Tag: standard equation of ellipse

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Questions Related to standard equation of ellipse

Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

The equation $\dfrac {x^{2}}{2 - \lambda} + \dfrac {y^{2}}{\lambda - 5} - 1 = 0$ represents an ellipse, if

  1. $\lambda < 5$
  2. $\lambda < 2$
  3. $2 < \lambda < 5$
  4. $\lambda < 2$ or $\lambda < 5$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
General equation of ellipse is $\dfrac {x^2}{a^2}+\dfrac {y^2}{b^2}=1$

So both denominator should be positive as they are squares
 
In the given equation

$\dfrac {x^2}{2-\lambda} +\dfrac {y^2}{\lambda -5}-1=0$

So,
 
$2-\lambda > 0, +(\lambda -5) >o$

$\Rightarrow \ \lambda < 2, \lambda > 5$

$\Rightarrow \ 2 < \lambda < 5$
Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

An ellipse has its centre at $(1, -1)$ and semi-major axis $= 8$ and it passes through the point $(1, 3)$. The equation of the ellipse is

  1. $\dfrac {(x + 1)^{2}}{64} + \dfrac {(y + 1)^{2}}{16} = 1$
  2. $\dfrac {(x - 1)^{2}}{64} + \dfrac {(y + 1)^{2}}{16} = 1$
  3. $\dfrac {(x - 1)^{2}}{16} + \dfrac {(y + 1)^{2}}{64} = 1$
  4. $\dfrac {(x + 1)^{2}}{64} + \dfrac {(y - 1)^{2}}{16} = 1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given that

centre is at $(1, -1)$

semi major axis $(a)=8$

so, equation of ellipse can be written as

$\dfrac {(x-1)^2}{a^2} +\dfrac {(y+1)^2}{b^2} =1.....(1)$

It passes through point $(1,3)$

i.e, $x=1, y=3$

Putting these value in equation $(1)$ we get

$\dfrac {(1-1)^2}{a^2} +\dfrac {(3+1)^2}{b^2}=1$

$\dfrac {16}{b^2}=1$

$b^2=16\ \Rightarrow b=4$

Substituting the values of $a$ and $b$ in equation $(1)$ we get

$\dfrac {(x-1)^2}{64}+\dfrac {(y+1)^2}{16}=1$

This is the required equation of ellipse
Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

If $F _{1}=\left ( 3, 0 \right )$, $F _{2}=\left ( -3, 0 \right )$ and $P$ is any point on the curve $16x^{2}+25y^{2}=400$, then $PF _{1}+PF _{2}$ equals to:

  1. $8$
  2. $6$
  3. $10$
  4. $12$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The equation of the ellipse can be written as $\displaystyle \frac{x^{2}}{25}+\frac{y^{2}}{16}=1$

Here $a^{2}=25$, $b^{2}=16$

But $b^{2}=a^{2}\left ( 1-e^{2} \right )$

$\Rightarrow $   $16=25\left ( 1-e^{2} \right )$   $\Rightarrow $   $e=\dfrac35$

So that foci of the ellipse are $\left ( \pm ae, 0 \right )$ i.e. $\left ( \pm 3, 0 \right )$ or $F _{1}$ and $F _{2}.$

By definition of the ellipse, since $P$ is any point on the ellipse

$PF _{1}+PF _{2}=2a=2\times 5=10$

Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

The equation $\displaystyle \frac {x^2}{8-t}\, +\, \displaystyle \frac {y^2}{t-4}\, =\, 1$ will represent an ellipse if

  1. $t\, \in\, (1,\, 5)$
  2. $t\, \in\, (2,\, 8)$
  3. $t\, \in\, (4,\, 8)\, -\, \{6\}$
  4. $t\, \in\, (4,\, 10)\, -\, \{6\}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Consider Equation, $\displaystyle\frac{x^2}{a^2}+\displaystyle\frac{y^2}{b^2}=1$ to represent an ellipse equation.
$a>0,b>0,a\neq b$
Given,equation $\displaystyle\frac{x^2}{(8-t)}+\displaystyle\frac{y^2}{(t-4)}=1$
$\Rightarrow (8-t)>0\;$ and $\;(t-4)>0,(8-t)\neq(t-4)$
$\Rightarrow t\in(-\infty,8) \cap (4,\infty) \cap$ {$t\neq6$}
$\Rightarrow t\in(4,8)-${$6$}

Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

The total number of real tangents that can be drawn to the ellipse $3x^{2}+5y^{2}=32$ and $25x^{2}+9y^{2}=450$ passing through $(3,5)$ is

  1. $0$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$(3,5)$ lies on $25x^2+9y^2=450$

Therefore, one tangent can be drawn

and $(3,5)$ lies outside $3x^2+5y^2=32$ because $S _1>0$

Therefore, two tangents can be drawn.
So total 3 tangents

Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

$\mathrm{S}$ and $\mathrm{S}^{'}$ are the foci of the ellipse $25x^{2}+16y^{2}=1600$, then the sum of the distances from $\mathrm{S}$ and $\mathrm{S}'$ to the point $(4\sqrt{3},5)$ is:

  1. $20$
  2. $15$
  3. $40$
  4. $30$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given ellipse is $25{ x }^{ 2 }+16{ y }^{ 2 }=1600$

This can be written as $\displaystyle \frac { { x }^{ 2 } }{ 64 } +\frac { { y }^{ 2 } }{ 100 } =1$
Compare it to standard form of ellipse to get $a=8,b=10$
here $a<b$
So eccentricity e$=\sqrt { 1-\displaystyle \frac { { a }^{ 2 } }{ { b }^{ 2 } }  } =\sqrt { 1-\displaystyle\frac { 64 }{ 100 }  } =\displaystyle \frac { 3 }{ 5 } $
Now the foci are given by $\left( 0,\pm \sqrt { { b }^{ 2 }-{ a }^{ 2 } }  \right) $
So $foci:\left( 0,\pm \sqrt { 36 }  \right) $
$S:\left( 0,6 \right) ,{ S } _{ 1 }:\left( 0,-6 \right) $
Now find distance of $S$ and ${ S } _{ 1 }$ from the given point $\left( 4\sqrt { 3 } ,5 \right) $
So sum$=\sqrt { { \left( 4\sqrt { 3 }  \right)  }^{ 2 }+1 } +\sqrt { { \left( 4\sqrt { 3 }  \right)  }^{ 2 }+{ \left( 11 \right)  }^{ 2 } } =7+13=20$

Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

The length of the latusrectum of the parabola $169\left{ { \left( x-1 \right)  }^{ 2 }+{ \left( y-3 \right)  }^{ 2 } \right} ={ \left( 5x-12y+17 \right)  }^{ 2 }$

  1. $\cfrac { 14 }{ 13 } $
  2. $\cfrac { 28 }{ 13 } $
  3. $\cfrac { 12 }{ 13 } $

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

Here ${ \left( x-1 \right)  }^{ 2 }+{ \left( y-3 \right)  }^{ 2 }={ \left{ \cfrac { 5x-12y+17 }{ \sqrt { { 5 }^{ 2 }+{ \left( -12 \right)  }^{ 2 } }  }  \right}  }^{ 2 }$


$\therefore$ The focus is $(1,3)$ and the directrix is $5x-12y+17=0$

The distance of the focus from the directrix

$=\left| \cfrac { 5\times 1-12\times 3+17 }{ \sqrt { { 5 }^{ 2 }+{ \left( -12 \right)  }^{ 2 } }  }  \right| =\cfrac { 14 }{ 13 } $

$\therefore$ Length of latusrectum $=2\times \cfrac { 14 }{ 13 } =\cfrac { 28 }{ 13 } $

Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

The equation of the ellipse having vertices at $\displaystyle \left( \pm 5,0 \right) $ and foci $\displaystyle \left( \pm 4,0 \right) $ is

  1. $\displaystyle \frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 16 } =1$
  2. $\displaystyle 9{ x }^{ 2 }+25{ y }^{ 2 }=225$
  3. $\displaystyle \frac { { x }^{ 2 } }{ 9 } +\frac { { y }^{ 2 } }{ 25 } =1$
  4. $\displaystyle 4{ x }^{ 2 }+5{ y }^{ 2 }=20$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The vertices and foci of an ellipse are $\displaystyle \left( \pm 5,0 \right) $ and $\displaystyle \left( \pm 4,0 \right) $ respectively.
$\displaystyle \therefore \quad a=5$ and $\displaystyle ae=4$
$\displaystyle \Rightarrow \quad e=\frac { 4 }{ 5 } $
We know that,
$\displaystyle e=\sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } }  } $
$\displaystyle \Rightarrow \quad \frac { 16 }{ 25 } =1-\frac { { b }^{ 2 } }{ 25 } \Rightarrow { b }^{ 2 }=9$
Hence, equation of an ellipse is
$\displaystyle \frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 9 } =1\Rightarrow 9{ x }^{ 2 }+25{ y }^{ 2 }=225$