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Special cases of an ellipse - class-XII

Description: special cases of an ellipse
Number of Questions: 35
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Tags: two dimensional analytical geometry-ii circles and conics section maths conic sections conic section ellipse
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Eccentricity of the conic $3x^{2}+2xy-3y^{2}+x+y-2=0$

  1. 2

  2. $\sqrt{2}$

  3. 3

  4. none


Correct Option: A

If circle whose diameter is major axis of ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ meets minor axis at point P and orthocentre of $\Delta PF _{1}F _{2}$ lies on ellipse where $F _{1}$  and $F _{2}$ are foci of ellipse, then square of eccentricity of ellipse, is 

  1. $2 sin\frac{\pi }{10}$

  2. $2 sin\frac{\pi }{12}$

  3. $2 sin\frac{\pi }{4}$

  4. $2 sin\frac{\pi }{2}$


Correct Option: A

If (3,4), (5,12) are two foci of the ellipse passing thrpsough the origin. Then the eccentricity of the ellipse is 

  1. $\frac{\sqrt{17}}{3}$

  2. $\frac{\sqrt[2]{17}}{3}$

  3. $\frac{\sqrt{34}}{3}$

  4. $\frac{\sqrt{34}}{9}$


Correct Option: B

An ellipse has foci (3, 1), (1, 1) and it passes through point (1, 3). Its eccentricity is equal to 

  1. $\sqrt { 2 } -1$

  2. $\sqrt { 3 } -1$

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

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


Correct Option: A

The ellipse $E _1:\dfrac{x^2}{9}+\dfrac{y^2}{4}=1$ is inscribed in a rectangle R whose sides are parallel to the coordinates axis. Another ellipse $E _2$ passing through the point $(0, 4)$ circumscribes the rectangle R. The eccentricity of the ellipse $E _2$ is?

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

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

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

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


Correct Option: A

The accentricity of the ellipse $4x^{2}+9y^{2}+8x+36y+4=0$ is

  1. $\dfrac{5}{6 }$

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

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

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


Correct Option: A

Eccentricity of the ellipse $5x^{2}+6xy+5y^{2}=8$ is

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

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

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

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


Correct Option: A

An ellipse has $OB$ as its semi-minor axis. $F _{1}$ and $F _{2}$ are its foci and angle $F _{1}BF _{2}$ is a right angle. The eccentricity of the ellipse is 

  1. $1/\sqrt{2}$

  2. $1/2$

  3. $1/\sqrt{3}$

  4. $2/\sqrt{3}$


Correct Option: A

An ellipse having foci $(3,1)$ and $(1,1)$ passes through the point $(1,3)$ ha the eccentricity 

  1. $\sqrt {2}-1$

  2. $\sqrt {3}-1$

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

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


Correct Option: A

The tangent at any point $P\left(a\cos\theta,b\sin\theta\right)$ on the ellipse $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$ meets the auxiliary circle at two points which subtend a right angle at the center ,then eccentricity is 

  1. $\dfrac{1}{\sqrt{1+\sin^{2}\theta}}$

  2. $\dfrac{1}{\sqrt{2-\cos^{2}\theta}}$

  3. $\dfrac{1}{\sqrt{1+\tan^{2}\theta}}$

  4. $none\ of\ these$


Correct Option: A

If S and S' are the foci of an ellipse of major axis of length 10 units and P is any point on the ellipse such that the perimeter of triangle PSS' is 15 units, then the eccentricity of the ellipse is 

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

  2. $\dfrac{1}{4}$

  3. $\dfrac{7}{25}$

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


Correct Option: A

If normal at any point P on the ellipse $\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1(a>b>0)$ meet the major and minor axes at Q and R respectively so that 3PQ = @PR, then the eccentricity of ellipse is equal to

  1. $\frac { 1 }{ \sqrt { 3 } } $

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

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

  4. $\frac { 1 }{ \sqrt { 2 } } $


Correct Option: A

Find the length of the semi-axes, coordinates of foci, length of latus rectum, eccentricity and equation direction for the ellipse given by the equations :-  (i) $25{ x }^{ 2 }-150x+16{ y }^{ 2 }=175$ (ii) The eccentricity of the ellipse $9{ x }^{ 2 }+4{ y }^{ 2 }30y=0$ is 

  1. 1/2

  2. 2/3

  3. 3/4

  4. None of these


Correct Option: A

If normal to the ellipse $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$ at $\left(ae,\dfrac{b^{2}}{a}\right)$ is passing throught $\left(0,-2b\right)$, then $c=$

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

  2. $2\left(\sqrt{2}-1\right)$

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

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


Correct Option: A

An ellipse having foci $\left(3,1\right)$ and $\left(1,1\right)$ passes through the point $\left(1,3\right)$ has the eccentricity

  1. $\sqrt{2}-1$

  2. $\sqrt{3}-1$

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

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


Correct Option: A

The eccentricity of ellipse whose line joining foci substends an angle of ${90} _{o}$ at an xtremity of minor axis is 

  1. $\dfrac{1}{\sqrt{6}}$

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

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

  4. $\sqrt{6}$


Correct Option: A

If the roots of the equation $x^2 - 4x + 1 = 0$ are the lengths of the semi-major axis and semi-minor axis of an ellipse, then the eccentricity of the ellipse lies between

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

  2. $\dfrac{1}{4}$ and $\dfrac{1}{3}$

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

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


Correct Option: A

If $\alpha,\beta$ are the eccentric of the extremities of a focal chord of an ellipse, then eccentricity of the ellipse is

  1. $\dfrac{sin\alpha+sin\beta}{sin(\alpha+\beta)}$

  2. $\dfrac{cos\alpha+cos\beta}{cos(\alpha+\beta)}$

  3. $\dfrac{(\alpha+\beta)}{sin\alpha+sin\beta}$

  4. none of these


Correct Option: A

(-4,1) and (6,1) are the vertices of an ellipse. If one of the foci of the ellipse. If one of the foci of the ellipse lies on x -2y = 2 then its eccentricity is

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

  2. $\dfrac{4}{5}$

  3. $\dfrac{2}{5}$

  4. $\dfrac{1}{5}$


Correct Option: A

An ellipse whose foci and $(2,4)$ and $(14,9)$ touches the x-axis then the eccentricity of the ellipse is $\dfrac{P}{\sqrt{q}}$ (when p,q an comprise) then the units place of $p+4q$ is 

  1. $1$

  2. $5$

  3. $9$

  4. $2$


Correct Option: A

The eccentricity of the ellipse $4x^{2}+16y^{2}=576$ is

  1. $\dfrac{\sqrt{7}}{2}$

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

  3. $\dfrac{7}{12}$

  4. $\dfrac{\sqrt{7}}{4}$


Correct Option: A

Eccentricity of the ellipse $5{ x }^{ 2 }+6xy+5{ y }^{ 2 }=8$ is.

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

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

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

  4. $\frac { 1 }{ \sqrt { 3 } } $


Correct Option: A

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


Correct Option: A

If  $( 5,12 )$  and  $( 24,7 )$  are the focii of a conic passing through the origin, then the eccentricity of conic is -

  1. $\sqrt { 386 } / 12$

  2. $\sqrt { 386 } / 13$

  3. $\sqrt { 386 } / 25$

  4. $\sqrt { 386 } / 38$


Correct Option: A

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


Correct Option: A

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


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.

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

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

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

  4. None of these


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 $x^2\, -\, y^2\,- \, 4x\, +\, 4y\, +\, 16\, =\, 0$

  1. $\sqrt2$

  2. $\sqrt {3}$

  3. $- \sqrt {2}$

  4. $- \sqrt {3}$


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

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


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 $

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

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

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

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


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

  1. $\sqrt{2}$

  2. 2

  3. 4

  4. 8


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?

  1. $\dfrac13$

  2. $\dfrac12$

  3. $\dfrac14$

  4. $\dfrac15$


Correct Option: B

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. $4 x - 9 = 0$


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

  1. $0$

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

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

  4. $\sqrt{2}$


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:

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

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

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

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


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

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

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

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

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

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