Tag: hyperbola

Questions Related to hyperbola

A hyperbola passes through the points $(3, 2)$ and $(-17, 12)$ and has its centre at origin and transverse axis is along $x-axis$. The length of its transverse axis is:

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

  1. $2$

  2. $4$

  3. $6$

  4. $None\ of\ these$

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Correct Option: A

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

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

  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) $

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Correct Option: A
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.

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

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

  1. $1$

  2. $3$

  3. $2$

  4. $4$

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Correct Option: C
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 $=2a=2$

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

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

  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

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Correct Option: B
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.

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

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

  1. $\displaystyle \frac{5}{4}$

  2. $\displaystyle \frac{\sqrt{41}}{5}$

  3. $\displaystyle \frac{25}{16}$

  4. $\displaystyle \frac{\sqrt{41}}{4}$

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Correct Option: D
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 $= \sqrt{1+\cfrac{25}{16}}=\cfrac{\sqrt{41}}{4}$

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 :

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

  1. $e$

  2. $\displaystyle \frac{e}{\sqrt{e^{2} - 1}}$

  3. $\displaystyle e \sqrt{e^{2} - 1}$

  4. $\displaystyle e^{2} - e$

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Correct Option: B
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}}$

Let $P(6, 3)$ be a point on the hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$. If the normal at the point P intersects the x-axis at $(9, 0)$, then the eccentricity of the hyperbola is?

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

  1. $\sqrt{\dfrac{5}{2}}$

  2. $\sqrt{\dfrac{3}{2}}$

  3. $\sqrt{2}$

  4. $\sqrt{3}$

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Correct Option: A

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

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

  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. $\displaystyle x^{2} \cos ^{2}\theta -y^{2}\sin ^{2}\theta=1$

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Correct Option: A
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.

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

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

  1. $its centre is \left(4,2\right)$

  2. $its centre is \left(2,4\right)$

  3. $length of latus rectum = 24$

  4. $length of latus rectum = 12$

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Correct Option: A

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

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

  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$

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Correct Option: A