Tag: mathematics and statistics

Questions Related to mathematics and statistics

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

Show answer
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. $e=2$

Show answer
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

Let the eccentricity of the hyperbola $  \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1  $ be reciprocal to that of the ellipse $  x^{2}+4 y^{2}=4 .  $ If thehyperbola passes through a focus of the ellipse, then __________________.

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

  1. (A) the equation of the hyperbola is $ \frac{x^{2}}{3}-\frac{y^{2}}{2}=1 $

  2. (B) a focus of the hyperbola is $ (2,0) $

  3. (C) the eccentricity of the hyperbola is $ \sqrt{\frac{5}{3}} $

  4. (D) the equation of the hyperbola is $ x^{2}-3 y^{2}=3 $

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Correct Option: D

The eccentricity of the hyperbola $\displaystyle \dfrac { \sqrt { 1999 }  }{ 3 } \left( { x }^{ 2 }-{ y }^{ 2 } \right) =1$ is:

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

  1. $\sqrt { 2 } $

  2. $2$

  3. $2\sqrt { 2 } $

  4. $\sqrt { 3 } $

Show answer
Correct Option: A
Explanation:

Equation of hyperbola is $\displaystyle \frac { { x }^{ 2 } }{ 3/\sqrt { 1999 }  } -\frac { { y }^{ 2 } }{ 3/\sqrt { 1999 }  } =1$

Here $\displaystyle { a }^{ 2 }={ b }^{ 2 }=\frac { 3 }{ \sqrt { 1999 }  } $
$\therefore$ Eccentricity $\displaystyle e=\sqrt { 1+\frac { { b }^{ 2 } }{ { a }^{ 2 } }  } =\sqrt { 1+1 } =\sqrt { 2 } $