Tag: ellipse

Questions Related to ellipse

If $(2,4)$ and $( 10,10)$ are the ends of a latus - rectum of an ellipse with eccentricity $\dfrac 12$, then the length of semi - major axis is 

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

  1. $\dfrac{20}{3}$

  2. $\dfrac {15}{3}$

  3. $\dfrac {40}{3}$

  4. None of these

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Correct Option: A
Explanation:
Given $(2,4)$ and $(10,10)$ are the ends of the latusrectum and eccentricity is $\dfrac{1}{2}$

We know that length of the latus rectum is $\dfrac{2b^{2}}{a}$

We know that the distance between the two points

$(x _{1}, y _{1})$ and $(x _{2}, y _{2})$ is

$\sqrt{(x _{1}-x _{2})^{2}+(y _{1}-y _{2})^{2}}$

$\Rightarrow \dfrac{2b^{2}}{a^{2}}=\sqrt{(2-10)^{2}+(4-10)^{2}}$

$\Rightarrow \dfrac{2b^{2}}{a}=\sqrt{(-8)^{2}+(-6)^{2}}=\sqrt{64+36}=\sqrt{100}$

$\Rightarrow 36^{2}=10a$

$\Rightarrow b^{2}=5a$

we know that $b^{2}=a^{2}(1-e^{2})$

$a^{2}(1-e^{2})=5a$

$a\left(1-\left(\dfrac{1}{2}\right)^{2}\right)=5$

$a=\dfrac{5}{1-\dfrac{1}{4}}=\dfrac{5}{\dfrac{3}{4}}$

$a=\dfrac{20}{3}$

Thus the length of semi major axis is $\dfrac{20}{3}$

The equation $\dfrac{x^2}{1-r}-\dfrac{y^2}{1+r}=1, |r| < 1$ represents?

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

  1. An ellipse

  2. A hyperbola

  3. A circle

  4. None of these

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Correct Option: A

Find the  Lactus Rectum of  $\displaystyle 9y^{2}-4x^{2}=36$ 

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

  1. $ 9$

  2. $6$

  3. $11$

  4. $15$

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Correct Option: A
Explanation:

$\displaystyle \frac{y^{2}}{4}-\frac{x^{2}}{9}= 1.$ 
Here the coefficient of $\displaystyle y^{2}$ is + ive and that of $\displaystyle x^{2}$ is -ive and hence it represents a hyperbola whose transerse axis is vertical, i.e.
$\displaystyle a^{2}=4, b^{2}=9.$
$\displaystyle b^{2}= a^{2}\left ( e^{2}-1 \right )$
or $\displaystyle \frac{9}{4}+1=e^{2}\therefore e= \frac{\sqrt{13}}{2}$ 
Foci lie on y-axis $\displaystyle \left ( 0, \pm ae \right )$ i.e $\displaystyle \left ( 0, \pm ae \sqrt{13} \right )$ 
$\displaystyle L.R.= \frac{2b^{2}}{a}= 2.\frac{9}{2}= 9$

The difference between the lengths of the major axis and the latus-rectum of an ellipse is

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

  1. $ae$

  2. $2ae$

  3. $ae^{2}$

  4. $2ae^{2}$

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Correct Option: D
Explanation:

We know that the length of major axis is $2a$ and latus rectum is $\dfrac {2b^2}{a}$

for the ellipse

$\dfrac {x^2}{a^2}+\dfrac {y^2}{b^2}=1$

Let $d$ be the difference

$d=2a-\dfrac {2b^2}{a}$

$d=\dfrac {2a^2 -2b^2}{a}$

We know thta $b^2 =a^2 (1-e^2)$

$d=\dfrac {2a^2 e^2}{a}$

$d=2ae^2$

The latus-rectum of the conic $3x^{2} + 4y^{2} - 6x + 8y - 5 = 0$ is

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

  1. $3$

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

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

  4. None of these

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Correct Option: A
Explanation:
Given equation of conic is:

$3x^2+4y^2-6x+8y-5=0$

$3(x^2-2x)+4(y^2+2y)=5$

$3(x^2-2x+1)+4(y^2+2y+1)=5+3+4$

$3(x-1)^2 +4(y+1)^2 =12$

$\dfrac {3(x-1)^2}{12}+\dfrac {4(y+1)^2}{12}=1 \Rightarrow \dfrac {(x-1)^2}{4}+\dfrac {(y+1)^2}{3}=1$

so, $a=2, b=\sqrt 3$

$\therefore \ $ Latus rectum $=\dfrac {2b^2}{a}$

$=\dfrac {2[\sqrt 3]^2}{2}$

$=3$

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 $\lambda < 5$

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

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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. $12$

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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. $4 \sqrt{5}$

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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 $\;(t-4)>0,(8-t)\neq(t-4)$
$\Rightarrow t\in(-\infty,8) \cap (4,\infty) \cap$ {$t\neq6$}
$\Rightarrow t\in(4,8)-${$6$}