Tag: ellipse

Questions Related to ellipse

The focus of extremities of the latus rectum of the family of the ellipse  ${b^2}{x^2} + {a^2}{y^2} = {a^2}{b^2}{\text{ is }}\left( {b \in R} \right)$ 

mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

  1. ${x^2} - ay = {a^3}$

  2. ${x^2} - ay - {e^2}$

  3. ${x^2} \pm ay = {a^2}$

  4. ${x^2} + ay - {b^2}$

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Correct Option: A

The equation of the latusrecta of the ellipse $9x^{2}+4^{2}-18x-8y-23=0$ are 

mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

  1. $y=\pm \sqrt {5}$

  2. $x=\pm \sqrt {5}$

  3. $y=1 \pm \sqrt {5}$

  4. $x=1 \pm \sqrt {5}$

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Correct Option: C
Explanation:

equation of ellipse is ${ 9x }^{ 2 }+{ 4y }^{ 2 }-18x-8y-23=0$

$\Rightarrow (9x^{ 2 }-18x+9)+(4y^{ 2 }-8y+4)-23-9-4=0$
$ \Rightarrow 9(x-1)^{ 2 }+4(y-1)^{ 2 }=36$
$ \Rightarrow \cfrac { (x-1)^{ 2 } }{ 4 } +\cfrac { (y-1)^{ 2 } }{ 9 } =1$
So, equation of latus recta is $(y-1)=\pm be$
$y=1\pm \sqrt { b^{ 2 }-a^{ 2 } } \Rightarrow y=1\pm \sqrt { 5 } $

The foci of the ellipse $\dfrac{x^{2}}{16} + \dfrac{y^{2}}{b^{2}} =1$ and the hyperbola $\dfrac{x^{2}}{144} - \dfrac{y^{2}}{81} =\dfrac{1}{25}$ coincide, then the value of $b^{2}$ is:

mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

  1. $5$

  2. $7$

  3. $9$

  4. $4$

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Correct Option: B
Explanation:

The foci of the ellipse are also the foci of an hyperbola,
then we have, for the ellipse,

$a^2 -c^2 = b^2$
so
$16 -c^2 = b^2...............(1) $
 
Equation of Hyperbola can also be written as $\dfrac{x^2}{\dfrac{144}{25}}-\dfrac{y^2}{\dfrac{81}{25}}=1$

For the hyperbola, which must have its transverse axis on the x-axis, the equation
$c^2 - a^2 = b^2\Rightarrow c^2-\dfrac{144}{25}=\dfrac{81}{25}\Rightarrow c^2=\dfrac{225}{25}=9$

Putting this value in equation (1)
$16-9=b^2\Rightarrow b^2=7$

If foci are points $(0,1)(0,-1)$ and minor axis is of length $1$, then equation of ellipse is

mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

  1. $\dfrac { { x }^{ 2 } }{ 1/4 } +\dfrac { { y }^{ 2 } }{ 5/4 } =1$

  2. $\dfrac { { x }^{ 2 } }{ 5/4 } +\dfrac { { y }^{ 2 } }{ 1/4 } =1$

  3. $\dfrac { { x }^{ 2 } }{ 3/4 } +\dfrac { { y }^{ 2 } }{ 1/4 } =1$

  4. $\dfrac { { x }^{ 2 } }{ 1/4 } +\dfrac { { y }^{ 2 } }{ 3/4 } =1$

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Correct Option: A
Explanation:

Given that focii are $(0,1), (0,-1)$


Axis lies along $y-axis$

Distance between focii$\rightarrow 2be=2$

$\Rightarrow be=1$

Given that $2a=1$

$\Rightarrow a=\dfrac{1}{2}$

We know that $\Rightarrow e^2=1-\dfrac{a^2}{b^2}$

$\Rightarrow b^2e^2=b^2-a^2$

Substituting above obtained values in this expression we get,

$\Rightarrow 1=b^2-(\dfrac{1}{2})^2$

$\Rightarrow 1=b^2-\dfrac{1}{4}$

$\Rightarrow b^2=\dfrac{5}{4}$

Thus equation of ellipse$\Rightarrow \dfrac{x^2}{\dfrac{1}{4}}+\dfrac{y^2}{\dfrac{5}{4}}=1$

The equation of the ellipse with its focus at $(6, 2)$, centre at $(1, 2)$ and which passes through the point $(4, 6)$ is?

mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

  1. $\dfrac{(x-1)^2}{25}+\dfrac{(y-2)^2}{16}=1$

  2. $\dfrac{(x-1)^2}{25}+\dfrac{(y-2)^2}{20}=1$

  3. $\dfrac{(x-1)^2}{45}+\dfrac{(y-2)^2}{20}=1$

  4. $\dfrac{(x-1)^2}{45}+\dfrac{(y-2)^2}{16}=1$

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Correct Option: C

The equation of the tangent to the ellipse such that sum of perpendiculars dropped from foci is 2 units, is

mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

  1. $y cos3\pi/ 4 - x sin 3\pi /4=1$

  2. $y sin \frac{3\pi}{8}- x cos \frac{3\pi}{8}=1$

  3. $x cos \pi /8 - sin \pi /8=1$

  4. $y cos \frac{5\pi}{8}+x sin \frac{5\pi}{8}=1$

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Correct Option: A

An ellipse $\cfrac { { x }^{ z } }{ 4 } +\cfrac { { y }^{ z } }{ 3 } =1$ confocal with hyperbola $\cfrac { { x }^{ 2 } }{ \cos ^{ 2 }{ \theta  }  } -\cfrac { { y }^{ 2 } }{ \sin ^{ 2 }{ \theta  }  } =1$ then the set of value of $'0'$

mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

  1. $R$

  2. $R-\left{ n\pi ,n\epsilon z \right} $

  3. $R-\left{ \left( 2n+1 \right) \cfrac { \pi }{ 2 } ,n\epsilon z \right} $

  4. $R-\left{ \cfrac { n\pi }{ 2 } ,n\epsilon z \right} $

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Correct Option: A
Explanation:
Focus of ellipse$=ae=a\sqrt { 1-\cfrac { { b }^{ 2 } }{ { a }^{ 2 } }  } $
$=\sqrt { { a }^{ 2 }-{ b }^{ 2 } } $
$=\sqrt { 1 } $
$=1$
Focus of hyperbola$=a\sqrt { 1+\cfrac { { b }^{ 2 } }{ { a }^{ 2 } }  } $
$=\sqrt { { a }^{ 2 }+{ b }^{ 2 } } $
$=\sqrt { \sin ^{ 2 }{ \theta  } +\cos ^{ 2 }{ \theta  }  } $
$=\sqrt { 1 } $
$=1$
$\therefore $The ellipse and hyperbola will be confocal for $\theta \epsilon R$.

Equation of the ellipse whose axes are the axes of coordinates and which passes through the point $ (-3,1)$ and has eccentricity $\sqrt {\frac{2}{5}} $ is 

mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

  1. $5x^3+3y^2-48=0$

  2. $3x^2+5y^2-15=0$

  3. $5x^2+3y^2-32=0$

  4. $3x^2+5y^2-32=0$

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Correct Option: C
Explanation:

We know that equation of ellipse is

 

  $ \dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1 $       …….(1)

Given that

  $ e=\sqrt{\dfrac{2}{5}} $

 $ \sqrt{\dfrac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}}}=\sqrt{\dfrac{2}{5}} $

 

Taking square both side and solving , we get


  $ 5{{a}^{2}}-5{{b}^{2}}=2{{a}^{2}} $

 $ {{a}^{2}}=\dfrac{5{{b}^{2}}}{3} $    …….(2)

$\because $ ellipse pass through (-3,1)

Then $x=-3, y=1$

Put in equation (1) we get

  $ \dfrac{{{\left( -3 \right)}^{2}}}{{{a}^{2}}}+\dfrac{{{1}^{2}}}{{{b}^{2}}}=1 $

 $ {{a}^{2}}+9{{b}^{2}}={{a}^{2}}{{b}^{2}} $

 $ \dfrac{5{{b}^{2}}}{3}+9{{b}^{2}}=\dfrac{5{{b}^{2}}}{3}.{{b}^{2}} $

 $ {{b}^{2}}=\dfrac{32}{5} $    (From equation (1) and (2)  )

Put in equation (2) , we get ${{a}^{2}}=\dfrac{32}{3}$

the value of a and b put in equation (1), we get


  $ \dfrac{{{x}^{2}}}{\dfrac{32}{3}}+\dfrac{{{y}^{2}}}{\dfrac{32}{5}}=1 $

 $ 3{{x}^{2}}+5{{y}^{2}}=32 $

This is required equation

S and S' foci of an ellipse. B is one end of the minor axis. If $\angle{SBS'}$ is a right angled isosceles triangle, then e$=?$

mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

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

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

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

  4. $\dfrac{3}{4}$

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Correct Option: A
Explanation:
We have
$S=(ae,0)\quad S'(-ae,0)and B=(0,b)$
Since it is given that $\angle SBS'=90^o$
Slope of SB$\times$ Slope of S'B$=-1$
$\left(\dfrac{b-0}{0-ae}\right)\times\left(\dfrac{b-0}{b+ae}\right)=-1$
$\left(\dfrac{-b}{ae}\right)\left(\dfrac{b}{ae}\right)=-1$
$b^2=a^2e^2$
But, $b^2=a^2(1-e^2)$
So,
$a^2(1-e^2)=a^2e^2$
$1-e^2=e^2$
$2e^2=1$
$e^2=\dfrac{1}{2}$
$e=\dfrac{1}{\sqrt2}$

The eccentricity of an ellipse is $\dfrac {\sqrt {3}}{2}$ its length of latus reetum is

mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

  1. $\dfrac {1}{2}$ (length of major axis)

  2. $\dfrac {1}{3}$ (length of major axis)

  3. $\dfrac {1}{4}$ (length of major axis)

  4. $\dfrac {2}{3}$ (length of major axis)

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Correct Option: C