Tag: parametric equation of the hyperbola

Questions Related to parametric equation of the 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

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

The focal length of the hyperbola $x^2-3y^2-4x-6y-11=0$, is?

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

  1. $4$

  2. $6$

  3. $8$

  4. $10$

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Correct Option: A

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

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

  1. Its centre is (4, 2)

  2. its centre is (2, 4)

  3. Length of latus rectum= 24

  4. length of latus rectum=12

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Correct Option: B

Find Directrix, foci and eccentricity of the conics:

$\displaystyle x^{2}+2x-y^{2}+5= 0$

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

  1. Directrices $\displaystyle y= \pm \sqrt{2}$

  2. foci $\displaystyle \left ( -1,\pm 2\sqrt{2} \right )$ 

  3. $e= \sqrt{2}$

  4. $e=2$

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Correct Option: A,B,C
Explanation:

$\displaystyle x^{2}+2x-y^{2}+5= 0$
$\displaystyle x^{2}+2x+1-y^{2}= -4$
$\displaystyle (x+1)^2-y^{2}= -4$
$\displaystyle \frac{y^2}{4}-\frac{(x+1)^2}{4}= 1$
Clearly this equation of rectangular hyperbola with $y-$axis as major axis
eccentricity $e = \sqrt{2}$ directrices $:y = \cfrac{a}{e}=\pm \sqrt{2}$ foci $(-1,\pm ae)=(-1,\pm 2\sqrt{2})$

The equation of a hyperbola whose directrix is $2x+y=1$ and focus is at $(1,2)$ with $e=\sqrt{3}$ is :

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

  1. $7x^2+12xy+2y^2-2x+14y-22=0$

  2. $7x^2+12xy-2y^2-2x+14y-22=0$

  3. $7x^2+12xy-2y^2-2x-14y-22=0$

  4. $7x^2+12xy+2y^2+2x+14y-22=0$

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Correct Option: A