Tag: equations of hyperbola

Questions Related to equations of hyperbola

For the hyperbola $16x^2\, -\, 9y^2\, +\, 32x\, +\, 36y\,-\, 164\, =\, 0$, find $2(a+b)$.

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

  1. $8$

  2. $6$

  3. $14$

  4. $12$

Show answer
Correct Option: C
Explanation:

Given hyperbola can be written as


$16x^2+32x+16-9y^2+36y-36=144$

$\displaystyle\, \frac{(x\, +\, 1)^2}{9}\, -\, \frac{(y\, -\, 2)^2}{16}\, =\, 1$

$\Rightarrow a =3, b = 4$

Length of major axis $=2\times 4=8$ and length of minor axis $=2\times 3 = 6$. 

Hence sum is $2(a+b)=2(3+4)=14.$

A hyperbola having the transverse axis of length $\sqrt{2}$ is confocal with $3x^2 + 4y^2 = 12$, then its equation is:

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

  1. $2x^2-2y^2=1$

  2. $2x^2+2y^2=1$

  3. $x^2+y^2=2$

  4. $x^2-y^2=2$

Show answer
Correct Option: A
Explanation:

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,
$\displaystyle \frac {x^2}{1/2}-\frac {y^2}{1/2}=1\Rightarrow 2x^2-2y^2=1$

Find the equation to the hyperbola, the distance between whose foci is $16$ and whose eccentricity is $\sqrt{2}$.

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

  1. $x^2\, -\, y^2\, =\, 32$

  2. $x^2\, -\, y^2\, =\, 18$

  3. $x^2\, -\, y^2\, =\, 64$

  4. $x^2\, -\, y^2\, =\, 48$

Show answer
Correct Option: A
Explanation:

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, $x^2-y^2=a^2=32$

A parabola is drawn with its vertex at $(0,-3)$, the axis of symmetry along the conjugate axis of the hyperbola $\displaystyle \frac { { x }^{ 2 } }{ 49 } -\frac { { y }^{ 2 } }{ 9 } =1$ and passing through the two foci of the hyperbola. The coordinates of the focus of the parabola are :

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

  1. $\displaystyle \left( 0,\frac { 11 }{ 6 }  \right) $

  2. $\displaystyle \left( 0,-\frac { 11 }{ 6 }  \right) $

  3. $\displaystyle \left( 0,\frac { 11 }{ 12 }  \right) $

  4. $\displaystyle \left( 0,-\frac { 11 }{ 12 }  \right) $

Show answer
Correct Option: A
Explanation:

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

Its conjugate axis is y-axis
Also, $\displaystyle e=\sqrt { 1+\frac { { b }^{ 2 } }{ { a }^{ 2 } }  } =\sqrt { 1+\frac { 9 }{ 49 }  } =\frac { \sqrt { 58 }  }{ 7 } $
$\therefore$ Foci of hyperbola is $\left( \pm ae,0 \right) \Rightarrow \left( \pm \sqrt { 58 } ,0 \right) $
Now equation of parabola with vertex at $(0,-3)$ and axis along y-axis is ${ x }^{ 2 }=l\left( y+3 \right) $
It passes through $\left( \pm \sqrt { 58 } ,0 \right) $
$\displaystyle \therefore 58=l\left( 0+3 \right) \Rightarrow l=\frac { 58 }{ 3 } $
$\therefore$ parabola is $\displaystyle { x }^{ 2 }=\frac { 58 }{ 3 } \left( y+3 \right) $
Its focus is $\displaystyle \left( 0,-3+\frac { 58 }{ 4.3 }  \right) \equiv \left( 0,\frac { 11 }{ 6 }  \right) $

Which of the following is true for the hyperbola $9x^2\, -\, 16y^2\, -\, 18x\, +\, 32y\, -\, 151\, =\, 0$?

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

  1. The length of the transverse axes is $4$

  2. Length of latus rectum is $9$

  3. Equation of directrix is $x\, =\, \displaystyle \frac{21}{5}$ and $x\, =\, - \displaystyle \frac{11}{5}$

  4. None of these

Show answer
Correct Option: C
Explanation:

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

$\therefore$ 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.

An ellipse intersects the hyperbola $\displaystyle 2x^{2}-2y^{2}=1$ orthogonally at point $P$. The eccentricity of the ellipse is reciprocal to that of the hyperbola. If the axes of the ellipse are along the co-ordinate axes and product of focal distances of $P$ is $x$ then $2x$ is:

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

  1. $1$

  2. $2$

  3. $3$

  4. $5$

Show answer
Correct Option: C
Explanation:

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)
$\displaystyle (1)^{2}-(2)^{2}\Rightarrow 4PS'PS=4(a^{2}-a'^{2})=4\left [ \left ( \frac{a'e}{e} \right )^{2} -a'^{2}\right ]=4\left ( 2-\frac{1}{2} \right )=6$
$\displaystyle \Rightarrow 2x = 3$

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