\Rightarrow \cfrac{b^2}{a^2}=e^2-1$
For hyperbola $\displaystyle \frac{x^{2}}{b^{2}} - \frac{y^{2}}{a^{2}} = 1$
Required eccentricity $ e'=\sqrt{1+\cfrac{a^2}{b^2}}=\sqrt{1+\cfrac{1}{e^2-1}}=\displaystyle \cfrac{e}{\sqrt{e^{2} - 1}}$

Given ellipse may be written as $\cfrac{x^2}{4}+\cfrac{y^2}{3}=1$
$\Rightarrow a^2=4, b^2=3$

$\displaystyle x^{2}+2x-y^{2}+5= 0$
$\displaystyle x^{2}+2x+1-y^{2}= -4$
$\displaystyle (x+1)^2-y^{2}= -4$
$\displaystyle \frac{y^2}{4}-\frac{(x+1)^2}{4}= 1$
Clearly this equation of rectangular hyperbola with $y-$axis as major axis
eccentricity $e = \sqrt{2}$ directrices $:y = \cfrac{a}{e}=\pm \sqrt{2}$ foci $(-1,\pm ae)=(-1,\pm 2\sqrt{2})$

Equation of hyperbola is $\displaystyle \frac { { x }^{ 2 } }{ 3/\sqrt { 1999 }  } -\frac { { y }^{ 2 } }{ 3/\sqrt { 1999 }  } =1$

Centre of the ellipse $=$ mid point of foci $=(1,5)$

$4x^2\, -\, 9y^2\, -\, 8x\, =\, 32$
$\Rightarrow 4(x^2-2x)-9y^2=32$
$\Rightarrow 4(x^2-2x+1)-9y^2=32+4=36$
$\Rightarrow \cfrac{(x-1)^2}{9}-\cfrac{y^2}{4}=1$
$\Rightarrow a^2=9, b^2=4$
$\therefore e=\sqrt{1+\cfrac{b^2}{a^2}}=\cfrac{\sqrt{13}}{3} $
Hence, option 'B' is correct.

Centre of hyperbola is $(5, 0)$, so equation is
$\displaystyle \frac{(x\, -\, 5)^2}{a^2}\, -\, \frac{y^2}{b^2}\, =\, 1$
$a\, =\, 5,\, ae\, -\, a\, =\, 8\, \Rightarrow\, e\, =\, \displaystyle \frac{13}{5}$

$b^2\, =a^2(e^2-1)=\, 144$
So required equation is,  $\displaystyle \frac{(x\, -\, 5)^2}{25}\, -\, \frac{y^2}{144}\, =\, 1$
Hence, option 'B' is correct.

Given hyperbolas may be written as,  $\displaystyle \frac{x^2}{9}-\frac{y^2}{4}=1$
$\Rightarrow a^2 =9, b^2=4$
$\therefore$ Eccentricity is, $\displaystyle e= \sqrt{1+\frac{b^2}{a^2}}=\frac{\sqrt{13}}{3}$
Thus, length of axes are $2a$ and $2b \Rightarrow 6 $ and $4$
Focus $\equiv (\pm ae, 0) =(\pm \sqrt{13},0)$
And length of latus rectum $=\cfrac{2b^2}{a}=\cfrac{8}{3}$

Using Hyperbola definition, $PS^2=e^2.PM^2$
$\displaystyle (x-a)^2+(y-0)^2=\frac{25}{16}\left| \frac{4x-3y-a}{\sqrt{3^2+4^2}} \right|^2$
$\Rightarrow 16(x^2+y^2-2ax+a^2)=16x^2+9y^2+a^2-24xy+6ya-8ax$
$\Rightarrow 7y^2\, +\, 24xy\, -\, 24ax\, -\, 6ay\, +\, 15a^2\, =\, 0$

Given $a=4, ae=6\Rightarrow e=\cfrac{3}{2}\Rightarrow \cfrac{b^2}{a^2}=e^2-1=\cfrac{9}{4}-1=\cfrac{5}{4}\Rightarrow b^2=20$
Therefore, required hyperbola is, $\cfrac{x^2}{16}-\cfrac{y^2}{20}=1$
$\Rightarrow 5x^2-4y^2=80$
Hence, option 'C' is correct.

The equation of hyperbola is $\displaystyle \frac { { x }^{ 2 } }{ 144 } -\frac { { y }^{ 2 } }{ 81 } =\frac { 1 }{ 25 } $

Point of intersection of lines
$7x + 13y - 87 = 0$ & $5x - 8y + 7 = 0$ is $(5, 4)$.
Also this point lies on the given hyperbola
$\therefore \displaystyle \frac{25}{a^2}\, -\, \frac{16}{b^2}\, =\, 1$ ......(1)
Also latus rectum $\displaystyle LR\, =\, \frac{2b^2}{a}\, =\, \frac{32 \sqrt{2}}{5}$
$\displaystyle \Rightarrow\, b^2\, =\, \frac{16 \sqrt{2}a}{5}$ .....(ii)
From (i) & (ii) $\displaystyle a^2\, =\, \frac{25}{2},\, b^2\, =\, 16$.

Given hyperbola can be written as

Given ellipse may be written as, $\displaystyle \frac {x^2}{4}+\frac {y^2}{3}=1$
$\displaystyle \Rightarrow a^2 = 4, b^2 = 3\therefore e = \sqrt{1-\frac{3}{4}}=\frac{1}{2}$
Thus foci of ellipse $(\pm 1, 0)$
Hence foci of the hyperbola is $(\pm 1, 0)$
And semi-major axis $a =\cfrac{1}{\sqrt{2}}$
$\Rightarrow \pm 1=\sqrt {a^2+b^2}\Rightarrow b^2=\frac {1}{2}$
or $b^2=1-a^2=1-\frac {1}{2}=\frac {1}{2}$
The required equation of hyperbola is,
$\displaystyle \frac {x^2}{1/2}-\frac {y^2}{1/2}=1\Rightarrow 2x^2-2y^2=1$

Given eccentricity of the hyperbola is $\sqrt{2}$
Hence hyperbola is rectangular $a=b$
Also given distance between focii is 16. $\Rightarrow 2ae = 16\Rightarrow a =4\sqrt{2}$
Hence required hyperbola is given by, $x^2-y^2=a^2=32$

Equation of hyperbola is $\displaystyle \frac { { x }^{ 2 } }{ 49 } -\frac { { y }^{ 2 } }{ 9 } =1$

$9x^2\, -\, 16y^2\, -\, 18x\, +\, 32y\, -\, 151\, =\, 0$
$9(x^2-2x)-16(y^2-2y)=151$
$9(x^2-2x+1)-16(y^2-2y+1)=151-7=144$
$9(x-1)^2-16(y-1)^2=144$
$\cfrac{(x-1)^2}{16}-\cfrac{(y-1)^2}{9}=1$
$\Rightarrow a^2=16, b^2=9$
$\therefore e=\sqrt{1+\dfrac{b^2}{a^2}}=\cfrac{5}{4}$

$\therefore$ The length of the transverse axis is $=2a=8$
Length of latus rectum is $=2\cfrac{b^2}{a}=\cfrac{9}{2}$
Equation of directrix is, $x=1\pm \cfrac{a}{e}=1 \pm \cfrac{16}{5}=\cfrac{21}{5}$ or $-\cfrac{11}{5}$
Hence, option 'C' is correct.

As ellipse and hyperbola intersect orthogonally 
$\displaystyle \Rightarrow $ their foci are coincident
Now we have $\displaystyle SP+S,P=2a\Rightarrow $ length of major axis ellipse ..........(1)
And $\displaystyle \left | SP-S,P \right |=2a'\Rightarrow $ length of transverse axis of hyperbola ....... (2)
$\displaystyle (1)^{2}-(2)^{2}\Rightarrow 4PS'PS=4(a^{2}-a'^{2})=4\left [ \left ( \frac{a'e}{e} \right )^{2} -a'^{2}\right ]=4\left ( 2-\frac{1}{2} \right )=6$
$\displaystyle \Rightarrow 2x = 3$

It is given that $2a=\sqrt {2}$ which implies that $a=\dfrac { 1 }{ \sqrt { 2 }  }$.

(A) $\dfrac{x^{2}}{4}-y^{2}=1$

Let the equation of hyperbola is $\dfrac {x^{2}}{a^{2}} - \dfrac {y^{2}}{b^{2}} = 1$
Given, $e = 2, 2ae = 8$
$\Rightarrow ae = 4\Rightarrow a = 2$
Now, $b^{2} = a^{2} (e^{2} - 1)$
$\Rightarrow b^{2} = 4(4- 1)$
$\Rightarrow b^{2} = 12$
$\therefore$ Equation of hyperbola is
$\dfrac {x^{2}}{4} - \dfrac {y^{2}}{12} = 1$

• the equation of hyperbola is $\dfrac { { x }^{ 2 }+4x+4 }{ 25 } -\dfrac { { y }^{ 2 }-6x+9 }{ 16 } =1$
• $\dfrac { { (x+2) }^{ 2 } }{ 25 } -\dfrac { { (y-3) }^{ 2 } }{ 16 } =1$
• Therefore the center is $(-2,3)$

Given, hyperbola is conjugate hyperbola of $\dfrac { { (x-1) }^{ 2 } }{ 3 } -\dfrac { { (y+2) }^{ 2 } }{ 16 } =-1$

General equation of hyperbola is $\dfrac{y^2}{b^2}-\dfrac{x^2}{a^2}=1$

$Vertices(\pm 5,0)\quad Foci(\pm 7,0)$ 

$Given\quad :\quad 16{ x }^{ 2 }−9{ y }^{ 2 }=144\quad \quad \quad \ Or,\quad \frac { { x }^{ 2 } }{ 9 } -\frac { { y }^{ 2 } }{ 16 } =1\ We\quad know,\ be=\sqrt { { a }^{ 2 }+{ b }^{ 2 } } \quad \quad \quad \quad \quad \because (b>a)\ Or,\quad be=\sqrt { 9+16 } \ Or,\quad be=\pm 5\ Or,\quad \frac { b }{ e } \quad =\frac { 16 }{ \pm 5 } \ We\quad know\quad equation\quad of\quad directrix\quad is\quad y=\frac { b }{ e } \ \therefore \quad 5y\pm 16=0$

$Given\quad :\quad 16{ x }^{ 2 }−9{ y }^{ 2 }=144\quad \quad \quad \ Or,\quad \frac { { x }^{ 2 } }{ 9 } -\frac { { y }^{ 2 } }{ 16 } =1\ We\quad know,\ be=\sqrt { { a }^{ 2 }+{ b }^{ 2 } } \quad \quad \quad \quad \quad \because (b>a)\ Or,\quad be=\sqrt { 9+16 } \ Or,\quad be=\pm 5\ \therefore \quad Focii\quad is\quad (0,\pm 5)\quad $

$\displaystyle \frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 9 } =1$
which is an ellipse.
For ellipse, $a=5$ , $b=3$ 
$\Rightarrow \displaystyle e=\sqrt {\frac {25-9}{25}}=\frac {4}{5}$
$\therefore ae=4$
Hence, the foci are $(-4, 0)$ and $(4, 0)$.
For the hyperbola,
$ae=4, e=2$
$\Rightarrow a=2$
$b^2=4(4-1)=12$
$\Rightarrow b=\sqrt {12}$
$\Rightarrow a^2+b^2=16$
Since $b^2>a^2$, so there is no director circle to the hyperbola.
Length of latus rectum $\displaystyle=\frac{2b^2}{a}=12$

Let $S\equiv 14x^2-4xy+11y^2-44x-58y+71$
To know centre of this hyperbola, $\cfrac{\partial S}{\partial x}=0$ and $\cfrac{\partial S}{\partial y}=0 $
$\Rightarrow \cfrac{\partial S}{\partial x}=28x-4y-44=0\Rightarrow 7x-y-11=0  ..(1)$
and $\cfrac{\partial S}{\partial y}=-4x+22y-58=0\Rightarrow 2x-11y+29=0  ..(2)$
Solving $(1)$ and $(2)$ we get required centre $(2,3)$
Hence, option 'A' is correct.

$\displaystyle a= 5, ae = 7\Rightarrow a^{2}e^{2}= 49$
or $\displaystyle a^{2}\left ( 1+\frac{b^{2}}{a^{2}} \right )= 49$ 

Eccentricity of the hyperbola is $\sqrt{2}$ as it is a rectangular hyperbola.