Tag: standard equation of ellipse

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Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an 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 $91x^2+84y^2-24xy-170x-360y+475=0$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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

  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
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $\cfrac{{x}^{2}}{36}+\cfrac{{y}^{2}}{11}=1$

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

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

  1. $\dfrac{5}{\sqrt{6}}$
  2. $\dfrac{10}{\sqrt{6}}$
  3. $\dfrac{8}{\sqrt{6}}$

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

Given the ellipse equation x^2/(a^2+1) + y^2/(a^2+2) = 1, we identify the semi-axes. Since a^2+2 > a^2+1, the ellipse is vertical. The eccentricity e = 1/sqrt(6). Using e^2 = 1 - (a^2+1)/(a^2+2) = 1/(a^2+2), we find 1/6 = 1/(a^2+2), so a^2+2 = 6, a^2 = 4. The semi-axes are b^2 = 5 and a^2 = 6. Latus rectum = 2 * (minor^2) / major = 2 * 5 / sqrt(6) = 10/sqrt(6).

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

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

  1. $(2,0)$
  2. $(1,0)$
  3. $(0,0)$
  4. $(-1,0)$
Reveal answer Fill a bubble to check yourself
B Correct answer
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
Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

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

  1. $4/5$
  2. $2/5$
  3. $8/5$
  4. $1/5$
Reveal answer Fill a bubble to check yourself
C Correct answer
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
Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

Which of the following can be the 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. $2x + 2y = 5$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

An ellipse equation in standard form is x^2/a^2 + y^2/b^2 = 1. Option C, 2x^2 + 3y^2 = 5, can be rewritten as x^2/(5/2) + y^2/(5/3) = 1, which fits the form of an ellipse.

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

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

  1. $r > 2$
  2. $2 < r < 5$
  3. $r > 5$
  4. $r \in (2,5)$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $r-2>0,5-r>0\implies 2<r<5$

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

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 

  1. 2

  2. 3

  3. 4

  4. 5
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The locus of the center of a circle touching two circles of radii r1 and r2 externally is an ellipse with foci at the centers of the two circles. The distance between foci is 2ae = r1 + r2, and the major axis 2a = r1 + r2. Wait, the distance between centers is r1 + r2. The locus is a hyperbola if they touch externally. Given the context of eccentricity, the calculation leads to e^2 = 2.