Tag: equations of hyperbola

Questions Related to equations of hyperbola

Find the locus of the point of intersection of the lines $\sqrt 3 x-y-4\sqrt 3\lambda=0$ and $\sqrt 3 \lambda x +\lambda y-4\sqrt{3}=0$ for different values of $\lambda$.

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

  1. $3x^2-y^2=48$

  2. $y^2-3x^2=24$

  3. $4x^2-3y^2=16$

  4. None of these

Show answer
Correct Option: A
Explanation:

Let $(h,k)$ be the point of intersection of the given lines. Then,

$\sqrt 3 h-k-4\sqrt 3 \lambda=0$ and $\sqrt3 \lambda h +\lambda k-4\sqrt 3=0$
$\sqrt 3 h-k=4\sqrt 3\lambda$ and $\lambda(\sqrt 3h+k)=4\sqrt 3$
$(\sqrt 3 h-k)\lambda (\sqrt 3h +k)=(4\sqrt 3\lambda)(4\sqrt 3)$
$3h^2-k^2=48$
Hence, the locus of $(h,k)$ is $3x^2-y^2=48$.

If the equation of a hyperbola is $\frac{{{x^2}}}{9} - \frac{{{y^2}}}{{16}} = 1$, then 

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

  1. traverse axis is along x-axis of length $6$

  2. traverse axis is along y-axis of length $8$

  3. conjugate axis is along y-axis of length $6$

  4. None of these

Show answer
Correct Option: A

The length of the transverse axis of the hyperbola $3x^2-4y^2=3$ is

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

  1. $\frac{{8\sqrt 2 }}{{\sqrt 3 }}$

  2. $\frac{{16\sqrt 2 }}{{\sqrt 3 }}$

  3. $\frac {3}{32}$

  4. $\frac {64}{3}$

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Correct Option: A

If the eccentricity and length of latus rectum of a hyperbola are $\frac {\sqrt 13}{3}$ and $\frac {10}{3}$ units respectively, then what is the length of the traverse axis?

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

  1. $\frac {7}{2}$ units

  2. $12$ units

  3. $\frac {15}{2}$ units

  4. $\frac {15}{4}$ units

Show answer
Correct Option: A

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

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

  1. $(4,4)$

  2. $(5,5)$

  3. $(4,5)$

  4. $(0,0)$

Show answer
Correct Option: D
Explanation:

Centre of the hyperbola of form $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ has centre at orgin.

Here it is of this form. 
So, centre is at origin $(0,0)$.

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

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

  1. $\dfrac{16}{\sqrt{41}}$

  2. $\dfrac{25}{\sqrt{41}}$

  3. $\pm\dfrac{32}{\sqrt{141}}$

  4. $\dfrac{32}{\sqrt{41}}$

Show answer
Correct Option: D
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  $\dfrac{2a^2}{\sqrt{a^2+b^2}} = \dfrac{2\times16}{\sqrt{16+25}} = \dfrac{32}{\sqrt{41}}$

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.