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

  1. A circle for all $\lambda$
  2. An ellipse if $\lambda < 0$
  3. The empty set if $\lambda > 0$
  4. A-point if $\lambda = 0$
Reveal answer Fill a bubble to check yourself
D Correct answer
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’.

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

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

  1. $36x^2+20y^2=405$
  2. $20x^2+36y^2=405$
  3. $30x^2+22y^2=411$
  4. $22x^2+32y^2=409$
Reveal answer Fill a bubble to check yourself
B Correct answer
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$
Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

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 

  1. $\frac{2}{{\sqrt 3 }}$
  2. $\frac{4}{{\sqrt 3 }}$
  3. $2\sqrt 3 $
  4. $\frac{{\sqrt 3 }}{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 }  } $
Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

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

  1. $\pi / 2$
  2. $\pi / 6$
  3. $\pi / 3$
  4. $5$ $\pi/ 12$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 
$\phi = \dfrac{ 5 \pi}{12}$

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

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

  1. Ellipse

  2. Parabola

  3. Circle

  4. Straight line
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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.

  1. Straight line

  2. Ellipse

  3. Parabola

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
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
Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

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 

  1. $a:b$
  2. $a^2:b^2$
  3. $b^2:a^2$
  4. $b:a$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For an ellipse x^2/a^2 + y^2/b^2 = 1, the normal at P(x1, y1) meets the axes at G and g. The ratio PG:Pg is a:b.

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

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

  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
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The normal to 4x^2 + 9y^2 = 36 (x^2/9 + y^2/4 = 1) parallel to 2x + y = 3 (slope -2). The normal slope is -2. Using the normal equation y = mx - (a^2-b^2)m/sqrt(a^2+b^2m^2), we find the point.