Tag: hyperbola

Questions Related to hyperbola

For hyperbola  $\dfrac{x^2}{16}-\dfrac{y^2}{25}=1$
vertices are 

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

  1. $(4,4)$

  2. $(\pm4,0)$

  3. $(\pm4,4)$

  4. $(0,\pm4)$

Show answer
Correct Option: B
Explanation:

For the hyperbola of the form $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$

Vertices are $(\pm a,0)$
Given equation is $\dfrac {x^2}{16}-\dfrac {y^2}{25}=1$
So, here $a^2=16 \Rightarrow  a=4$
So, the vertices are $(\pm 4,0)$.

For hyperbola  $-\dfrac{x^2}{16}+\dfrac{y^2}{25}=1$ equation of directrices are

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

  1. $y=\pm\dfrac{16}{\sqrt{41}}$

  2. $y=\pm\dfrac{5}{\sqrt{41}}$

  3. $y=\pm\dfrac{2}{\sqrt{41}}$

  4. $y=\pm\dfrac{25}{\sqrt{41}}$

Show answer
Correct Option: D
Explanation:

$\dfrac{y^2}{a^2}-\dfrac{x^2}{b^2}=1$


Directrix is at $y=\pm \dfrac{b}{e}$ where $e=\sqrt{\dfrac{b^2}{a^2}+1}$

Here $a^2=25,b^2=16 \implies e=\sqrt{\dfrac{16}{25}+1}=\sqrt{\dfrac{41}{25}}$ 

So equation of directrix is $y=\pm\dfrac{5\sqrt{25}}{\sqrt{41}}=\pm\dfrac{25}{\sqrt{41}}$

For hyperbola  $\dfrac{-x^2}{9}+\dfrac{y^2}{16}=1$, centre is 

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

  1. $(3,3)$

  2. $(5,5)$

  3. $(0,0)$

  4. $(4,5)$

Show answer
Correct Option: C
Explanation:

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. 


By comparing it with it's standard form $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = -1$

We can know $a = 3$ and $b = 4$

For a standard form of a conjugate hyperbola, the center lies at origin at $(0,0)$. Hence the correct option is $C$.

For hyperbola  $\dfrac{x^2}{16}-\dfrac{y^2}{25}=1$, focus is is on 

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

  1. x-axis

  2. y-axs

  3. z-axis

  4. none

Show answer
Correct Option: A
Explanation:

The given hyperbola $\dfrac{x^2}{16} - \dfrac{y^2}{25} = 1$ is a standard form of hyperbola. 


By comparing it with standard form $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$

We can know $a = 4$ and $b = 5$

For a standard form of hyperbola, the foci lie on the transverse axis. $i.e.$ $x$ - axis. one each side of the hyperbola, at $(ae,0)$ and $(-ae,0)$ respectively.

For hyperbola  $-\dfrac{x^2}{16}+\dfrac{y^2}{25}=1$ vertices are 

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

  1. $(\pm4,0)$

  2. $(4,\pm5)$

  3. $(0,\pm5)$

  4. $(5,\pm5)$

Show answer
Correct Option: C
Explanation:

Since  constant term is negative, so its major axis will be on $y$ axis.

Now vertices are $(0,b)$ and $(0.-b)$.  
Hence, answer is  $(0,5)$ and $(0,-5)$.

For hyperbola  $-\dfrac{x^2}{9}+\dfrac{y^2}{16}=1$, focus is is on 

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

  1. x-axis

  2. y-axis

  3. z-axis

  4. none

Show answer
Correct Option: B
Explanation:

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. 


By comparing it with it's standard form $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = -1$

We can know $a = 3$ and $b = 4$

For a standard form of a conjugate hyperbola, the foci lie on its transverse axis. $i.e.$ $y$ - axis. one each side of the hyperbola, at $(0, be)$ and $(0,-be)$ respectively.

The foci of the hyperbola $4{ x }^{ 2 }-9{ y }^{ 2 }-1=0$ are

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

  1. $\left( \pm \sqrt { 13 } ,0 \right) $

  2. $\left( \pm \dfrac { \sqrt { 13 } }{ 6 } ,0 \right) $

  3. $\left( 0,\pm \dfrac { \sqrt { 13 } }{ 6 } \right) $

  4. None of the above

Show answer
Correct Option: B
Explanation:

The given equation of hyperbola can be re-written as,

$\dfrac { x^{ 2 } }{ \left( \dfrac { 1 }{ 2 }  \right) ^{ 2 } } -\dfrac { { y }^{ 2 } }{ \left( \dfrac { 1 }{ 3 }  \right) ^{ 2 } } =1$
So, the hyperbola has x-axis as the transverse axis, with $a=\dfrac { 1 }{ 2 } $
and $b=\dfrac { 1 }{ 3 }$ 

And, $e=\displaystyle \sqrt { 1+\dfrac { b^{ 2 } }{ a^{ 2 } }  }$

 $ =\displaystyle \sqrt { 1+\dfrac { 4 }{ 9 }  } =\dfrac { \sqrt { 13 }  }{ 3 }$ 

$\therefore$ Focii $=(\pm ae,0)=\left(\pm \dfrac { \sqrt { 13 }  }{ 6 } ,0\right)$
Hence, B is correct.

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. $(2,3)$

  2. $(\pm\sqrt3,3)$

  3. $(2,\pm3)$

  4. $(1,2)$,$(1,-6)$

Show answer
Correct Option: D
Explanation:

Hyperbola $\dfrac{y^2}{a^2}-\dfrac{x^2}{b^2}=1$

It has center at $(0,0)$ and vertex at $(0,\pm a)$.
For hyperbola $\dfrac{(y+2)^{2}}{16}-\dfrac{(x-1)^{2}}{3}=1$
Its center shifted to $(1,-2)$.
So, vertex will also shift by $(1,-2)$. 
So, vertex is $(0,\pm a)+(1,-2)=(0,\pm 4)+(1,-2) $
$\Rightarrow  (1,2)$ and $(1,-6)$

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

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

  1. $(1,-2)$

  2. $(0,0)$

  3. $(1,-1)$

  4. $(2,-2)$

Show answer
Correct Option: A
Explanation:

The given hyperbola is $-\dfrac{(x-1)^2}{3} + \dfrac{(y+2)^2}{16} = 1$ is a conjugate hyperbola.


It can be written as $\dfrac{(x-1)^2}{3} - \dfrac{(y+2)^2}{16} = -1$  ......$(1)$

Now let $x-1 = X$ and $y+2 = Y$

Putting values of $x-1$ and $y+2$ in eq. $(2)$ we get,

$\rightarrow  \ \dfrac{(X)^2}{3} - \dfrac{(Y)^2}{16} = -1$  ....$(2)$

We can see the eq$(2)$ is a standard form of conjugate hyperbola and it's center lies at origin $(0,0)$

So $X =0 , Y = 0$ is the center of the hyperbola given in eq.$(2)$

$\rightarrow$ $ X = x-1 = 0$ or $x = 1$

$\rightarrow$ $ Y = y+2 = 0$ or $y = -2$

Hence center of the given hyperbola in eq. $(1)$ is $(1,-2)$. So correct option is $A$.

Find the equation to the hyperbola of given length of transverse axis $6$ and the join of centre and focus is bisected by vertex.

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

  1. $3x^{2} - y^{2} = 27$.

  2. $3x^{2} + y^{2} = 27$.

  3. $x^{2} - y^{2} = 27$.

  4. None of these

Show answer
Correct Option: A