Tag: introduction to ellipse

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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}.$

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

For the ellipse 4x2+y2−8x+2y+1=04x2+y2−8x+2y+1=0 which of the following statements are correct:

  1. Foci are $\displaystyle \left ( -1, -1\pm \sqrt{3} \right ),$ Directrices are $\displaystyle y=-1\pm \frac{4}{\sqrt{3}}$
  2. Foci are $\displaystyle \left ( 1, 1\pm \sqrt{3} \right ),$ Directrices are $\displaystyle y=1\pm \frac{4}{\sqrt{3}}$
  3. Foci are $\displaystyle \left ( 1, -1\pm \sqrt{3} \right ),$ Directrices are $\displaystyle y=-1\pm \frac{4}{\sqrt{3}}$
  4. Foci are $\displaystyle \left ( 1, -1\pm \sqrt{3} \right ),$ Directrices are $\displaystyle y=1\pm \frac{4}{\sqrt{3}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\displaystyle 4x^{2}+y^{2}-8x+2y+1=0$
$\displaystyle 4(x^2-2x)+(y^2+2y)=-1$
$\displaystyle 4(x^2-2x+1)+(y^2+2y+1)=-1+5=4$
$\displaystyle 4(x-1)^2+(y+1)^2=4$
$\displaystyle \frac{\left ( x-1 \right )^{2}}{1}+\frac{\left (

y+1 \right )^{2}}{4}=1$
$\displaystyle \Rightarrow a^2 =1, b^2 = 4$ Clearly here $a^2< b^2$ so the axis of the ellipse is parallel to y-axis
Now,  $\displaystyle e=\sqrt{1-\frac{a^2}{b^2}}=\frac{\sqrt{3}}{2} \therefore $ Foci $\displaystyle \left

( 1, -1\pm \sqrt{3} \right ),$ Directrices  $\displaystyle y=-1\pm

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

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

Find the length of latus rectum of the ellipse $4x^2\, +\, 9y^2\, \,+ 8x\, \,+ 36y\, +\, 4\, =\, 0$.

  1. $\displaystyle \frac{1}{3} $
  2. $\displaystyle \frac{2}{3} $
  3. $\displaystyle \frac{4}{3} $
  4. $\displaystyle \frac{8}{3} $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$4x^2+9y^2+8x+36y+4=0$
$\Rightarrow 4(x^2+2x)+9(y^2+4y)=-4$
$\Rightarrow 4(x^2+2x+1)+9(y^2+4y+4)=-4+4+36$
$\Rightarrow 4(x+1)^2+9(y+2)^2=36$
$\Rightarrow \dfrac{(x+1)^2}{9}+\dfrac{(y+2)^2}{4}=1$
$\therefore a^2=9,b^2=4$
Hence length of latus rectum is $=\dfrac{2b^2}{a}=\dfrac{8}{3}$