Tag: eccentricity

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Questions Related to eccentricity

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

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

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

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

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