Tag: special cases of an ellipse

Questions Related to special cases of an 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

maths ellipse special cases of an ellipse eccentricity equation of ellipse

  1. $\sqrt{2}$

  2. 2

  3. 4

  4. 8

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Correct Option: B
Explanation:
Focus $=(\pm ae, o)$
directive $x \Rightarrow \pm a/e$ 
$\left(ae- \dfrac{a}{e} \right)= 3 \left(ae+ \dfrac{a}{e} \right)=5$
$\dfrac{a (e^{2}-1)= 3e}{a (e^{2}+1)= 5e} \Rightarrow 5e^{2}-5 =-3 e^{2}+3$
$2 e^{2}=8$
$e^{2}= 4$
$e=2$

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

maths ellipse special cases of an ellipse eccentricity equation of ellipse

  1. $\dfrac13$

  2. $\dfrac12$

  3. $\dfrac14$

  4. $\dfrac15$

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Correct Option: B

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

maths ellipse special cases of an ellipse eccentricity equation of ellipse

  1. $4 x + 1 = 0$

  2. $4 x - 1 = 0$

  3. $4 x + 9 = 0$

  4. $4 x - 9 = 0$

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Correct Option: A
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}\$

$4x+1=0$

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

maths ellipse special cases of an ellipse eccentricity equation of ellipse

  1. $0$

  2. $\dfrac{1}{2}$

  3. $\dfrac{1}{\sqrt{2}}$

  4. $\sqrt{2}$

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Correct Option: C
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
$e = \sqrt {1 - \dfrac{b^2}{a^2}} = \sqrt {1 - \dfrac8{16}} = \dfrac1{\sqrt 2}$

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

maths ellipse special cases of an ellipse eccentricity equation of ellipse

  1. $\cfrac { 1 }{ 3 } $

  2. $\cfrac { 2 }{ 3 } $

  3. $\cfrac { 4 }{ 9 } $

  4. $\cfrac { 2 }{ \sqrt { 5 } } $

Show answer
Correct Option: B
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 $=\sqrt{1-\dfrac{b^2}{a^2}}$

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

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

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

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