Tag: introduction to ellipse

Questions Related to introduction to ellipse

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

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

  1. $\lambda < 5$

  2. $\lambda < 2$

  3. $2 < \lambda < 5$

  4. $\lambda < 2$ or $\lambda < 5$

Show answer
Correct Option: C
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,
 
$2-\lambda > 0, +(\lambda -5) >o$

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

$\Rightarrow \ 2 < \lambda < 5$

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

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

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

Show answer
Correct Option: B
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

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:

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

  1. $8$

  2. $6$

  3. $10$

  4. $12$

Show answer
Correct Option: C
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

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

For a parabola whose focus is $(1, 1)$ and whose vertex is $(2, 1)$, the latus rectum is

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

  1. $ \sqrt{5}$

  2. $2 \sqrt{5}$

  3. $4$

  4. $4 \sqrt{5}$

Show answer
Correct Option: C
Explanation:

The length of latus rectum$(l)=4 \times SC$
where S is the focus and C is the vertex
Therefore, $l=4$

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

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

  1. $t\, \in\, (1,\, 5)$

  2. $t\, \in\, (2,\, 8)$

  3. $t\, \in\, (4,\, 8)\, -\, {6}$

  4. $t\, \in\, (4,\, 10)\, -\, {6}$

Show answer
Correct Option: C
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 $\;(t-4)>0,(8-t)\neq(t-4)$
$\Rightarrow t\in(-\infty,8) \cap (4,\infty) \cap$ {$t\neq6$}
$\Rightarrow t\in(4,8)-${$6$}

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

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

  1. $0$

  2. $2$

  3. $3$

  4. $4$

Show answer
Correct Option: C
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

$\mathrm{S}$ 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:

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

  1. $20$

  2. $15$

  3. $40$

  4. $30$

Show answer
Correct Option: A
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$=\sqrt { { \left( 4\sqrt { 3 }  \right)  }^{ 2 }+1 } +\sqrt { { \left( 4\sqrt { 3 }  \right)  }^{ 2 }+{ \left( 11 \right)  }^{ 2 } } =7+13=20$

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

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

  1. $\cfrac { 14 }{ 13 } $

  2. $\cfrac { 28 }{ 13 } $

  3. $\cfrac { 12 }{ 13 } $

  4. None of these

Show answer
Correct Option: B
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 } $

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

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

  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. $\displaystyle 4{ x }^{ 2 }+5{ y }^{ 2 }=20$

Show answer
Correct Option: B
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
$\displaystyle \frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 9 } =1\Rightarrow 9{ x }^{ 2 }+25{ y }^{ 2 }=225$

The sum of the focal distances of any point on the conic $\dfrac {x^{2}}{25} + \dfrac {y^{2}}{16} = 1$ is

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

  1. $10$

  2. $9$

  3. $41$

  4. $18$

Show answer
Correct Option: A
Explanation:

We know, if P is any point on the curve, then Sum of focal distances $=$ length of major axis
i.e., $SP + S'P = 2a$
$= 2(5) [\because a^{2} = 5^{2}]$
$= 10$