Tag: position of point wrt ellipse

Questions Related to position of point wrt 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

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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.

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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

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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.

If $P=(x, y), F _1=(3, 0), F _2=(-3, 0)$ and $  16x^2+25y^2=400$, then $PF _1+PF _2$ equals

position of point wrt ellipse ellipse maths

  1. $8$

  2. $6$

  3. $10$

  4. $12$

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Correct Option: C
Explanation:

Given equation is $16x^2+25y^2=400$
The ellipse can be written as, $\displaystyle \frac{x^2}{25}+\frac{y^2}{16}=1$
Here $a^2=25, b^2=16$,  but  $b^2=a^2(1-e^2)$
$\Rightarrow \dfrac {16}{25}=1-e^2$
$\Rightarrow e^2=1-\dfrac {16}{25}=\dfrac {9}{25} $
$ \Rightarrow e=\pm \dfrac {3}{5}$
Foci of the ellipse are $(+ae, 0)$ $=$ $ (\pm 3, 0)$, i.e., $F _1$ and $F _2$.
We have $PF _1+PF _2=2a=10$ for every point $P$ on the ellipse.

The position of the point $(1, 2)$ relative to the ellipse $2x^{2} + 7y^{2} = 20$ is

position of point wrt ellipse ellipse maths

  1. outside the ellipse

  2. inside the ellipse but not at the focus

  3. on the ellipse

  4. at the focus

Show answer
Correct Option: A
Explanation:

$f\left( x,y \right) ={ 2x }^{ 2 }+7{ y }^{ 2 }-20$

Point will be outside of ellipse if $f\left( 1,2 \right) >0$

on ellipse if $f\left( 1,2 \right) =0$

will be inside if $f\left( 1,2 \right) <0$

$ f\left( 1,2 \right) ={ 2\times 1 }^{ 2 }+7{ \times 2 }^{ 2 }-20=10$

Here we see that $f\left( 1,2 \right) >0$ so point will be outside of ellipse

So correct answer will be option A

The minimum distance of origin from the curve $\frac{a^2}{x^2}+\frac{b^2}{y^2}=1$ is $(a>0,b>0)$

position of point wrt ellipse ellipse maths

  1. a-b

  2. a+b

  3. 2a+2b

  4. 2(a-b)

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Correct Option: B
Explanation:
Let $(a sec\theta,b cosec \theta)$ be a point on the curve, then its diastance  from the origin
$=\sqrt{a^2 sec^2 \theta + b^2 cosec ^2 \theta}$
$\therefore f(\theta)=a^2 sec^2 \theta +b^2 cosec ^2 \theta$
$=a^2+a^2 tan ^2 \theta + b^2 + b^2 cot ^ 2 \theta$
$=a^2+b^2 + a^2 tan ^2 \theta + b^2 cot ^2 \theta$
$\geq a^2+b^2+2 \sqrt{a^2 b^2}=(a+b)^2$
$\therefore$ minimum value of $f(\theta)=(a+b)^2$
$\therefore$ minimum value of $\sqrt{a^2 sec^2 \theta + b^2 cosec^2 \theta }$ is $a+b$

If $a$ and $c$ positive real number and the ellipse $\dfrac { { x }^{ 2 } }{ { 4c }^{ 2 } } +\dfrac { { y }^{ 2 } }{ { c }^{ 2 } } =1$ has four distinet points in common with the circle ${ x }^{ 2 }+{ y }^{ 2 }=9{ a }^{ 2 }$, then

position of point wrt ellipse ellipse maths

  1. $6ac+9{ a }^{ 2 }-2{ c }^{ 2 }>0$

  2. $6ac+9{ a }^{ 2 }-2{ c }^{ 2 }<0$

  3. $9ac-9{ a }^{ 2 }-2{ c }^{ 2 }<0$

  4. $9ac-9{ a }^{ 2 }-2{ c }^{ 2 }>0$

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Correct Option: C

An ellipse is inscribed in a circle and a point within the circle is chosen at random. If the probability that this point lies outside the ellipse is $2/3$ then the eccentricity of the ellipse is:

position of point wrt ellipse ellipse maths

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

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

  3. $\dfrac{8}{9}$

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

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Correct Option: A