Tag: maths

Questions Related to maths

$(3,2)$ 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.

Let $5x^2+7y^2=140$, then $(3,-4)$ is:

position of point wrt ellipse ellipse maths

  1. Outside the ellipse

  2. Inside the ellipse

  3. On the ellipse

  4. Data insufficient

Show answer
Correct Option: A
Explanation:

Equation of ellipse :   ${ 5x }^{ 2 }+{ 7y }^{ 2 }=140$

putting point $(3,-4)$ in equation of ellipse
     $=5\times { 3 }^{ 2 }+7{ \left( -4 \right)  }^{ 2 }-140$
     $=45+112-140$
     $=17(>0)$
Hence, $(3,-4)$ is outside the ellipse.

Let $5x^2+7y^2=140$, then Position of $(4,-3)$ relative to the ellipse is

position of point wrt ellipse ellipse maths

  1. Inside the ellipse.

  2. Outside the ellipse.

  3. On the ellipse.

  4. None of the above.

Show answer
Correct Option: B
Explanation:

Equation of ellipse : ${ 5x }^{ 2 }+{ 7y }^{ 2 }=140$

putting point (4,3) in equation
     $=5\times { \left( 4 \right)  }^{ 2 }+7{ \left( -3 \right)  }^{ 2 }-140$
     $=80+63-140$
     $=3\left( >0 \right) $
Hence, point is outside the ellipse.

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

position of point wrt ellipse ellipse maths

  1. Inside the ellipse

  2. Outside the ellipse

  3. On the ellipse

  4. Centre of the ellipse

Show answer
Correct Option: A,D
Explanation:

If equation of ellipse is $\dfrac { { \left( x-h \right)  }^{ 2 } }{ { a }^{ 2 } } +\dfrac { { \left( y-k \right)  }^{ 2 } }{ { b }^{ 2 } } =1$,  centre $\left( h,k \right) $

for ellipse   $\dfrac { { \left( x-3 \right)  }^{ 2 } }{ 9 } +\dfrac { { \left( y-4 \right)  }^{ 2 } }{ 16 } =1$
            Centre of ellipse (3,4)
and centre is inside ellipse only.

Let $5x^2+7y^2=140$, then $(0,0)$ is: 

position of point wrt ellipse ellipse maths

  1. Inside the ellipse

  2. On the ellipse

  3. Outside the ellipse

  4. Centre of the ellipse

Show answer
Correct Option: A,D
Explanation:

Equation of ellipse    ${ 5x }^{ 2 }+{ 7y }^{ 2 }=140$

   Dividing both sides by $140$
           $\dfrac { { x }^{ 2 } }{ 28 } +\dfrac { { y }^{ 2 } }{ 20 } =1$
Comparing with $\dfrac { { \left( x-h \right)  }^{ 2 } }{ { a }^{ 2 } } +\dfrac { { \left( y-k \right)  }^{ 2 } }{ { b }^{ 2 } } =1$
          $(h,k)\ =\ (0,0)$
Hence (0,0) is centre of given ellipse and it inside the ellipse.

Let $5x^2+7y^2=140$, then $(\sqrt {14},\sqrt {10})$ is:

position of point wrt ellipse ellipse maths

  1. Outside the ellipse

  2. Inside the ellipse

  3. On the ellipse

  4. Centre of the ellipse

Show answer
Correct Option: C
Explanation:

          ${ 5x }^{ 2 }+{ 7y }^{ 2 }=140$          (equation of ellipse)

putting $\left( \sqrt { 14 } ,\sqrt { 10 }  \right) $ in equation
            $=5\times { \left( \sqrt { 14 }  \right)  }^{ 2 }+7\times { \left( \sqrt { 10 }  \right)  }^{ 2 }-140$
            $=70+70-140$
            $=0$
Hence, point lie on the ellipse.