Tag: hyperbola

Questions Related to hyperbola

The eccentricity the hyperbola $x=\left( t+\dfrac { 1 }{ t }  \right) ,y=\dfrac { a }{ 2 } \left( t-\dfrac { 1 }{ t }  \right) $ is ____________.

mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

  1. $\sqrt { 2 } $

  2. $\sqrt { 3 } $

  3. $2\sqrt { 3 } $

  4. $3\sqrt { 2 } $

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Correct Option: A

The equation $ \displaystyle 3x^{2}-2xy+y^{2}=0 $ represents:

mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

  1. a circle

  2. hyperbola

  3. a pair of lines

  4. none of these

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Correct Option: D
Explanation:

Given expression,$\displaystyle 3{ x }^{ 2 }-2xy+{ y }^{ 2 }=0$ 
As Coefficient of $\displaystyle xy$ is not zero,It will not be a circle and hyperbola.
Let $\displaystyle \frac { y }{ x } =m$

We get $\displaystyle { m }^{ 2 }-2m+3=0$ will not have any real solutions as discriminant is less than zero.
$\displaystyle \therefore $ They will not be pair of lines too.

The graph between $\log {(\theta-{\theta} _{0})}$ and time $(t)$ is a straight line in the experiment based on Newton's law cooling. What is the shape of graph between $\theta$ and $t$?

mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

  1. A straight line

  2. A parabola

  3. A hyperbola

  4. A circle

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Correct Option: C
Explanation:

It is given that
$log(\theta-\theta _{0})=kt $ where k is a constant of proportionality.
Hence
$e^{kt}=\theta-\theta _{0}$
Or
$\triangle \theta=e^{kt}$
The graph of $\triangle\theta$ vs $t$ will therefore be a hyperbola.

Equation $(2\, +\, \lambda)x^2\, -\, 2 \lambda xy\, +\, (\lambda\, -\, 1)y^2\, -\, 4x\, -\, 2\, =\, 0$ represents a hyperbola if

mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

  1. $\lambda\, =\, 4$

  2. $\lambda\, =\, 1$

  3. $\lambda\, =\, \dfrac43$

  4. $\lambda\, =\, 3$

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Correct Option: B
Explanation:

Given equation will represent hyperbola if
$h^2> ab$
$\Rightarrow \lambda^2\, >\, (\lambda\, +\, 2)\, (\lambda\, -\, 1)$
$\Rightarrow\, \lambda\, <\, 2$
Also $\Delta\, \neq\, 0$
$\Rightarrow\, -2(\lambda^2\, +\, \lambda\, -\, 2)\, -\, 4(\lambda\, -\, 1)\, +\, 2 \lambda^2\, \neq\, 0$
$\Rightarrow\, \lambda\, \neq\, \displaystyle \frac{4}{3}$.
Hence option 'B' is correct.

The difference between the length $2a$ of the transverse axis of a hyperbola of eccentricity $e$ and the length of its latus rectum is :
mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

  1. $2a\left| 3-{ e }^{ 2 } \right| $

  2. $2a\left| 2-{ e }^{ 2 } \right| $

  3. $2a\left( { e }^{ 2 }-1 \right) $

  4. $a\left( 2{ e }^{ 2 }-1 \right) $

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Correct Option: B
Explanation:

Let the equation of hyperbola be $\displaystyle \frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1$

Length of transverse axis is $2a$ and 
Length of latus rectum is $\displaystyle \frac { 2{ b }^{ 2 } }{ a } $
Now, difference $\displaystyle =\left| 2a-\frac { 2{ b }^{ 2 } }{ a }  \right| =\frac { 2 }{ a } \left| 2{ a }^{ 2 }-{ a }^{ 2 }{ e }^{ 2 } \right| $
$\therefore$ Difference $=2a\left| 2-{ e }^{ 2 } \right| $

Assertion(A): The difference of the focal distances of any point on the hyperbola $\displaystyle \frac{x^{2}}{36}-\frac{y^{2}}{9}=1$ is 12.
Reason(R): The difference of the focal distances of any point on the hyperbola is equal to the length of it transverse axis

mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

  1. Both A and R are true and R is the correct

    explanation of A.

  2. Both A and R are true but R is not the correct

    explanation of A.

  3. A is true but R is false.

  4. A is false but R is true.

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Correct Option: A
Explanation:

Clearly, $|SP-S'P|=2a=12$
Thus statement 1 is correct.
Also statement 2 is correct and followed by statement 1.

The asymptotes of a hyperbola $4x^2 - 9y^2=36$ are

mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

  1. $2x \pm 3y = 1$

  2. $2x \pm 3y = 0$

  3. $3x \pm 2y = 1$

  4. None

Show answer
Correct Option: B
Explanation:

The equation of hyperbola is $\displaystyle \frac{x^{2}}{9}+\frac{y^{2}}{4}=1$

So the equation of asymptotes is $\displaystyle \frac{x^{2}}{9}-\frac{y^{2}}{4}=0$
$\Rightarrow 4x^{2}-9y^{2}=0$
$\Rightarrow 2x \pm 3y=0$
Therefore option $B$ is correct

The equation of hyperbola whose coordinates of the foci are $(\pm8,0)$ and the lenght of latus rectum is $24$ units, is

mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

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

  2. $4{ x }^{ 2 }-{ y }^{ 2 }=48$

  3. ${ x }^{ 2 }-3{ y }^{ 2 }=48$

  4. ${ x }^{ 2 }-4{ y }^{ 2 }=48$

Show answer
Correct Option: A
Explanation:

The Foci of hyperbola are $(\pm8,0)$, hence Foci lie on $x$ - axis. 


We know that foci of hyperbola lie at $(\pm ae,0)$, So $ae = 8$ ...$(1)$

squaring both sides of equation $(1)$, we get,

$\Rightarrow a^2e^2 = 64$

Eccentricity of hyperbola $e^2 =1 + \dfrac {b^2}{a^2}$  

$\Rightarrow a^2(1+\dfrac{b^2}{a^2}) = 64$

$\Rightarrow a^2 + b^2 = 64$ ...$(2)$

Now the length of latus rectum is given as 24 units.

length of latusrectum of hyperbola $ = \dfrac{2b^2}{a} = 24$

$\Rightarrow b^2 = 12a$ ...$(3)$

putting value of $b^2$ in eq. $(2)$, we get,

$\Rightarrow a^2 +12a -64 = 0$

Hence $a  = 4, -16$

As $a$ is always taken as positive value so $a =4$ 

from eq. $(3)$,  $b = \sqrt{48}$

Hence equation of hyperbola is $\dfrac{x^2}{16} -\dfrac {y^2}{48} = 1$

Or $3x^2 -y^2 = 48$, So correct option is $A$.

If $ e$ and $e'$ be the eccentricities of a hyperbola and its conjugate, them $ \dfrac {1}{e^2} + \dfrac {1}{e'^{2}} $ is equal to :

mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

  1. 0

  2. 1

  3. 2

  4. None of these

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Correct Option: B
Explanation:

$ e^2 = \dfrac {a^2 + b^2}{a^2} $ and $ e'^{2} = \dfrac {a^2 + b^2}{b^2} $

$ \Rightarrow \dfrac {1}{e^2} + \dfrac {1}{e'^2} = \dfrac {a^2}{a^2+b^2} + \dfrac {b^2}{a^2 + b^2} = 1 $

The equation of the hyperbola whose foci are $(6, 5), (-4, 5)$ and eccentricity $5/4$ is?

mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

  1. $\displaystyle\frac{(x-1)^2}{16}-\frac{(y-5)^2}{9}=1$

  2. $\displaystyle\frac{x^2}{16}-\frac{y^2}{9}=1$

  3. $\displaystyle\frac{(x-1)^2}{16}-\frac{(y-5)^2}{9}=-1$

  4. $\displaystyle\frac{(x-1)^2}{4}-\frac{(y-5)^2}{9}=1$

Show answer
Correct Option: A
Explanation:

Let the centre of hyperbola be $(\alpha , \beta)$

As $y=5$ line has the foci, it also has the major axis.
$\therefore \dfrac{(x-\alpha)^2}{a^2}-\dfrac{(y-\beta)^2}{b^2}=1$
Midpoint of foci = centre of hyperbola
$\therefore \alpha =1, \beta =5$
Given, $e=\dfrac{5}{4}$.
We know that foci is given by $(\alpha \pm ae, \beta)$
$\therefore \alpha +ae=6$
$\Rightarrow 1+\dfrac{5}{4}a=6\Rightarrow a=4$
Using $b^2=a^2(e^2-1)$
$\Rightarrow b^2=16\left(\dfrac{25}{16}-1\right)=9$
$\boxed{\therefore\ Equation\ of\ hyperbola \Rightarrow\ \dfrac{(x-1)^2}{16}-\dfrac{(y-5)^2}{9}=1}$