Tag: hyperbola

Questions Related to hyperbola

The equations of the transverse and conjugate axes of a hyperbola are respectively $x + 2y - 3 = 0, 2x - y + 4 = 0$ and their respective lengths are $\displaystyle \sqrt{2}$ 2/$\displaystyle \sqrt{2}$. The equation of the hyperbola is 

mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

  1. $\displaystyle \frac{2}{5}(x+2y-3)^{2}-\frac{3}{5}(2x-y+4)^{2}=1$

  2. $\displaystyle \frac{2}{5}(2x+y-4)^{2}-\frac{3}{5}(x+2y3-4)^{2}=1$

  3. $\displaystyle 2(2x-y+4)^{2}-3(x+2y-3)^{2}=1$

  4. $\displaystyle 2(2x+2y-3)^{2}-3(2x-y+4)^{2}=1$

Show answer
Correct Option: B
Explanation:

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


Also, $2b=\dfrac { 2 }{ \sqrt { 3 }  } \Rightarrow b=\dfrac { 1 }{ \sqrt { 3 }  }$ 

If we take the two axes as the new coordinate system and the point of intersection of the axes of the new origin, then in the new coordinate system, equation of the hyperbola will be:

$\dfrac { { X }^{ 2 } }{ a^{ 2 } } -\dfrac { { Y }^{ 2 } }{ b^{ 2 } } =1\ \Rightarrow \dfrac { { X }^{ 2 } }{ \left( \dfrac { 1 }{ \sqrt { 2 }  }  \right) ^{ 2 } } -\dfrac { { Y }^{ 2 } }{ \left( \dfrac { 1 }{ \sqrt { 3 }  }  \right) ^{ 2 } } =1\ \Rightarrow \dfrac { { X }^{ 2 } }{ \dfrac { 1 }{ 2 }  } -\dfrac { { Y }^{ 2 } }{ \dfrac { 1 }{ 3 }  } =1\ \Rightarrow 2{ X }^{ 2 }-3{ Y }^{ 2 }=1\quad ....(1)$

Let $P(x,y)$ be the coordinates of a point on the hyperbola in original x-y system, then 

$X=\dfrac { \left| 2x-y+4 \right|  }{ \sqrt { 5 }  } ,\quad Y=\dfrac { \left| x+2y-3 \right|  }{ \sqrt { 5 }  } $

($\because$ $X$ is the distance of a point on hyperbola from $2x-y+4=0$ and $Y$ is the distance of a point on hyperbola from $x+2y-3=0$)

Therefore, equation 1 becomes:

$2{ \left( \dfrac { \left| 2x-y+4 \right|  }{ \sqrt { 5 }  }  \right)  }^{ 2 }-3{ \left( \dfrac { \left| x+2y-3 \right|  }{ \sqrt { 5 }  }  \right)  }^{ 2 }=1\ \Rightarrow \dfrac { 2 }{ 5 } { \left( 2x-y+4 \right)  }^{ 2 }-\dfrac { 3 }{ 5 } { \left( x+2y-3 \right)  }^{ 2 }=1$

Hence, the equation of the hyperbola is $\dfrac { 2 }{ 5 } { \left( 2x-y+4 \right)  }^{ 2 }-\dfrac { 3 }{ 5 } { \left( x+2y-3 \right)  }^{ 2 }=1$.

For different values of k if the locus of point of intersection of the lines $\sqrt{3}x-y-4\sqrt{3}k=0,\ \sqrt{3}kx+ky-4\sqrt{3}=0$ represents the hyperbola then the equations of latusrectam are

mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

  1. $x=\pm 8$

  2. $x=\pm\sqrt{2}$

  3. $y=\pm 8$

  4. $y=\pm 4\sqrt{2}$

Show answer
Correct Option: A

MATCH THE FOLLOWING
Hyperbola                                                   Length of latusrectum
A}$x^{2}-4y^{2}=4$                                               1. 1
B}$25x^{2}-16y^{2}=400$                                     2.12
C}$ 2x^{2}-y^{2}-4x-4y-20=0$                   3.9/2
D)$9x^{2}-16y^{2}+72x-32y-16=0$           4. 25/2

The correct match is

mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

  1. I II III IV

    1 2 3 4

  2. 1 4 2 3

  3. 3 1 2 4

  4. 2 3 4 1

Show answer
Correct Option: B
Explanation:

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


$LR=\dfrac{2b^{2}}{a}$

$=\dfrac{2}{2}$

$=1$

(B)$\dfrac{x^{2}}{16}-\frac{y^{2}}{25}=1$

$b=5, a=4$

$LR=\dfrac{2.25}{4}$

$=\dfrac{25}{2}$

(C) $2(x-1)^{2}-(y-2)^{2}=18$

$\dfrac{(x-1)^{2}}{9}-\frac{(y-2)^{2}}{18}=1$

$b^{2}=18, a=3$

$LR=\dfrac{2\times 18}{3}$

$=12$

(D) $\dfrac{(x+4)^{2}}{16}-\frac{(y+1)^{2}}{9}=1$

$b^{2}=9$

$a=4$

$LR=2\times \dfrac{9}{4}$

$=\dfrac {9}{2}$

The equation to the hyperbola having its eccentricity $2$ and the distance between its foci is $8$, is

mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

  1. $\dfrac {x^{2}}{12} - \dfrac {y^{2}}{4} = 1$

  2. $\dfrac {x^{2}}{4} - \dfrac {y^{2}}{12} = 1$

  3. $\dfrac {x^{2}}{8} - \dfrac {y^{2}}{2} = 1$

  4. $\dfrac {x^{2}}{16} - \dfrac {y^{2}}{9} = 1$

Show answer
Correct Option: B
Explanation:

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 centre of the hyperbola $\dfrac {x^{2} + 4x + 4}{25} - \dfrac {y^{2} - 6x + 9}{16} = 1$ is: 

mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

  1. $(-4, -9)$

  2. $(-2, 3)$

  3. $(2, -3)$

  4. $(5, 4)$

  5. $(25, 16)$

Show answer
Correct Option: B
Explanation:
  • 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 $(-2,3)$

For hyperbola  $-\dfrac{(x-1)^2}{3}+\dfrac{(y+2)^2}{16}=1$ distance between directrices is ?

mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

  1. $\dfrac{2}{\sqrt{19}}$

  2. $\dfrac{3}{\sqrt{19}}$

  3. $\dfrac{4}{\sqrt{19}}$

  4. $\dfrac{32}{\sqrt{19}}$

Show answer
Correct Option: D
Explanation:
Comparing the equation of given hyperbola with the standard equation
$\dfrac{(y-k)^2}{a^2}-\dfrac{(x-h)^2}{b^2}=1$
$h=1,k=-2,a^2=16,b^2=3$

$e=\sqrt{1+\dfrac{b^2}{a^2}}=\dfrac{\sqrt{19}}{4}$

Distance between the directrices $=\dfrac{2a}{e}=\dfrac{32}{\sqrt{19}}$

For hyperbola  $-\dfrac{(x-1)^2}{3}+\dfrac{(y+2)^2}{16}=1$ vertices are

mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

  1. $(\pm\sqrt3,0)$

  2. $(\pm\sqrt3+1,-2)$

  3. $(\pm1,-2)$

  4. $(0,0)$

Show answer
Correct Option: B
Explanation:

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

So the vertices of given hyperbola are
${ (x-1) }^{ 2 }=3,{ (y+2) }^{ 2 }=0\ \Rightarrow x-1=\pm \sqrt { 3 } ,y+2=0\ \Rightarrow x=1\pm \sqrt { 3 } ,y=-2\ \Rightarrow \left( 1\pm \sqrt { 3 } ,-2 \right) $
So, option B is correct.

Find the equation to the hyperbola, referred to its axes as axes of coordinates, whose transverse axis is $7$ and which passes through the point $\left( 3,-2 \right) $.

mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

  1. $65y^2-16x^2=196$

  2. $65y^2-14x^2=196$

  3. $85y^2-16x^2=196$

  4. $85y^2-16x^2=147$

Show answer
Correct Option: C
Explanation:

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

Length of transverse axis is $2a$.
So, $2a=7$
$\Rightarrow a=\dfrac{7}{2}$
Equation becomes,
$\dfrac{y^2}{b^2}-\dfrac{4x^2}{49}=1$
It passes through $(3,-2)$, so it should satisfy the parabola,
$\dfrac{4}{b^2}-\dfrac{36}{49}=1$
On solving, we get 
$85y^2-16x^2=196$

Equation of the hyperbola with vertices at $(\pm 5, 0)$ and foci at $(\pm 7, 0)$ is

mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

  1. $24x^2-25y^2=600$

  2. $25x^2-24y^2=600$

  3. $\displaystyle \frac{x^2}{25}-\frac{y^2}{24}=1$

  4. $\displaystyle \frac{x^2}{24}-\frac{y^2}{25}=1$

Show answer
Correct Option: A,C
Explanation:

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

$a=\pm 5$ and $ae=\pm 7$ 
And $e=\dfrac { 7 }{ 5 } $ 
We know $e=\sqrt { 1+\frac { { b }^{ 2 } }{ { a }^{ 2 } }  }$ 
On squaring both sides we get:
${ e }^{ 2 }=\quad 1+\frac { { b }^{ 2 } }{ { a }^{ 2 } }$ 
Or $\dfrac { 49 }{ 25 } =1+\frac { { b }^{ 2 } }{ 25 }$ 
Or $\dfrac { { b }^{ 2 } }{ 25 } =\frac { 24 }{ 25 } $
${ b }^{ 2 }=24$ 
The equation of hyperbola is 
$\dfrac { { x }^{ 2 } }{ 25 } -\dfrac { y^{ 2 } }{ 24 } =1$ 
$24{ x }^{ 2 }-25y^{ 2 }=600$

Hence, Option [A] and [C] are correct.

The equation of a hyperbola is given in its standard form as $16x^2-9y^2=144$.Equations of directrices is

mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

  1. $5x \pm 16=0$

  2. $5y \pm 16=0$

  3. $5x \pm 12=0$

  4. $5y \pm 12=0$

Show answer
Correct Option: B
Explanation:

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


Option [B]