Tag: mathematics and statistics

Quizzes (38) Questions (955)

Questions Related to mathematics and statistics

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

The eccentricity of the ellipse $9x^2+5y^2-30 y=0$ is=

  1. $\dfrac{1}{3}$
  2. $\dfrac{2}{3}$
  3. $\dfrac{3}{4}$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$9x^2+5(y^2-6y)=0$
$9x^2+5(y^2-6y+9)=45$
$9x^2+5(y-3)^2=45$
$\dfrac { x^2 }{ 5 }+\dfrac { (y-3)^2 }{ 9 }=1$  
$a^2 < b^2$
$a^2=b^2(1-e^2)$
$5=9(1-e^2)$
$\dfrac { 5 }{ 9 }=1-e^2$
$e^2=\dfrac{4}{9}$ 
$e=\dfrac{2}{3}$

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

The equation of the ellipse whose vertices are $\left (2,-2\right),\left (2,4\right)$ and eccentricity is $a/3$ is- 

  1. $\dfrac { { \left( x-2 \right) }^{ 2 } }{ 9 } +\dfrac { { \left( y-1 \right) }^{ 2 } }{ 8 } =1$
  2. $\dfrac { { \left( x-2 \right) }^{ 2 } }{ 8 } +\dfrac { { \left( y-1 \right) }^{ 2 } }{ 9 } =1$
  3. $\dfrac { { \left( x+2 \right) }^{ 2 } }{ 8 } +\dfrac { { \left( y+1 \right) }^{ 2 } }{ 9 } =1$
  4. $\dfrac { { \left( x-2 \right) }^{ 2 } }{ 9 } +\dfrac { { \left( y+1 \right) }^{ 2 } }{ 8 } =1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The vertices are (2, -2) and (2, 4), so the center is (2, 1) and the major axis is vertical with length 2a = 6, so a = 3. Given e = a/3 = 1, this is a degenerate case or the eccentricity is defined differently. Assuming standard ellipse properties, the vertical orientation and center (2, 1) match option B.

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

Equations of the ellipse with centre $(1,2),$ one focus at $(6,2)$ and passing through $(4,6)$ is:

  1. $\dfrac{{{\left( x+1 \right)}^{2}}}{45}+\dfrac{{{\left( y-2 \right)}^{2}}}{20}=1$
  2. $\dfrac{{{\left( x-1 \right)}^{2}}}{45}+\dfrac{{{\left( y+2 \right)}^{2}}}{20}=1$
  3. $\dfrac{{{\left( x-1 \right)}^{2}}}{45}+\dfrac{{{\left( y-2 \right)}^{2}}}{20}=1$
  4. $\dfrac{{{\left( x+1 \right)}^{2}}}{45}+\dfrac{{{\left( y+2 \right)}^{2}}}{20}=1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We know that,

The equation of ellipse whose centre $(h, k)$.

$\dfrac{{{\left( x-h \right)}^{2}}}{{{a}^{2}}}+\dfrac{{{\left( y-k \right)}^{2}}}{{{b}^{2}}}=1$


Given centre of ellipse $\left( h,k \right)=\left( 1,2 \right)$

Then, equation of ellipse $\dfrac{{{\left( x-1 \right)}^{2}}}{{{a}^{2}}}+\dfrac{{{\left( y-2 \right)}^{2}}}{{{b}^{2}}}=1$          ……. (1)


But, the ellipse passes through given point $\left( x,y \right)=\left( 4,6 \right)$


By equation (1), we get

$ \dfrac{{{\left( 4-1 \right)}^{2}}}{{{a}^{2}}}+\dfrac{{{\left( 6-2 \right)}^{2}}}{{{b}^{2}}}=1 $

$ \dfrac{{{3}^{2}}}{{{a}^{2}}}+\dfrac{{{4}^{2}}}{{{b}^{2}}}=1 $

$ \Rightarrow 9{{b}^{2}}+16{{a}^{2}}={{a}^{2}}{{b}^{2}} $

$\Rightarrow 16{{a}^{2}}+9{{b}^{2}}={{a}^{2}}{{b}^{2}}$        …… (2)


Now, distance between focus and centre is $c=\sqrt{{{a}^{2}}-{{b}^{2}}}$

So,

$ c=\sqrt{{{\left( 1-6 \right)}^{2}}+{{\left( 2-2 \right)}^{2}}} $

$ c=\sqrt{{{5}^{2}}} $

$ c=5 $

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


On squaring both sides, we get,

${{a}^{2}}-{{b}^{2}}=25$          …… (3)


By equation (2) and (3), we get

$ 9{{b}^{2}}+400+16{{b}^{2}}=25{{b}^{2}}+{{b}^{4}} $

$ 25{{b}^{2}}+400=25{{b}^{2}}+{{b}^{4}} $

$ 400={{b}^{4}} $

$ {{b}^{4}}-400=0 $

$ \left( {{b}^{2}}-20 \right)\left( {{b}^{2}}+20 \right)=0 $

${{b}^{2}}-20=0\,$            and        ${{b}^{2}}+20=0\,$

${{b}^{2}}=20$                and        ${{b}^{2}}=-20$      (Rejected)


Put the value of ${{b}^{2}}$ in equation (3) and we get,

${{a}^{2}}-{{b}^{2}}=25$

$ {{a}^{2}}-20=25 $

$ {{a}^{2}}=45 $


Now, put the value of ${{a}^{2}}$ and ${{b}^{2}}$ in equation (1), we get,

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

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

Show that the equation $(10x-5)^2+(10y-5)^2=(3x+4y-1)^2$ represents an ellipse. Find the length of its latus rectum.

  1. $\dfrac{5}{2}$
  2. $\dfrac{1}{2}$
  3. $\dfrac{3}{2}$
  4. $-\dfrac{1}{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$(10x-5)^{2}+(10y-5)^{2}=(3x+4y-1)^{2}$
$25(2x-1)^{2}+25(2y-1)^{2}=(3x+4y-1)^{2}$
$\therefore (2x-1)^{2}+(2y-1)^{2}=\left(\dfrac{3x+4y-1}{5}\right)^{2}$
$4\left(x-\dfrac{1}{2}\right)^{2}+4\left(y-\dfrac{1}{2}\right)^{2}=\dfrac{1}{4}\left(\dfrac{3x+4y-1}{5}\right)^{2}$
$\left(x-\dfrac{1}{2}\right)^{2}+\left(y-\dfrac{1}{2}\right)^{2}=\dfrac{1}{4}\left(\dfrac{3x+4y-1}{5}\right)^{2}$
By observing equation $(1)$, we infer that 
$\sqrt{\left(x-\dfrac{1}{2}\right)^{2}+\left(y-\dfrac{1}{2}\right)^{2}}=\dfrac{1}{2}\left(\dfrac{3x+4y-1}{5}\right)$
Centre of ellipse: $\left(\dfrac{1}{2}, \dfrac{1}{2}\right)$
$\left(e^{2}=1-\dfrac{b^{2}}{a^{2}}\right)$
Equation of directrix $=(3x+4y-1=0)$
Also $d$ (Centre, directrix) $=\dfrac{3\left(\dfrac{1}{2}\right)+4\left(\dfrac{1}{2}\right)-1}{5}$
$=\dfrac{\dfrac{5}{2}+1}{5}=\boxed{\dfrac{1}{2}}$
$\therefore \boxed{a+a _{e}=\dfrac{1}{2}; a+\dfrac{a}{2}=\dfrac{1}{2}}$
$\therefore \dfrac{3a}{2}=\dfrac{1}{2}; \boxed{a\dfrac{1}{3}}$
$1-\dfrac{b^{2}}{a^{2}}=\dfrac{1}{4}; \dfrac{b^{2}}{a^{2}}=\dfrac{3}{4}; b=\dfrac{1}{\sqrt{12}}$
Length $=\dfrac{2b^{2}}{a}=2\left(\dfrac{1}{2}\right)\times \dfrac{1}{1/3}=\boxed{\dfrac{1}{2}}$




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

If the equation of the ellipse is $3x^2+2y^2+6x-8y+5=0$, then which of the following is/are true?

  1. $e=\dfrac {1}{\sqrt 3}$
  2. Center is $(-1, 2)$
  3. Foci are $(-1, 1)$ and $(-1, 3)$
  4. Directrices are $y=2\pm \sqrt 3$
Reveal answer Fill a bubble to check yourself
A,B,C Correct answer
Explanation

$3x^2+2y^2+6x-8y+5=0$
$\Rightarrow \dfrac {(x+1)^2}{2}+\dfrac {(y-2)^2}{3}=1$
Therefore, centre is $(-1, 2)$ and ellipse is vertical
$(\because b > a)$
$a^2=2, b^2=3$
Now $2=3(1-e^2)$
$\Rightarrow e=\dfrac {1}{\sqrt 3}$
Foci are $(-1, 2\pm be)$ and $(-1, 2\pm 1)$.
Hence, the foci are $(-1, 3)$ and $(-1, 1)$
The equations of the directrices are $y=2\pm \dfrac {b}{e}\Rightarrow y=5$ and $y=-1$

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

The eccentricity of the ellipse $\displaystyle 9x^{2}+4y^{2}-30y=0$ is $\displaystyle \frac{1}{p}\sqrt{q}$. Find the value $p $ and $q.$

  1. $p=2,q=2$
  2. $p=3,q=5$
  3. $p=2,q=5$
  4. $p=4,q=5$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given ellipse may be written as,
$\displaystyle9x^{2}+4\left ( y^{2}-\frac{15}{2}y+\frac{225}{16} \right

)=\frac{225}{4}$
or $\displaystyle \frac{x^{2}}{225/36}+\frac{\left (

y-15/4 \right )^{2}}{225/16}=1$
$\displaystyle \therefore

b^{2}=\frac{225}{16}, a^{2}=\frac{225}{36}$
$\displaystyle \therefore

e^{2}=1-\frac{a^{2}}{b^{2}}=1-\frac{16}{36}=1-\frac{4}{9}=\frac{5}{9}$
$\displaystyle

\therefore e=\frac{1}{3}\sqrt{5}.$