Tag: maths

Questions Related to maths

The eccentricity of the ellipse $\dfrac {x^{2}}{a^{2}} + \dfrac {y^{2}}{b^{2}} = 1$ if its latus-rectum is equal to one half of its minor axis, is

maths ellipse special cases of an ellipse eccentricity equation of ellipse

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

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

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

  4. None of these

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Correct Option: B
Explanation:
The given equation of ellipse is:

$\dfrac {x^2}{a^2}+\dfrac {y^2}{b^2}=1$

According to equation

latus rectum $=\dfrac12 \times$ mirror axis

i.e. $\dfrac {2b^2}{a}=\dfrac 12 \times 2b$

$2b^2 =ab$

$a=2\ b$

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

$e=\sqrt {1-\dfrac {b^2}{4\ b^2}}$

$e=\sqrt {1-\dfrac {1}{4}}$

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

if the distance between the foci is equal to the length of the latus-rectum. Find the eccentricity of the ellipse.

maths ellipse special cases of an ellipse eccentricity equation of ellipse

  1. $\dfrac {\sqrt {5} - 1}{2}$

  2. $\dfrac {\sqrt {5} + 1}{2}$

  3. $\dfrac {\sqrt {5} - 1}{4}$

  4. None of these

Show answer
Correct Option: A
Explanation:
Given
Distance between the foci of an ellipse = length of latus rectum

i.e. $\dfrac {2b^2}{a}=2\ ae$

$e=b^2 /a^2$

But $e=\sqrt {1-b^2 /a^2}$

Then $e=\sqrt {1-e}$

Squaring both sides, we get

$e^2+e-1=0$

$e=\dfrac {-1\pm \sqrt {1+4}}{2}$ $(\because $ Eccentricity cannot be negative)

$e=\dfrac {\sqrt 5 -1}{2}$

Find the eccentricity of the conic represented by $x^2\, -\, y^2\,- \, 4x\, +\, 4y\, +\, 16\, =\, 0$

maths ellipse special cases of an ellipse eccentricity equation of ellipse

  1. $\sqrt2$

  2. $\sqrt {3}$

  3. $- \sqrt {2}$

  4. $- \sqrt {3}$

Show answer
Correct Option: A
Explanation:

Given conic can be written as,

$x^2-4x +4 -(y^2 -4y +4) +16 =0$
$(x-2)^2 - (y-2)^2 = -16$
$\displaystyle \frac{(x\, -\, 2)^2}{16}\, -\, \frac{(y\, -\, 2)^2}{16}\, =\, -1$
Which is a rectangular hyperbola so its eccentricity is $\sqrt2$

If $e _{1}$ is the eccentricity of the ellipse $\displaystyle \frac{x^{2}}{16}+\frac{y^{2}}{25}=1$ and $e _{2}$ is the eccentricity of the hyperbola passing through the foci of the ellipse and $e _{1}e _{2}=1$, then equation of the hyperbola is

maths ellipse special cases of an ellipse eccentricity equation of ellipse

  1. $\displaystyle \frac{x^{2}}{9}-\frac{y^{2}}{16}=1$

  2. $\displaystyle \frac{x^{2}}{16}-\frac{y^{2}}{9}=-1$

  3. $\displaystyle \frac{x^{2}}{9}-\frac{y^{2}}{25}=1$

  4. $\displaystyle \frac{x^{2}}{25}-\frac{y^{2}}{9}=1$

Show answer
Correct Option: B
Explanation:

We have ${ e } _{ 1 }=\sqrt { 1-\cfrac { 16 }{ 25 }  } =\cfrac { 3 }{ 5 } $
$\because \quad { e } _{ 1 }{ e } _{ 2 }=1\Rightarrow { e } _{ 2 }=\cfrac { 5 }{ 3 } $
Clearly y-axis is transverse axis of the ellipse.
Thus, coordinates of focii of the ellipse are $(0,\pm b e _1)$ or $\left( 0,\pm 3 \right) $. 
Let hyperbola is, $\cfrac{y^2}{b^2}-\cfrac{x^2}{a^2}=1..(1)$ 
given hyperbola passes through foci of the ellipse
$\Rightarrow b^2=9$ and also $a^2=b^2(e^2-1)=9(25/9-1)=16$
Therefore, required hyperbola is, $\cfrac{x^2}{16}-\cfrac{y^2}{9}=-1$
Hence, option 'A' is correct.

What is the eccentricity of the conic $4x^2 + 9 y^2 = 144 $

maths ellipse special cases of an ellipse eccentricity equation of ellipse

  1. $\dfrac{\sqrt{5}}{3}$

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

  3. $\dfrac{3}{\sqrt{5}}$

  4. $\dfrac{2}{3}$

Show answer
Correct Option: A
Explanation:
Given conic is $4{ x }^{ 2 }+9{ y }^{ 2 }=144$
$\Rightarrow \dfrac { { x }^{ 2 } }{ 36 } +\dfrac { { y }^{ 2 } }{ 16 } =1$ .... $(i)$ which is an equation of ellipse
Eccentricity of an ellipse $\dfrac { { x }^{ 2 } }{ a^2 } +\dfrac { { y }^{ 2 } }{ b^2 } =1$ is $e=\sqrt { 1-\dfrac { b^{ 2 } }{ { a }^{ 2 } }  } $
From $(i)$,
$a^{2}=36$ and $b^{2}=16$
So, eccentricity of given conic is $e=\sqrt { 1-\dfrac { 16 }{ 36 }  }= \sqrt { \dfrac { 20 }{ 36 }  } =\dfrac { \sqrt { 5 }  }{ 3 } $

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

Show answer
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. $\dfrac15$

Show answer
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. $4 x - 9 = 0$

Show answer
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. $\sqrt{2}$

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

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

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

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

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