Tag: mathematics and statistics

Questions Related to mathematics and statistics

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. $5$ $\pi/ 12$

Show answer
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

Show answer
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. $4x-2y=1$

Show answer
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

Show answer
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

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

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

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

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

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

  4. None of these

Show answer
Correct Option: B
Explanation:

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

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

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

  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. $\dfrac { { \left( x-2 \right) }^{ 2 } }{ 9 } +\dfrac { { \left( y+1 \right) }^{ 2 } }{ 8 } =1$

Show answer
Correct Option: B

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

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

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

Show answer
Correct Option: C
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,

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

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

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

  1. $\dfrac{5}{2}$

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

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

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

Show answer
Correct Option: C
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 $=\dfrac{2b^{2}}{a}=2\left(\dfrac{1}{2}\right)\times \dfrac{1}{1/3}=\boxed{\dfrac{1}{2}}$




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

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

  1. $e=\dfrac {1}{\sqrt 3}$

  2. Center is $(-1, 2)$

  3. Foci are $(-1, 1)$ and $(-1, 3)$

  4. Directrices are $y=2\pm \sqrt 3$

Show answer
Correct Option: A,B,C
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 $y=-1$

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

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

  1. $p=2,q=2$

  2. $p=3,q=5$

  3. $p=2,q=5$

  4. $p=4,q=5$

Show answer
Correct Option: B
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}.$