Tag: standard equation of an ellipse

Questions Related to standard equation of an ellipse

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

The length of latus rectum of $\dfrac {x^2}9+\dfrac {y^2}2=1$ is 

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

  1. $\dfrac 74$

  2. $\dfrac 34$

  3. $\dfrac 43$

  4. None.

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Correct Option: C
Explanation:

The length of latus Rectum of $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$

is $\dfrac{2b^2}{a}$

Here $a=3\quad b=\sqrt 2$

$\Rightarrow \dfrac{2b^2}{a}=\dfrac{2(\sqrt 2)^2}{3}=\dfrac{2(2)}{3}=\dfrac 43$

An ellipse of semi-axis $a,b,$ slides between two perpendicular lines, then the locus of its foci is, (the two lines being taken  as the axes of coordinates)

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

  1. $(x^{2}+y^{2})(x^{2}y^{2}+b^{2})=4a^{2}x^{2}y^{2}$

  2. $(x^{2}+y^{2})(x^{2}y^{2}+b^{2})=4b^{2}x^{2}y^{2}$

  3. $(x^{2}-y^{2})(x^{2}y^{2}+b^{2})=4b^{2}x^{2}y^{2}$

  4. $(x^{2}-y^{2})(x^{2}y^{2}+b^{2})=4a^{2}x^{2}y^{2}$

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Correct Option: A

If equation $(5x-1)^{2}+(5y-2)^{2}=(\lambda^{2}-2\lambda+1)(3x+4y-1)^{2}$ represents an ellipse, then $\lambda \in$

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

  1. $(0, 1)$

  2. $(0, 2)$

  3. $(1, 2)$

  4. $(0, 1)\cup (1, 2)$

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Correct Option: B

The number of parabolas that can be drawn if two ends of the latus rectum are given 

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

  1. 1

  2. 2

  3. 4

  4. 3

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Correct Option: B

The equation $\dfrac{{x}^{2}}{2-r}+\dfrac{{y}^{2}}{r-5}+1=0$ represents an ellipse if

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

  1. $r>1$

  2. $r>5$

  3. $2 < r< 5$

  4. $r<2$ or $r>5$

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Correct Option: C