Tag: ellipse

Quizzes (5) Questions (224)

Questions Related to ellipse

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

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

  1. ${x^2} - ay = {a^3}$
  2. ${x^2} - ay - {e^2}$
  3. ${x^2} \pm ay = {a^2}$
  4. ${x^2} + ay - {b^2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The extremities of the latus rectum for the ellipse x^2/a^2 + y^2/b^2 = 1 are (ae, b^2/a) and (ae, -b^2/a). With b^2 = a^2(1-e^2), the coordinates are (ae, a(1-e^2)). Eliminating e leads to the locus x^2 = ay.

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

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

  1. $y=\pm \sqrt {5}$
  2. $x=\pm \sqrt {5}$
  3. $y=1 \pm \sqrt {5}$
  4. $x=1 \pm \sqrt {5}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 } $

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

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:

  1. $5$
  2. $7$
  3. $9$
  4. $4$
Reveal answer Fill a bubble to check yourself
B Correct answer
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)
$16-9=b^2\Rightarrow b^2=7$

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

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

  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. $\dfrac { { x }^{ 2 } }{ 1/4 } +\dfrac { { y }^{ 2 } }{ 3/4 } =1$
Reveal answer Fill a bubble to check yourself
A Correct answer
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$\Rightarrow \dfrac{x^2}{\dfrac{1}{4}}+\dfrac{y^2}{\dfrac{5}{4}}=1$

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

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

  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. $\dfrac{(x-1)^2}{45}+\dfrac{(y-2)^2}{16}=1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Center (1, 2), Focus (6, 2). Distance ae = 5. Ellipse passes through (4, 6). Using the definition of an ellipse (sum of distances to foci = 2a), we can solve for a and b.

Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse 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

  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. $y cos \frac{5\pi}{8}+x sin \frac{5\pi}{8}=1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The product of perpendiculars from foci to any tangent of an ellipse is b^2. The sum of perpendiculars is a property related to the tangent equation. The given option A fits the standard form.

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

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 $'0'$

  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\} $
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $\theta \epsilon R$.
Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

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 

  1. $5x^3+3y^2-48=0$
  2. $3x^2+5y^2-15=0$
  3. $5x^2+3y^2-32=0$
  4. $3x^2+5y^2-32=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
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

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

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$=?$

  1. $\dfrac{1}{\sqrt{2}}$
  2. $\dfrac{1}{2}$
  3. $\dfrac{\sqrt{3}}{2}$
  4. $\dfrac{3}{4}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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,
$a^2(1-e^2)=a^2e^2$
$1-e^2=e^2$
$2e^2=1$
$e^2=\dfrac{1}{2}$
$e=\dfrac{1}{\sqrt2}$
Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

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

  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)
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Latus rectum L = 2b^2/a. Given e = sqrt(3)/2, e^2 = 3/4. Since b^2 = a^2(1-e^2), b^2 = a^2(1/4). So L = 2(a^2/4)/a = a/2. The length of the major axis is 2a. Thus, L = (a/2) / (2a) = 1/4 of the major axis.