Tag: hyperbola

Questions Related to hyperbola

If $x + 2 = 0$ and $y = 1$ are the equation of asymptotes of rectangular hyperbola passing through (1,0).Then which of the following is(are) not the equation(s) of hyperbola :

mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

  1. $xy + 2y -1 = 0$

  2. $xy - 2y + 1 = 0$

  3. $xy - 2y - 1 = 0$

  4. $xy-x+2y+1=0$

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Correct Option: A,B,C
Explanation:

Equation of hyperbola is of the form $(x+2)(y-1)=k$
Since, it passes through $(1,0)$
Therefore, $(1+2)(0-1)=k$
$\Rightarrow k=-3$
Therefore, equation of hyperbola is $xy-x+2y+1=0$

If ax + by + c = 0 and $\displaystyle \varphi \chi $ + my + n = 0 are asymptotes of a hyperbola, then: 

mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

  1. $\displaystyle am\neq b\varphi $

  2. $\displaystyle \frac{am+b\varphi }{a\varphi +bm}\neq 0$

  3. $\displaystyle a\varphi \neq bm$

  4. none of these

Show answer
Correct Option: A
Explanation:

Asymptotes of Hyperbola are Intersecting each other 


So, These line will be anti-parallel or intersecting

Condition for intersecting lines is $\dfrac{a}{b}\neq\dfrac{\varphi}{m}\Rightarrow am \neq b\varphi$

If $\theta$ is the angle between the asymptotes of the hyperbola $\displaystyle \frac{x^2}{a^2}\, -\, \displaystyle \frac{y^2}{b^2}\, =\, 1$ with eccentricity $e$, then $\sec \displaystyle  \frac{\theta}{2}$can be

mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

  1. $e$

  2. $\dfrac{e}2$

  3. $\dfrac{e}3$

  4. $\displaystyle \frac{e}{\sqrt{e^2\, -\, 1}}$

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Correct Option: A,D
Explanation:

$\tan\, \displaystyle \frac{\theta}{2}\, =\, \displaystyle \frac{b}{a}\,

\Rightarrow\, e^2\, -\, 1\, =\, \tan^2\, \displaystyle

\frac{\theta}{2}\, \Rightarrow\, \sec \displaystyle \frac{\theta}{2}\,

=\, e$
or $e^2\, -\, 1\, =\, \cot^2\, \displaystyle \frac{\theta}{2}\, \Rightarrow\, co\sec\, \displaystyle \frac{\theta}{2}\, =\, e$
$\Rightarrow\, \sec\, \displaystyle \frac{\theta}{2}\, =\, \displaystyle \frac{e}{\sqrt{e^2\, -\, 1}}$.

The asymptotes of a hyperbola are parallel to lines $2x + 3y = 0$ and $3x + 2y = 0.$ The hyperbola has its centre at $(1, 2)$ and it passes through $(5, 3).$ Find its equation.

mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

  1. $(2x\, +\, 3y\, -\, 8) (3y\, +\, 2y\, -\, 7)\, =\, 154$

  2. $(2x\, +\, 3y\, -\, 7) (3y\, +\, 2y\, -\, 8)\, =\, 154$

  3. $(2x\, +\, 3y\, -\, 7) (3y\, +\, 2y\, -\, 8)\, =\, 127$

  4. $(2x\, +\, 3y\, -\, 8) (3y\, +\, 2y\, -\, 7)\, =\, 127$

Show answer
Correct Option: A
Explanation:

let the equation of asymtotes be $2x+3y=a$ and $3x+2y=b$
both asymtotes intersect at centre $(1,2)$
Therefore, $a=2+3(2)=8$ and $b=3+2(2)=7$
now, the equation of hyperbola is of the form $(2x+3y-8)(3x+2y-7)=k$
It passes through $(5,3)$
Therefore, $(2(5)+3(3)-8)(3(5)+2(3)-7)=k$
Thus $k=154$

The asymptotes of the hyperbola $xy+3x+2y = 0$ are

mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

  1. $x - 2 = 0$ and $y - 3 = 0$

  2. $x - 3 = 0$ and $y - 2 = 0$

  3. $x + 2 = 0$ and $y + 3 = 0$

  4. $x + 3 = 0$ and $y + 2 = 0$

Show answer
Correct Option: C
Explanation:

Let equation of asymptotes be $xy\,+ \, 3x\, \, +\, 2y +\, \lambda$ = 0.
Then $abc\, +\, 2fgh\,-\, af^2\,-\, bg^2\,-\, ch^2\, =\, 0$
$\displaystyle \Rightarrow\, \frac{3}{2}\, -\, \frac{\lambda}{4}\, =\, 0\, \Rightarrow\, \lambda\, =\, 6$
$\therefore$ Equation of asymptotes is $xy +3x +2y + 6 = 0$
$\Rightarrow (x+ 2) (y +3) = 0\Rightarrow x+2=0$ and $y+3=0$

Find the asymptotes of the hyperbola $2x^2\, -\, 3xy\,- \, 2y^2\, +\, 3x\,- \, y\, +\, 8\, =\, 0$. Also find the equation to the conjugate hyperbola & the equation of the principal axes of the curve.

mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

  1. $x - 2y + 1 = 0; 2x + y + 1 = 0; 2x^2\,- \, 3xy\, -\, 2y^2\, +\, 3x\,- \, y\,- \, 6\, =\, 0; 3x y + 2 = 0; x - 3y = 0$

  2. $x + 2y - 1 = 0; 2x + y + 1 = 0; 2x^2\,- \, 3xy\, -\, 2y^2\, +\, 3x\,- \, y\,+\, 6\, =\, 0; 3x y + 2 = 0; x + 3y = 0$

  3. $x - 2y + 1 = 0; 2x + y + 1 = 0; 2x^2\,- \, 3xy\, -\, 2y^2\, +\, 3x\,- \, y\,- \, 6\, =\, 0; 3x y + 2 = 0; x + 3y = 0$

  4. $x - 2y + 1 = 0; 2x - y + 1 = 0; 2x^2\,- \, 3xy\, -\, 2y^2\, +\, 3x\,- \, y\,+ \, 6\, =\, 0; 3x y - 2 = 0; x - 3y = 0$

Show answer
Correct Option: C
Explanation:

Let equation of asymptotes are 
$2x^2\, \, -3xy\, \,- 2y^2\, +\, 3x\, \, -y\, +\, 8\, +\, \lambda\,=\, 0$
As it represents two straight lines
$\displaystyle \therefore\, -4(8\, +\, \lambda)\, +\, \frac{9}{4}\, -\, \frac{1}{2}\, +\, \frac{9}{2}\, -\, (8\, +\, \lambda) \frac{9}{4}\, =\, 0$
$\Rightarrow\, \lambda\, =\, -7$
So asymptotes are $2x^2\, -\, 3xy\, -\, 2y^2\, +\, 3x\, -\, y\, +\, 1\, =\, 0$
$\Rightarrow$ 2y - x - 1 = 0 & 2x + y + 1 = 0
and the equation of conjugate hyperbola will be
$2x^2\, \, -3xy\, \, -2y^2\, +\, 3x\, \, -y\, +\, 8\, \, -14\, =\, 0$.

Any straight line parallel to an asymptote of a hyperbola intersects the hyperbola at 

mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

  1. one point

  2. two points

  3. three points

  4. four points

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Correct Option: A
Explanation:

One point. 

Assertion(A): The angle between the asymptotes of $3x^{2}-y^{2}=3$ is $120^{\circ}$
Reason(R): The angle between the asymptotes of $x^{2}-y^{2}=a^{2}$ is $90^{\circ}$

mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

  1. Both A and R are true and R is the correct

    explanation of A.

  2. Both A and R are true but R is not correct

    explanation of A.

  3. A is true but R is false.

  4. A is false but R is true.

Show answer
Correct Option: D
Explanation:

Asymptotes of a hyperbola is given by $y=\pm \frac { x }{ a } $


So for hyperbola $3{ x }^{ 2 }-{ y }^{ 2 }=3$
The asymptotes make an angle of ${ 60 }^{ o }$ and  ${ 120 }^{ o }$ with x-axis which means they make an angle of ${ 60 }^{ o } $ among themself.

Now for hyperbola ${ x }^{ 2 }-{ y }^{ 2 }={a}^{2}$

The asymptotes makes and angle of ${45}^{o}$ and  ${135}^{0}$ which means ${90}^{o}$ among themself.

If $e$ is the eccentricity of $\displaystyle \frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1$ and $\theta$ be the angle between the asymptotes then $\displaystyle \sec { \frac { \theta  }{ 2 }  } $ equals :

mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

  1. ${ e }^{ 2 }$

  2. $\displaystyle \frac { 1 }{ e } $

  3. $2e$

  4. $e$

Show answer
Correct Option: D
Explanation:

Equation of asymptotes to $\displaystyle \frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1$ are given by 

$\displaystyle y=-\frac { b }{ a } x$
$\displaystyle \therefore { m } _{ 1 }=-\frac { b }{ a } $
Similarly $\displaystyle y=\frac { bx }{ a } $
$\therefore \displaystyle { m } _{ 2 }=\frac { b }{ a } $
Now $\displaystyle \theta =2\tan ^{ -1 }{ \frac { b }{ a }  } $
$\displaystyle \Rightarrow \tan { \frac { \theta  }{ 2 }  } =\frac { b }{ a } \Rightarrow \tan ^{ 2 }{ \frac { \theta  }{ 2 }  } =\frac { { b }^{ 2 } }{ { a }^{ 2 } } ={ e }^{ 2 }-1$
$\displaystyle \Rightarrow \sec ^{ 2 }{ \frac { \theta  }{ 2 }  } ={ e }^{ 2 }\Rightarrow \sec { \frac { \theta  }{ 2 }  } =e$

The equation of hyperbola conjugate to the hyperbola $2x^2 + 3xy - 2y^2 - 5 + 5y + 2 = 0$ is

mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

  1. $2x^2 + 3xy - 2y^2 - 5x + 5y - 8 = 0$

  2. $x^2 + 3xy - 2y^2 - 5x + 5y + 8 = 0$

  3. $2x^2 + 3xy - 2y^2 + 5x - 5y - 8 = 0$

  4. None of these

Show answer
Correct Option: A
Explanation:

Let the given hyperbola be $ H =2x^{2}+3xy-2y^{2}-5x+5y+2=0$


Thus the pair of asymptotes be $ A = 2x^{2}+3xy-2y^{2}-5x+5y+\lambda =0 $

Pair of straight line has $ \Delta =0 $
$\Delta = abc+ 2fgh-af^{2}-bg^{2}-ch^{2}=0 $

where,

$a =2 $ 
$b=-2 $
$c=\lambda $
$f=\dfrac{5}{2}$
$g=-\dfrac{5}{2}$
$h=\dfrac{3}{2}$


Thus, $\lambda=-3 $

$A= 2x^{2}+3xy-2y^{2}-5x+5y-3 =0$

Since  $ H+C=2A $  Where $ C $ be the equation of conjugate hyperbola

$C= 2A-H $
So,the equation of conjugate hyperbola be  $ C=2x^{2}+3xy-2y^{2}-5x+5y-8 =0 $