Tag: ellipse

Questions Related to ellipse

The equation $2x^2+3y^2-8x-18y+35=\lambda$ represents?

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

  1. A circle for all $\lambda$

  2. An ellipse if $\lambda < 0$

  3. The empty set if $\lambda > 0$

  4. A-point if $\lambda = 0$

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Correct Option: D
Explanation:

Given:

$ 2x^{2} + 3y - 8x - 18y + 35 = \lambda  $

$ 2\left (x^{2} - 4x \right ) + 3 \left ( y^{2} - 6y + 35 \right ) = \lambda  $

$ 2\left (x - 2 \right )^{2} + 3 \left ( y - 3 \right )^{2} = \lambda  $

For $ \lambda = 0 $, then

$ 2\left (x - 2 \right )^{2} + 3 \left ( y - 3 \right )^{2} = 0  $

Thus, the point is $ \left ( 2,3 \right ) $.

Hence, the correct option is ‘d’.

The equation of ellipse whose major axis is along the direction of x-axis, eccentricity is $e=2/3$

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

  1. $36x^2+20y^2=405$

  2. $20x^2+36y^2=405$

  3. $30x^2+22y^2=411$

  4. $22x^2+32y^2=409$

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Correct Option: B
Explanation:
$\begin{array}{l} e=\frac { 2 }{ 3 } =\sqrt { \frac { { { a^{ 2 } }-{ b^{ 2 } } } }{ c }  }  \\ \Rightarrow { \left( { \frac { 2 }{ 3 }  } \right) ^{ 2 } }=\frac { { { a^{ 2 } }-{ b^{ 2 } } } }{ { { a^{ 2 } } } }  \\ \Rightarrow \frac { { 4{ a^{ 2 } } } }{ a } ={ a^{ 2 } }-{ b^{ 2 } } \\ \Rightarrow { b^{ 2 } }={ a^{ 2 } }-4{ a^{ 2 } }=\frac { { 5{ a^{ 2 } } } }{ 9 } \to \left( i \right)  \end{array}$
Equation of Ellipse are
$\begin{array}{l} \Rightarrow \frac { { { x^{ 2 } } } }{ { { a^{ 2 } } } } +\frac { { { y^{ 2 } } } }{ { { b^{ 2 } } } } =1 \\ \Rightarrow \frac { { { x^{ 2 } } } }{ { { a^{ 2 } } } } +\frac { { 9{ y^{ 2 } } } }{ { 5{ a^{ 2 } } } } =1\to \left( { ii } \right)  \\ Put\, \, { a^{ 2 } }=\frac { { 405 } }{ { 20 } } \, \, \left( { From\, \, option\, \, in\, \, equation\left( i \right)  } \right)  \\ Then,\, \, { b^{ 2 } }=\frac { { 405 } }{ { 360 } }  \end{array}$
Hence, equation of ellipse is
$ \Rightarrow \frac{{{x^2}}}{{\left( {\frac{{405}}{{20}}} \right)}} + \frac{{{y^2}}}{{\left( {\frac{{405}}{{36}}} \right)}} =  - 1,20{x^2} + 36{y^2} = 405$

Eccentricity of ellipse $\frac{{{x^2}}}{{{a^2} + 1}} + \frac{{{y^2}}}{{{a^2} + 2}} = 1\,is\,\frac{1}{{\sqrt 3 }}$ then length of Latus rectum is 

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

  1. $\frac{2}{{\sqrt 3 }}$

  2. $\frac{4}{{\sqrt 3 }}$

  3. $2\sqrt 3 $

  4. $\frac{{\sqrt 3 }}{2}$

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Correct Option: B
Explanation:
Let ${ A }^{ 2 }={ a }^{ 2 }+1$,  ${ B }^{ 2 }={ a }^{ 2 }+2$    {Here, ${ B }^{ 2 }>{ A }^{ 2 }$}
So, $e=\sqrt { 1-\dfrac { { A }^{ 2 } }{ { B }^{ 2 } }  } =\sqrt { 1-\dfrac { { a }^{ 2 }+1 }{ { a }^{ 2 }+2 }  } =\dfrac { 1 }{ \sqrt { { a }^{ 2 }+2 }  } =\dfrac { 1 }{ \sqrt { 3 }  } $
So,  $\sqrt { { a }^{ 2 }+2 } =\sqrt { 3 } \Rightarrow a=\pm 1$
So, length of lotus return $=2\dfrac { { A }^{ 2 } }{ B } $
Length $=\dfrac { 2\left( { a }^{ 2 }+1 \right)  }{ \sqrt { { a }^{ 2 }+2 }  } =\dfrac { 2\left( 2 \right)  }{ \sqrt { 3 }  } =\dfrac { 4 }{ \sqrt { 3 }  } $

If the latus rectum of an ellipse $x ^ { 2 } \tan ^ { 2 } \varphi + y ^ { 2 } \sec ^ { 2 } \varphi =$ $1$ is $1 / 2 ,$ then $\varphi$ is

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

  1. $\pi / 2$

  2. $\pi / 6$

  3. $\pi / 3$

  4. $5$ $\pi/ 12$

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Correct Option: D
Explanation:

Given $x^2 tan^2 \phi + y^2 \, sec^2 \phi = 1$

$\rightarrow \dfrac{x^2}{(1/tan^2 \phi)} + \dfrac{y^2}{(1/sec^2 \phi)} = 1$
$a = \pm \dfrac{1}{tan \phi} , b = \pm \dfrac{1}{sec \phi}$
and $\rightarrow e^2 = 1 - \dfrac{b^2}{a^2}$
$\rightarrow e^2 = 1 - \dfrac{1/sec^2 \phi}{1/tan^2 \phi} = 1 - \dfrac{tan^2 \phi}{sec^2 \phi}$
$\rightarrow e^2 = 1 - sin^2 \phi = cos^2 \phi$
length of latus rectum 
$(LL') = \dfrac{2 b^2}{a} = 2a (1 - e^2)$
$\rightarrow 2a (1 - cos^2 \phi) = 2a. sin^2 \phi = \dfrac{1}{2} $ (Given)
$\therefore 2. \dfrac{cos \phi}{sin \phi} sin^2 \phi = \dfrac{1}{2}$
$\rightarrow 2 cos \phi \, sin \phi = \dfrac{1}{2}$
$\rightarrow sin^2 \phi = \dfrac{1}{2} $
$\rightarrow 2 \phi = \dfrac{\pi}{6} , \dfrac{5 \pi}{6}$
$\therefore \phi = \dfrac{\pi}{12}$    or 
$\phi = \dfrac{ 5 \pi}{12}$

The curve represented by $Rs \left(\dfrac{1}{z}\right)=C$ is (where $C$ is a constant and $\neq 0$)

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

  1. Ellipse

  2. Parabola

  3. Circle

  4. Straight line

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Correct Option: C
Explanation:

$\begin{array}{l} { { Re } }\, \, \left( { \frac { 1 }{ z }  } \right) =c \ { { Re } }\, \, \left( { \frac { 1 }{ { x+iy } }  } \right) =c \ { { Re } }\, \, \left( { \frac { { x-iy } }{ { { x^{ 2 } }+{ y^{ 2 } } } }  } \right) =c \ \frac { x }{ { { x^{ 2 } }+{ y^{ 2 } } } } =c \ c\left( { { x^{ 2 } }+{ y^{ 2 } } } \right) -{ x }=0 . \end{array}$


Hence, this is represent circle.

The eccentricity of an ellipse whose centre is at the origin is $\frac{1}{2}$.If one of its directrices is $x=-4$, then the equation of the normal to it at $(1, \frac{3}{2})$ is:

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

  1. $4x+2y=7$

  2. $x+2y=4$

  3. $2y-x=2$

  4. $4x-2y=1$

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Correct Option: A

A point $(\alpha, \beta)$ lies on a circle $x^2+y^2=1$, then locus of the point $(3\alpha +2\beta)$ is a$/$an.

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

  1. Straight line

  2. Ellipse

  3. Parabola

  4. None of these

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Correct Option: B
Explanation:
Point will be $(3\alpha ,2\beta )$ not $( 3\alpha +2\beta )$
Now $ x^{2}+y^{2}=1 $
Radium is $1$ unit,hence parametric co - ordinate is 
$(\alpha ,\beta )= (1\cos\theta ,1\sin\theta )=(\cos\theta , \sin\theta )$
Hence
Point is $ (3\cos\theta ,2\sin\theta )$
Hence
$(x,y)= (3\cos\theta ,2\sin\theta )$
$x=3\cos\theta $
$ \Rightarrow \dfrac{x}{3}\cos\theta$    ...(i)
$ y=2\sin\theta $
$ \Rightarrow \dfrac{y}{2} = \sin \theta$   ...(ii)
$ (i)^{2} + (ii)^{2} $
$\dfrac{x^{2}}{9} + \dfrac{y^{2}}{4} \cos^{2}\theta + \sin^{2} \theta $
$ \dfrac{x^{2}}{9}+ \dfrac{y^{2}}{4} = 1 $
which is equation of ellipse

If a normal is drawn at point $P$ of ellipse $ \dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$, then the maximum distance from centre of ellipse will be $a-b$ 

maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

  1. True

  2. False

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Correct Option: A

If the normal at any point $P$ of the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ meets the axes in $G$ and $g$ respectively, then $|PG| : |Pg|$ is equal to 

maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

  1. $a:b$

  2. $a^2:b^2$

  3. $b^2:a^2$

  4. $b:a$

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Correct Option: A

One foot of normal of the ellipse $4x^2$ $+$ 9$y^2$ $= 36 $, that is parallel to the line $2x + y = 3 $, is

maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

  1. $\left ( \dfrac{9}{8}, \dfrac{5}{8} \right )$

  2. $\left ( \dfrac{9}{8}, \dfrac{8}{5} \right )$

  3. $\left ( \dfrac{8}{9}, \dfrac{8}{5} \right )$

  4. None

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Correct Option: B