Tag: special cases of an ellipse

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Questions Related to special cases of an ellipse

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

For all admissible values of the parameter $a$ the straight line $2ax+y\sqrt{1-a^2}=1$ will touch an ellipse whose eccentricity is equal to

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

The line 2ax + y*sqrt(1-a^2) = 1 is of the form x*cos(theta) + y*sin(theta) = p where cos(theta) = 2a and sin(theta) = sqrt(1-a^2). Squaring and adding gives cos^2(theta) + sin^2(theta) = 4a^2 + 1 - a^2 = 3a^2 + 1 = 1, which is not quite right; however, the condition for a line to touch an ellipse x^2/A^2 + y^2/B^2 = 1 is p^2 = A^2*cos^2(theta) + B^2*sin^2(theta). Comparing coefficients leads to A^2=1/4 and B^2=1, so e^2 = 1 - (1/4)/1 = 3/4, e = sqrt(3)/2.

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

If the focal chord of the ellipse  $\dfrac { x ^ { 2 } } { a ^ { 2 } } + \dfrac { y ^ { 2 } } { b ^ { 2 } } = 1 , ( a > b )$  is normal at  $( a \cos \theta , b \sin \theta )$  then eccentricity of the ellipse is (it is given that  $sin\theta \neq0)$

  1. $| \sec \theta |$
  2. $| \cos \theta |$
  3. $| \sin \theta |$

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

The normal at (a*cos(theta), b*sin(theta)) to the ellipse x^2/a^2 + y^2/b^2 = 1 is ax*sec(theta) - by*csc(theta) = a^2 - b^2. If this is a focal chord, it must pass through a focus (ae, 0). Substituting gives a*ae*sec(theta) = a^2 - b^2. Thus a^2*e*sec(theta) = a^2(1 - b^2/a^2) = a^2*e^2. Simplifying gives e = |sec(theta)|.

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

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

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

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

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

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

$e=\dfrac {\sqrt 3}{2}$
Multiple choice maths ellipse special cases of an ellipse eccentricity equation of ellipse

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

  1. $\dfrac {\sqrt {5} - 1}{2}$
  2. $\dfrac {\sqrt {5} + 1}{2}$
  3. $\dfrac {\sqrt {5} - 1}{4}$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
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)

$e=\dfrac {\sqrt 5 -1}{2}$
Multiple choice maths ellipse special cases of an ellipse eccentricity equation of ellipse

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

  1. $\sqrt2$
  2. $\sqrt {3}$
  3. $- \sqrt {2}$
  4. $- \sqrt {3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $\sqrt2$

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

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

  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. $\displaystyle \frac{x^{2}}{25}-\frac{y^{2}}{9}=1$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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

  1. $\dfrac{\sqrt{5}}{3}$
  2. $\dfrac{\sqrt{5}}{6}$
  3. $\dfrac{3}{\sqrt{5}}$
  4. $\dfrac{2}{3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 } $