Tag: standard equation of an ellipse

Questions Related to standard equation of an ellipse

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

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

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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}.$

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

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

  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  $\displaystyle y=1\pm \frac{4}{\sqrt{3}}$

Show answer
Correct Option: C
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}}.$

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

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

  1. $\displaystyle \frac{1}{3} $

  2. $\displaystyle \frac{2}{3} $

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

  4. $\displaystyle \frac{8}{3} $

Show answer
Correct Option: D
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 $=\dfrac{2b^2}{a}=\dfrac{8}{3}$