Tag: standard equation of ellipse

Questions Related to standard equation of ellipse

The equation of the ellipse whose equation of directrix is $3x+4y-5=0$, coordinates of the focus are $(1,2)$ and the eccentricity is $\dfrac{1}{2}$ is $91x^2+84y^2-24xy-170x-360y+475=0$

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

  1. True

  2. False

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Correct Option: A
Explanation:

Let $P(x,y)$ be any point on the ellipse and PM be the perpendicular from P upon the directrix $3x+4y-5=0$.

Then by the definition,
$\dfrac{SP}{PM}=e$

$SP=e.PM$
$\sqrt{(x-1)^2+(y-2)^2}=\dfrac{1}{2}|\dfrac{3x+4y-5}{\sqrt{3^2+4^2}}|$

$(x-1)^2+(y-2)^2=\dfrac{1}{4}. \dfrac{(3x+4y-5)^2}{25}$

$100(x^2+y^2-2x-4y+5)=9x^2+16y^2+24xy-30x-40y+25$
$91x^2+84y^2-24xy-170x-360y+475=0$ is the equation of the ellipse.

The equation of the ellipse whose foci are $(\pm5,0)$ and of the directrix is $5x=36$, is

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

  1. $\dfrac{x^2}{36}+\dfrac{y^2}{11}=1$

  2. $\dfrac{x^2}{6}+\dfrac{y^2}{\sqrt{11}}=1$

  3. $\dfrac{x^2}{6}+\dfrac{y^2}{11}=1$

  4. None of these

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Correct Option: A
Explanation:

Given foci $(\pm 5,0)$ and directrix $x=\cfrac{36}{5}$

Then $ae=5$ (focus coordinates ($\pm ae,0)]$....(1)
$\cfrac{a}{e}=\cfrac{36}{5}$ (directrix equation $x=\cfrac{a}{e}$]....(2)
From (1) and (2) ${a}^{2}=36\Rightarrow$ $a=6$
$e=\cfrac{5}{6}\Rightarrow $ $\sqrt { 1-\cfrac { { b }^{ 2 } }{ { a }^{ 2 } }  } =\cfrac { 5 }{ 6 } $
$1-\cfrac { { b }^{ 2 } }{ 36 } =\cfrac{25}{36}$
$b=\sqrt 11$
required equation $\cfrac{{x}^{2}}{36}+\cfrac{{y}^{2}}{11}=1$

If the eccentricity of the ellipse $\dfrac{x^2}{a^2 + 1} + \dfrac{y^2}{a^2 + 2 } = 1$ is $\dfrac{1}{\sqrt{6}}$, then the length of latusrectum is

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

  1. $\dfrac{5}{\sqrt{6}}$

  2. $\dfrac{10}{\sqrt{6}}$

  3. $\dfrac{8}{\sqrt{6}}$

  4. None of these

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Correct Option: B

If focus of the parabola is $(3,0)$ and length of latus rectum is $8$, then its vertex is

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

  1. $(2,0)$

  2. $(1,0)$

  3. $(0,0)$

  4. $(-1,0)$

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Correct Option: B
Explanation:
Given, focus $=(3,0)$ and Length of latus rectum $= 8$

$\Rightarrow 4a=8$ $\Rightarrow a=2$

$\Rightarrow$ Vertex = $(3-a,0)$ $=(1,0)$

$\therefore $ Option B is correct

If $(0,0)$ be the vertex and $3x-4y+2=0$ be the directrix of a parabola, then the length of its latus rectum is

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

  1. $4/5$

  2. $2/5$

  3. $8/5$

  4. $1/5$

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Correct Option: C
Explanation:
Distance of vertex from directrix = $\dfrac{\left | 3(0)-4(0)+2 \right |}{\sqrt{3^{2}+4^{2}}}= \dfrac{2}{5}=a$

Length of latus rectum = $4a= \dfrac{8}{5}$

$\therefore $ Option C is correct

Eccentricity of an ellipse is $\sqrt {\cfrac{2}{5}} $ and it passes through the point $(-3,1)$ then its equation is 

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

  1. $3{x^2} + 5{y^2} = 32$

  2. $2{x^2} + 3{y^2} = 33$

  3. $3{x^2} + 4{y^2} = 30$

  4. $2{x^2} + 3{y^2} = 34$

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Correct Option: B

If $P = (x, y), F _1 = (3, 0)$ and $16x^2 + 25y^2 = 400$, then $PF _1 + PF _2$ equals

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

  1. $8$

  2. $6$

  3. $10$

  4. $12$

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Correct Option: C
Explanation:
$PF _1+PF _2=2a$

$\dfrac{x^2}{5^2}+\dfrac{y^2}{4^2}=1\Rightarrow a=5,b=4$

$\therefore PF _1+PF _2=2(5)$

$=10$

Which of the following can be the equation of an ellipse?

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

  1. $x^{2} + y^{2} = 5$

  2. $\dfrac {x^{2}}{9} + \dfrac {x^{2}}{9} = 1$

  3. $2x^{2} + 3y^{2} = 5$

  4. $2x + 2y = 5$

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Correct Option: C

The equation $\dfrac {x^{2}}{2-r}+\dfrac {y^{2}}{r-5}+1=0$ represents an ellipse, if

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

  1. $r > 2$

  2. $2 < r < 5$

  3. $r > 5$

  4. $r \in (2,5)$

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Correct Option: B
Explanation:

Given $\dfrac{x^2}{2-r}+\dfrac{y^2}{r-5}+1=0$ represents a ellipse

$\implies \dfrac{x^2}{2-r}+\dfrac{y^2}{r-5}=-1$
$\implies \dfrac{x^2}{r-2}+\dfrac{y^2}{5-r}=1$
Since this equation is an ellipse so $r-2>0,5-r>0\implies 2<r<5$

The locus of center of a variable circle touching the circle of radius ${ r } _{ 1 }and{ r } _{ 2 }$ extemally which also touch each other externally , is a conic of the eccentricity $e$.If $\dfrac { { r } _{ 1 } }{ { r } _{ 2 } } =3+2\sqrt { 2 } $ then ${ e }^{ 2 }$ is 

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

  1. 2

  2. 3

  3. 4

  4. 5

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Correct Option: A