Forms of equations of a hyperbola - class-XI

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

For the hyperbola x^2/9 - y^2/16 = 1, a^2 = 9 and b^2 = 16. The transverse axis is along the x-axis with length 2a = 2 * 3 = 6.

The length of the latus rectum is given by 2b^2/a = 10/3, and eccentricity e = sqrt(13)/3, with b^2 = a^2(e^2 - 1). Solving these relations gives a = 15/4, and the length of the transverse axis is 2a = 15/2 units.

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.

Transverse axis 2a = 6, so a = 3. Vertex bisects center and focus means a = ae/2, so e = 2. Then b^2 = a^2(e^2 - 1) = 9(4 - 1) = 27. Equation is x^2/9 - y^2/27 = 1, or 3x^2 - y^2 = 27.

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

This is a standard property of conics where the chord of contact of tangents from a point on one conic to another conic touches the original conic under specific conditions.

The equation represents a hyperbola. Calculating the eccentricity involves finding the angle between the asymptotes or using the general conic eccentricity formula. The result sqrt(10)/3 is the standard derivation for this specific equation.

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.

Using the condition that the tangent passes through (0, -b) and the normal through (2a*sqrt(2), 0) for a point P on the hyperbola, one can derive the eccentricity e = 2.

Using the definition of a hyperbola (distance from focus = e * distance from directrix), the equation is (x-1)^2 + (y-2)^2 = 3 * (2x+y-1)^2 / (2^2 + 1^2). Expanding this yields the provided equation.

Solving the differential equation 2xy dy/dx = x^2 + y^2 leads to the hyperbola x^2 - y^2 = c. Passing through (2, 1) gives 4 - 1 = c, so c = 3. The hyperbola is x^2 - y^2 = 3. For a rectangular hyperbola, e = sqrt(2).

Given transverse axis is 2a, so the center is at the origin and vertex is at (a, 0). The focus is at (ae, 0). Since the vertex bisects the distance between the center and the focus, a = ae / 2, which means e = 2. Using b^2 = a^2(e^2 - 1) gives b^2 = 3a^2, yielding the hyperbola equation 3x^2 - y^2 = 3a^2.

The asymptotes are 3x - y - 1 = 0 and x + 3y - 3 = 0. The angle between them is 90 degrees because the product of their slopes is -1 (3 * -1/3 = -1). A hyperbola with perpendicular asymptotes is a rectangular hyperbola, which has an eccentricity of sqrt(2).

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 $\cfrac{x^2}{4}+\cfrac{y^2}{3}=1$
$\Rightarrow a^2=4, b^2=3$

Completing the square for 3x^2 - y^2 - 24x + 4y + 4 = 0: 3(x^2 - 8x + 16) - (y^2 - 4y + 4) = -4 + 48 - 4, resulting in 3(x-4)^2 - (y-2)^2 = 40. The center is (4, 2).

For line y=mx+c to be tangent to hyperbola x^2/a^2 - y^2/b^2 = 1, we need c^2 = a^2m^2 - b^2. For real c, we need a^2m^2 > b^2. When a > b, we can find m values satisfying this condition. The question is poorly phrased but option A correctly identifies the standard hyperbola condition.

$\displaystyle x^{2}+2x-y^{2}+5= 0$
$\displaystyle x^{2}+2x+1-y^{2}= -4$
$\displaystyle (x+1)^2-y^{2}= -4$
$\displaystyle \frac{y^2}{4}-\frac{(x+1)^2}{4}= 1$
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 $(-1,\pm ae)=(-1,\pm 2\sqrt{2})$

The definition of a hyperbola is PF = e * PM, where PM is the perpendicular distance to the directrix. Squaring both sides: (x-1)^2 + (y-2)^2 = 3 * ((2x+y-1)/sqrt(2^2+1^2))^2. Expanding this leads to 5(x^2 - 2x + 1 + y^2 - 4y + 4) = 3(4x^2 + y^2 + 1 + 4xy - 4x - 2y). Simplifying results in 7x^2 + 12xy + 2y^2 - 2x + 14y - 22 = 0.

Ellipse x^2/4 + y^2/1 = 1 has a^2=4, b^2=1. e_e = sqrt(1 - 1/4) = sqrt(3)/2. Hyperbola eccentricity e_h = 2/sqrt(3). Hyperbola x^2/a^2 - y^2/b^2 = 1 passes through focus (ae_e, 0) = (sqrt(3), 0). So 3/a^2 = 1 => a^2 = 3. Then b^2 = a^2(e_h^2 - 1) = 3(4/3 - 1) = 1. Equation: x^2/3 - y^2/1 = 1, or x^2 - 3y^2 = 3.

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 $\displaystyle=\frac{2b^2}{a}=12$

$\displaystyle a= 5, ae = 7\Rightarrow a^{2}e^{2}= 49$
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.