Tag: mathematics and statistics

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Questions Related to mathematics and statistics

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

For hyperbola  $\dfrac{x^2}{16}-\dfrac{y^2}{25}=1$ distances between two directrices are

  1. $\dfrac{16}{\sqrt{41}}$
  2. $\dfrac{25}{\sqrt{41}}$
  3. $\pm\dfrac{32}{\sqrt{141}}$
  4. $\dfrac{32}{\sqrt{41}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given equation of hyperbola $\dfrac {x^2}{16}-\dfrac {y^2}{25}=1$
Here $a^2=16, b^2=25$
Distance between directrix is  $\dfrac{2a^2}{\sqrt{a^2+b^2}} = \dfrac{2\times16}{\sqrt{16+25}} = \dfrac{32}{\sqrt{41}}$

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

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

  1. $(4,4)$
  2. $(\pm4,0)$
  3. $(\pm4,4)$
  4. $(0,\pm4)$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $(\pm 4,0)$.

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

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

  1. $y=\pm\dfrac{16}{\sqrt{41}}$
  2. $y=\pm\dfrac{5}{\sqrt{41}}$
  3. $y=\pm\dfrac{2}{\sqrt{41}}$
  4. $y=\pm\dfrac{25}{\sqrt{41}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $y=\pm\dfrac{5\sqrt{25}}{\sqrt{41}}=\pm\dfrac{25}{\sqrt{41}}$

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

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

  1. $(3,3)$
  2. $(5,5)$
  3. $(0,0)$
  4. $(4,5)$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $C$.

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

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

  1. x-axis

  2. y-axs

  3. z-axis

  4. none
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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

  1. x-axis

  2. y-axis

  3. z-axis

  4. none
Reveal answer Fill a bubble to check yourself
B Correct answer
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.