Tag: ellipse

Quizzes (5) Questions (224)

Questions Related to ellipse

Multiple choice maths ellipse special cases of an ellipse eccentricity equation of ellipse

If the distance of one of the focus of hyperbola from the two directrices of hyperbola are 5 and 3, then its eccentricity is

  1. $\sqrt{2}$

  2. 2

  3. 4

  4. 8
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Focus $=(\pm ae, o)$
directive $x \Rightarrow \pm a/e$ 
$\left(ae- \dfrac{a}{e} \right)= 3 \left(ae+ \dfrac{a}{e} \right)=5$
$\dfrac{a (e^{2}-1)= 3e}{a (e^{2}+1)= 5e} \Rightarrow 5e^{2}-5 =-3 e^{2}+3$
$2 e^{2}=8$
$e^{2}= 4$
$e=2$
Multiple choice maths ellipse special cases of an ellipse eccentricity equation of ellipse

The eccentricity of the conic represented by $\sqrt{(x+2)^2+y^2}+\sqrt{(x-2)^2+y^2}=8$ is?

  1. $\dfrac13$
  2. $\dfrac12$
  3. $\dfrac14$
  4. $\dfrac15$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The equation represents the sum of distances from (x,y) to (-2,0) and (2,0) being 8. This is an ellipse with 2a = 8 (so a = 4) and 2ae = 4 (distance between foci (-2,0) and (2,0)). Thus ae = 2, so 4e = 2, e = 1/2.

Multiple choice maths ellipse special cases of an ellipse eccentricity equation of ellipse

The parabola $( y + 1 ) ^ { 2 } = a ( x - 2 )$ passes through the point $( 1 , - 2 )$ then the equation of its directrix is

  1. $4 x + 1 = 0$
  2. $4 x - 1 = 0$
  3. $4 x + 9 = 0$
  4. $4 x - 9 = 0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equation of parabola is $(y+1)^2=a(x-2)$


it passes through $(1,-2)$

$\implies (-2+1)^2=a(1-2)\$

$(-1)^2=-a\$

$a=-1$

So the equation of a parabola is 

$(y+1)^2=-1(x-2)\$

$(y+1)^2=4\left(\dfrac{-1}{4}\right)(x-2)$

the directrix of parabola is $x=\dfrac{-1}{4}\$

$4x+1=0$

Multiple choice maths ellipse special cases of an ellipse eccentricity equation of ellipse

The eccentricity of the conic represented by the equation $x^{2} + 2y^{2} - 2x + 3y + 2 = 0$ is

  1. $0$
  2. $\dfrac{1}{2}$
  3. $\dfrac{1}{\sqrt{2}}$
  4. $\sqrt{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$x^2 + 2y^2 - 2x + 3y + 2 = 0$
$\Rightarrow (x - 1)^2 + 2 (y + \dfrac34)^2 = \dfrac{1}{8}$
$\Rightarrow \dfrac {(x - 1)^2}{1 / 8} + \dfrac {(y + 3 / 4)^2}{1 / 16} = 1$
It is an ellipse with $a^2 = 1/8 , b^2 = 1/16$ .Hence its eccentricity
$e = \sqrt {1 - \dfrac{b^2}{a^2}} = \sqrt {1 - \dfrac8{16}} = \dfrac1{\sqrt 2}$

Multiple choice maths ellipse special cases of an ellipse eccentricity equation of ellipse

The eccentricity of the conic $9{ x }^{ 2 }+5{ y }^{ 2 }-54x-40y+116=0$ is:

  1. $\cfrac { 1 }{ 3 } $
  2. $\cfrac { 2 }{ 3 } $
  3. $\cfrac { 4 }{ 9 } $
  4. $\cfrac { 2 }{ \sqrt { 5 } } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given conic is $9x^2+5y^2-54x-40y+116=0$


$\Rightarrow 9(x^2-6x)+5(y^2-8y)+116=0$

$\Rightarrow 9(x-3)^2+5(y-4)^2-81-80+116=0$

$\Rightarrow 9(x-3)^2+5(y-4)^2-45=0$

$\Rightarrow 9(x-3)^2+5(y-4)^2=45$

Divide both sides by $45$, we get

$\dfrac{(x-3)^2}{5}+\dfrac{(y-4)^2}{9}=1$ which is in the standard form $\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1$ of ellipse.

Thus $b^2=5, a^2=9$

Eccentricity $=\sqrt{1-\dfrac{b^2}{a^2}}$

$\Rightarrow e=\sqrt{1-\dfrac{5}{9}}$

$=\sqrt{\dfrac{9-5}{9}}$

$=\sqrt{\dfrac{4}{9}}$

$\therefore e=\dfrac{2}{3}$

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