Tag: hyperbola

Quizzes (10) Questions (269)

Questions Related to hyperbola

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

If a hyperbola passes through the focii of the ellipse$\dfrac { { x }^{ 2 } }{ 25 } +\dfrac { { y }^{ 2 } }{ 16 } =1.$ Its transverse and conjugate axes coincide respectively with the major and minor axes of the ellipse and if the product of eccentricities hyperbola and ellipse is 1, then

  1. the equation of hyperbola is $\dfrac { x^{ 2 } }{ 9 } -\dfrac { { y }^{ 2 } }{ 16 } =1\\ \quad \quad $
  2. the equation of hyperbola is $\dfrac { x^{ 2 } }{ 9 } -\dfrac { { y }^{ 2 } }{ 25 } =1\\ \quad \quad $

  3. focus of hyperbola is (5,0)
  4. focus of hyperbola is $\left( 5\sqrt { 3, } 0 \right) $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Formula,

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

$=1-\dfrac{16}{25}$

$\therefore e=\dfrac{3}{5}$

$e _2 \times e =1$

$\Rightarrow e _2=\dfrac{5}{3}$

Equation,

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

Given,

$\Rightarrow (3,0)$

$\dfrac{3^2}{a^2}=1$

$\Rightarrow a^2=9$

we have,

$e _2^2=1+\dfrac{b^2}{a^2}$

$\dfrac{25}{9}=1+\dfrac{b^2}{9}$

$\Rightarrow b^2=16$

$\dfrac{x^2}{9}-\dfrac{y^2}{16}=1$

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

The hyperbola $\displaystyle \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}}=1$ passes through the point $\displaystyle \left ( 2, : 3 \right )$ and has the eccentricity $2$. Then the transverse axis of the hyperbola has the length

  1. $1$
  2. $3$
  3. $2$
  4. $4$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given hyperbola is,  $\displaystyle \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}}=1$
It passes through $(2,3)$
$\cfrac{4}{a^{2}} - \cfrac{9}{b^{2}}=1 ..(1)$
Also eccentricity is $2$,
$\Rightarrow e^2=1+\cfrac{b^2}{a^2}=4\Rightarrow \cfrac{b^2}{a^2}=3   ..(2)$
Solving (1) and (2) we get $a=1, b=\sqrt{3}$
Hence, length of transverse axis is $=2a=2$

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

If in a hyperbola the eccentricity is $\displaystyle \sqrt{3}$, and the distance between the foci is $9$ then the equation of the hyperbola in the standard form is

  1. $\displaystyle \dfrac{x^{2}}{\left ( \dfrac{\sqrt{3}}{2} \right )^{2}} - \dfrac{y^{2}}{\left ( \sqrt{\dfrac{3}{2}} \right )^{2}} = 1$
  2. $\displaystyle \dfrac{x^{2}}{\left ( \dfrac{3 \sqrt{3}}{2} \right )^{2}} - \dfrac{y^{2}}{\left ( \dfrac{3\sqrt{3}}{\sqrt{2}} \right )^{2}} = 1$
  3. $\displaystyle \dfrac{x^{2}}{\left ( \dfrac{3\sqrt{3}}{\sqrt{2}} \right )^{2}} - \dfrac{y^{2}}{\left ( \dfrac{3\sqrt{2}}{2} \right )^{2}} = 1$

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

Given eccentricity of the hyperbola $e=\sqrt{3}$
and distance between focii is 9. $\Rightarrow 2ae=9\Rightarrow a=\cfrac{3\sqrt{3}}{2}$
also $b^2=a^2(e^2-1)=\cfrac{27}{4}(3-1)=\cfrac{27}{2}\Rightarrow b=\cfrac{3\sqrt{3}}{\sqrt{2}}$
Hence equation of required hyperbola is,
$\cfrac{x^2}{\left(\cfrac{3\sqrt{3}}{2}\right )^2}-\cfrac{y^2}{\left (\cfrac{3\sqrt{3}}{\sqrt{2}}\right )^2}=1$
Hence, option 'B' is correct.

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

If any point on a hyperbola has the coordinates $\displaystyle \left ( 5 \tan \phi , : 4 \sec \phi \right )$ then the ecentricity of the hyperbola is

  1. $\displaystyle \frac{5}{4}$
  2. $\displaystyle \frac{\sqrt{41}}{5}$
  3. $\displaystyle \frac{25}{16}$
  4. $\displaystyle \frac{\sqrt{41}}{4}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We have   $x=5\tan\phi, y=4\sec\phi$
Eliminating $\phi$ we get
$\cfrac{y^2}{16}-\cfrac{x^2}{25}=\sec^2\phi-\tan^2\phi=1$
This is a hyperbola. Therefore, its eccentricity $= \sqrt{1+\cfrac{25}{16}}=\cfrac{\sqrt{41}}{4}$

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

If the eccentricity of the hyperbola $\displaystyle \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ is $e$ then the eccentricity of the hyperbola $\displaystyle \frac{y^{2}}{b^{2}} - \frac{x^{2}}{a^{2}} = 1$ is :

  1. $e$
  2. $\displaystyle \frac{e}{\sqrt{e^{2} - 1}}$
  3. $\displaystyle e \sqrt{e^{2} - 1}$
  4. $\displaystyle e^{2} - e$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For hyperbola $\displaystyle \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$
eccentricity $\Rightarrow e=\sqrt{1+\cfrac{b^2}{a^2}}\Rightarrow \cfrac{b^2}{a^2}=e^2-1$
For hyperbola $\displaystyle \frac{x^{2}}{b^{2}} - \frac{y^{2}}{a^{2}} = 1$
Required eccentricity $ e'=\sqrt{1+\cfrac{a^2}{b^2}}=\sqrt{1+\cfrac{1}{e^2-1}}=\displaystyle \cfrac{e}{\sqrt{e^{2} - 1}}$

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

A hyperbola, having the transverse axis of length $\displaystyle 2\sin \theta$, is confocal with the ellipse $\displaystyle 3x^{2}+4y^{2}=12$, then its equation is

  1. $\displaystyle x^{2}\text{cosec} ^{2}\theta -y^{2}\sec ^{2}\theta=1$
  2. $\displaystyle x^{2} \sec ^{2}\theta -y^{2}\text{cosec}^{2}\theta=1$
  3. $\displaystyle x^{2} \sin ^{2}\theta -y^{2}\cos ^{2}\theta=1$
  4. $\displaystyle x^{2} \cos ^{2}\theta -y^{2}\sin ^{2}\theta=1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given ellipse may be written as $\cfrac{x^2}{4}+\cfrac{y^2}{3}=1$
$\Rightarrow a^2=4, b^2=3$

$\Rightarrow e= \sqrt{1-\dfrac{3}{4}}=\cfrac{1}{2}$
$\therefore $ Focus of the ellipse $=(\pm ae,0)=(\pm 1, 0)$
Given required hyperbola is confocal to the ellipse
Let $a',b',e'$ are transverse axis, conjugate axis an eccentricity of the hyperbola
$a'e'=1\Rightarrow \sin\theta. e'=1\Rightarrow e'=\cfrac{1}{\sin\theta}$
Using $b'^2=a'^2(e^2-1)\Rightarrow b'^2=1-\sin^2\theta=\cos^2\theta$
Therefore required hyperbola is $\cfrac{x^2}{a'^2}-\cfrac{y^2}{b'^2}=1$
$\Rightarrow \cfrac{x^2}{\sin^2\theta}-\cfrac{y^2}{\cos^2\theta}=1$
$\Rightarrow x^2 \text{cosec}^2\theta-y^2\sec^2\theta=1$

Hence, option 'A' is correct.

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

Consider the hyoerbola ${ 3x^{2} }-{ y }^{ 2 }-{ 24x } + { 4y } { 4 } = 0$

  1. $its centre is \left(4,2\right)$
  2. $its centre is \left(2,4\right)$
  3. $length of latus rectum = 24$
  4. $length of latus rectum = 12$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Completing the square for 3x^2 - y^2 - 24x + 4y + 4 = 0: 3(x^2 - 8x + 16) - (y^2 - 4y + 4) = -4 + 48 - 4, resulting in 3(x-4)^2 - (y-2)^2 = 40. The center is (4, 2).

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

$y=mx+c$ is tangent to hyperbola find $c$ if hyperbola eqn is

  1. $\dfrac{{x}^{2}}{{a}^{2}}-\dfrac{{y}^{2}}{{b}^{2}}=1$$a>b$
  2. $\dfrac{{x}^{2}}{{a}^{2}}-\dfrac{{y}^{2}}{{b}^{2}}=1$$b>a$
  3. $\dfrac{{-x}^{2}}{{a}^{2}}-\dfrac{{y}^{2}}{{b}^{2}}=1$$a>b$
  4. $\dfrac{{-x}^{2}}{{a}^{2}}-\dfrac{{y}^{2}}{{b}^{2}}=1$$b>a$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For line y=mx+c to be tangent to hyperbola x^2/a^2 - y^2/b^2 = 1, we need c^2 = a^2m^2 - b^2. For real c, we need a^2m^2 > b^2. When a > b, we can find m values satisfying this condition. The question is poorly phrased but option A correctly identifies the standard hyperbola condition.