Tag: hyperbola

Quizzes (10) Questions (269)

Questions Related to hyperbola

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

  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. $\displaystyle 2(2x+2y-3)^{2}-3(2x-y+4)^{2}=1$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $\dfrac { 2 }{ 5 } { \left( 2x-y+4 \right)  }^{ 2 }-\dfrac { 3 }{ 5 } { \left( x+2y-3 \right)  }^{ 2 }=1$.

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

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

  1. I II III IV

    1 2 3 4

  2. 1 4 2 3

  3. 3 1 2 4

  4. 2 3 4 1
Reveal answer Fill a bubble to check yourself
B Correct answer
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) $\dfrac{(x+4)^{2}}{16}-\frac{(y+1)^{2}}{9}=1$

$b^{2}=9$

$a=4$

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

$=\dfrac {9}{2}$

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

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

  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. $\dfrac {x^{2}}{16} - \dfrac {y^{2}}{9} = 1$
Reveal answer Fill a bubble to check yourself
B Correct answer
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
$\dfrac {x^{2}}{4} - \dfrac {y^{2}}{12} = 1$

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

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

  1. $(-4, -9)$
  2. $(-2, 3)$
  3. $(2, -3)$
  4. $(5, 4)$
  5. $(25, 16)$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $(-2,3)$
Multiple choice mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

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

  1. $\dfrac{2}{\sqrt{19}}$
  2. $\dfrac{3}{\sqrt{19}}$
  3. $\dfrac{4}{\sqrt{19}}$
  4. $\dfrac{32}{\sqrt{19}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $=\dfrac{2a}{e}=\dfrac{32}{\sqrt{19}}$

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

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

  1. $(\pm\sqrt3,0)$
  2. $(\pm\sqrt3+1,-2)$
  3. $(\pm1,-2)$
  4. $(0,0)$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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) $.

  1. $65y^2-16x^2=196$
  2. $65y^2-14x^2=196$
  3. $85y^2-16x^2=196$
  4. $85y^2-16x^2=147$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 
$85y^2-16x^2=196$

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

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

  1. $24x^2-25y^2=600$
  2. $25x^2-24y^2=600$
  3. $\displaystyle \frac{x^2}{25}-\frac{y^2}{24}=1$
  4. $\displaystyle \frac{x^2}{24}-\frac{y^2}{25}=1$
Reveal answer Fill a bubble to check yourself
A,C Correct answer
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.

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

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

  1. $5x \pm 16=0$
  2. $5y \pm 16=0$
  3. $5x \pm 12=0$
  4. $5y \pm 12=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
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]