Tag: position of point wrt ellipse

Questions Related to position of point wrt ellipse

$\mathrm{A}$ssertion ($\mathrm{A}$): The point $(5,-2)$ lies outside the ellipse $24x^{2}+7y^{2}=12$.
Reason (R): lf the point $(x _{1},y _{1})$ lie outside the ellipse $\mathrm{S}=0$ then $S _{11}>0$ 

position of point wrt ellipse ellipse maths

  1. Both A and R are true and R is the correct explanation of A

  2. Both A and R are true but R is not coorect explanation of A

  3. A is true but R is false

  4. A is false but R is true

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Correct Option: A
Explanation:

$24x^{2}+7y^{2}-12=5$

$S(5,-2)=24\times 25+7\times 4-12$

$S(5,-2)>0$

$\therefore $ lies outside the ellipse.
$S=0$
$S(x _{1}y _{1})>0$ the point is outside the ellipse.

The point $(2\cos \theta , 3\sin \theta)$ lies ____________ the ellipse $\dfrac{x^2}{4}+\dfrac{y^2}{9}=1$.

position of point wrt ellipse ellipse maths

  1. outside

  2. inside

  3. on the periphery of

  4. on the auxillary of

Show answer
Correct Option: C
Explanation:

Given point is $\left ( 2\cos\theta, 3\sin\theta  \right )$


Substituting given point in the ellipse equation:
$\dfrac{x^{2}}{4}+\dfrac{y^{2}}{9}-1$
$= \dfrac{4\cos^{2}\theta}{4}+\dfrac{9\sin^{2}\theta }{9}-1$
$= \cos^{2}\theta + \sin^{2}\theta -1$
$= 1-1=0$

$\therefore$ Given point satisfies the ellipse 
$\Rightarrow$ point lie on periphery of ellipse.

The distance of point '$\theta$' on the ellipse $\dfrac {x^2}{a^2} + \dfrac {y^2}{b^2}=1$ from a focus is:

position of point wrt ellipse ellipse maths

  1. $a(e + \cos \theta)$

  2. $a(e - \cos \theta)$

  3. $a(1 + e \cos \theta)$

  4. $a(1 + 2e \cos \theta)$

Show answer
Correct Option: C
Explanation:
Given equation of ellipse is
$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$
Any point on the ellipse will be $(a cos\theta, b sin \theta)$
$P=(acos\theta, b sin \theta)$
Centre of the ellipse is $(0,0)$
Ellipse is parallel to horizontal axis
Foci of the ellipse is
$F=(h-ae,k)$ if $(h,k) $ is the centre
$F=(0-ae,0)=(-ae,o)$
Distance $FP=\sqrt{(-ae-cos\theta)^2+(bsin\theta-0)^2}$
$=a \sqrt{(e^2+cos^2 \theta+2ecos\theta+(1-e^2)sin \theta-0)^2}$
$a\sqrt{(e^2+cos^2 \theta+2ecos\theta+sin^2 \theta-e^2sin^2\theta)}$
$a\sqrt{(1+2ecos\theta+e^2(1-sin^2\theta)}$
$a\sqrt{(1+2ecos \theta+e^2cos^2\theta)}$
$a\sqrt{(1+ecos\theta)^2}$
$FP=a(1+ecos\theta)$



$(2,3)$ lies _______  the ellipse $16 x^{2} + 9y^{2} - 16x - 32 = 0$

position of point wrt ellipse ellipse maths

  1. Inside

  2. Outside

  3. On

  4. None of the above

Show answer
Correct Option: B
Explanation:
Given ellipse $16x^2+9y^2-16x-32=0$
Let $S=16x^2+9y^2-16x-32=0$
Put $(2,3) $ in $S$ we get
$S(2,3)=16(2)^2+9(3)^2-16(2)-32=81 $
$S(2,3)>0$
So it $(2,3) $ lies outside the ellipse.

The point $(4\cos \theta , 4\sin \theta)$ lies ____________ the ellipse $\dfrac{x^2}{16}+\dfrac{y^2}{9}=1$

position of point wrt ellipse ellipse maths

  1. outside

  2. inside

  3. on the periphery

  4. on the auxiliary circle

Show answer
Correct Option: D
Explanation:

Equation of ellipse is $\dfrac { { x }^{ 2 } }{ 16 } +\dfrac { { y }^{ 2 } }{ 9 } =1$

Equation of auxiliary circle of ellipse is
${ x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }\ { x }^{ 2 }+{ y }^{ 2 }=16$
Substiuting $(4\cos\theta,4\sin\theta$) in the equation, we get
${ (4\cos { \theta  } ) }^{ 2 }+{ (4\sin { \theta  } ) }^{ 2 }=16\ 16(\cos ^{ 2 }{ \theta  } +\sin ^{ 2 }{ \theta  } )=16\ 16=16$
The point satisfies the equation of auxiliary circle.
So, option D is the correct.

$(3,2)$ lies _______  the ellipse $16 x^{2} + 9y^{2} - 16x - 32 = 0$

position of point wrt ellipse ellipse maths

  1. Inside

  2. Outside

  3. On

  4. None of the above

Show answer
Correct Option: B
Explanation:

Ellipse : $16{ x }^{ 2 }+9{ y }^{ 2 }-16x-32=0$

Let $S=16{ x }^{ 2 }+9{ y }^{ 2 }-16x-32$
Putting point (3,2) in S we get
         S(3,2) $=16\times { (3) }^{ 2 }+9\times { (2) }^{ 2 }-16\times 3-32$
                   $=144+36-48-32$
                   $=180-80$
                   $=100$
          $S(3,2) > 0$
Hence, point lies outside ellipse.

The point $(1,1)$ lies ____________ the ellipse $\dfrac{x^2}{4}+\dfrac{y^2}{9}=1$

position of point wrt ellipse ellipse maths

  1. outside

  2. inside

  3. on the periphery

  4. on the auxillary circle

Show answer
Correct Option: B
Explanation:

We will substitute the given point in ellipse equation

$\Rightarrow$ $\dfrac{x^{2}}{4}+\dfrac{y^{2}}{9}-1$
$=\dfrac{1}{4}+\dfrac{1}{9}-1$
$\Rightarrow  \dfrac{13}{36}-1$
$\Rightarrow \dfrac{-23}{36}<0$

By Substituting the point, we are getting less than $0$ 

$\therefore $ point lies inside the ellipse .

The distance of a point $(\sqrt 6 \cos \theta, \sqrt 2 \sin \theta)$ on the ellipse $\dfrac {x^2}{6} + \dfrac {y^2}{2}=1$ from the centre is $2$, if:

position of point wrt ellipse ellipse maths

  1. $\theta =\dfrac {\pi}{2}$

  2. $\theta =\dfrac {3\pi}{2}$

  3. $\theta =\dfrac {5\pi}{2}$

  4. $\theta =  \dfrac{\pi}{4}$

Show answer
Correct Option: D
Explanation:

$P(\sqrt { 6 } \cos { \theta  } ,\sqrt { 2 } \sin { \theta  } )$

$Centre(0,0)$
$PC=\sqrt { 6\cos ^{ 2 }{ \theta  } +2\sin ^{ 2 }{ \theta  }  } $
$2=\sqrt { 4\cos ^{ 2 }{ \theta  } +2 } $
$2=4\cos ^{ 2 }{ \theta  } $
$\cos { \theta  } =\pm \dfrac { 1 }{ \sqrt { 2 }  } $
$\theta =\dfrac { \pi  }{ 4 } $

State the following statement is True or False
Let $\dfrac {x^2}{9}+\dfrac{y^2}{16}=1$ then
$(\sqrt 2,\sqrt {24})$ is inside the given ellipse.

position of point wrt ellipse ellipse maths

  1. True

  2. False

Show answer
Correct Option: B
Explanation:

Equation of ellipse,  $\dfrac { { x }^{ 2 } }{ 9 } +\dfrac { { y }^{ 2 } }{ 16 } =1$

Putting point $\left( \sqrt { 2 } ,\sqrt { 24 }  \right) $ in equation of ellipse
     $=\dfrac { 2 }{ 9 } +\dfrac { 24 }{ 16 } -1$
     $=\left( \dfrac { 31 }{ 18 } -1 \right) >0$
Hence point lies outside ellipse.
False

Let $\dfrac {(x-3) ^2}9+\dfrac {(y-4) ^2}{16}=1$ then   $(0,0)$ is

position of point wrt ellipse ellipse maths

  1. On the ellipse.

  2. Outside the ellipse.

  3. Inside the ellipse.

  4. None of the above.

Show answer
Correct Option: B
Explanation:

Equation of ellipse : $\dfrac { { \left( x-3 \right)  }^{ 2 } }{ 9 } +\dfrac { { \left( y-4 \right)  }^{ 2 } }{ 16 } -1=0$

putting point (0,0) in above ellipse,
          $=\dfrac { { \left( -3 \right)  }^{ 2 } }{ 9 } +\dfrac { { \left( -4 \right)  }^{ 2 } }{ 16 } -1$
          $=1+1-1$
          $=1>0$
Hence, point lies outside ellipse.