Tag: hyperbola

Quizzes (10) Questions (269)

Questions Related to hyperbola

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

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

  1. $\sqrt { 2 } $
  2. $\sqrt { 3 } $
  3. $2\sqrt { 3 } $
  4. $3\sqrt { 2 } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The parametric equations x = a*sec(t) and y = b*tan(t) define a hyperbola. The given equations are a variation of this, leading to an eccentricity of sqrt(2).

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

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

  1. a circle

  2. hyperbola

  3. a pair of lines

  4. none of these
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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

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 $t$?

  1. A straight line

  2. A parabola

  3. A hyperbola

  4. A circle
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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

  1. $\lambda\, =\, 4$
  2. $\lambda\, =\, 1$
  3. $\lambda\, =\, \dfrac43$
  4. $\lambda\, =\, 3$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections
The difference between the length $2a$ of the transverse axis of a hyperbola of eccentricity $e$ and the length of its latus rectum is :
  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) $
Reveal answer Fill a bubble to check yourself
B Correct answer
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| $

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

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

  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.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

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

  1. $2x \pm 3y = 1$
  2. $2x \pm 3y = 0$
  3. $3x \pm 2y = 1$

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

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

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

  1. $3{ x }^{ 2 }-{ y }^{ 2 }=48$
  2. $4{ x }^{ 2 }-{ y }^{ 2 }=48$
  3. ${ x }^{ 2 }-3{ y }^{ 2 }=48$
  4. ${ x }^{ 2 }-4{ y }^{ 2 }=48$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $A$.

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

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 :

  1. 0

  2. 1

  3. 2

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $

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

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

  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. $\displaystyle\frac{(x-1)^2}{4}-\frac{(y-5)^2}{9}=1$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $b^2=a^2(e^2-1)$
$\Rightarrow b^2=16\left(\dfrac{25}{16}-1\right)=9$
$\boxed{\therefore\ Equation\ of\ hyperbola \Rightarrow\ \dfrac{(x-1)^2}{16}-\dfrac{(y-5)^2}{9}=1}$