Tag: standard equation of ellipse

Questions Related to standard equation of ellipse

For the ellipse $ {12x}^{2} +{4y}^{2} +24x-16y+25=0 $

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

  1. centre is $(-1,2) $

  2. Length of axes are $ {\sqrt {3}} and 1 $

  3. eceentricity is $ \sqrt {\cfrac {2} {3}} $

  4. All of these

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Correct Option: B
Explanation:
Given,

$12x^2+4y^2+24x-16y+25=0$

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

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

$\therefore a=\dfrac{1}{2},b=\dfrac{\sqrt 3}{2}$

⇒ Centre $ = (-1, 2)$

Here $b^2>a^2$

⇒ eccentricity$(e) = \sqrt {\dfrac {b^2-a^2}{b^2}} $

$= \sqrt {\dfrac {\dfrac 3 4 - \dfrac 1 4}{\dfrac 3 4}}=\sqrt{\dfrac 2 3}$

Length of arcs,

length of major arc $=2b=2\left ( \dfrac{\sqrt 3}{2} \right )=\sqrt 3$

length of minor arc $=2a=2\left ( \dfrac{1}{2} \right )=1$

Option D is correct.

A point $P$ on the ellipse $\displaystyle \frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$ has the eccentric angle $\displaystyle \frac{\pi}{8}$. The sum of the distance of $P$ from the two foci is

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

  1. $5$

  2. $6$

  3. $10$

  4. $3$

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Correct Option: C
Explanation:

Given,ellipse equation as $\dfrac{x^2}{25}+\dfrac{y^2}{9}=1$
Length of major axis, $a=5$ and length of minor axis, $b=3$
$P$ is a point on the ellipse whose eccentricity is $\dfrac{\pi}{8}.$
We know that, sum of the distances of any point on the ellipse from its foci equal to twice the major axis.
Let $S,S'$ be foci of ellipse and $a,b$ as the length of major,minor axis respectively.
$\Rightarrow SP+S'P=2a$
$\Rightarrow SP+S'P=2 \times 5=10$

Axes are coordinates axes, the ellipse passes through the points where the straight line $\dfrac {x}{4}+\dfrac {y}{3}=1$  meets the coordinates axes. Then equation of the ellipses is 

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

  1. $\dfrac {x^{2}}{16}+\dfrac {y^{2}}{9}=1$

  2. $\dfrac {x^{2}}{64}+\dfrac {y^{2}}{36}=1$

  3. $\dfrac {x^{2}}{4}+\dfrac {y^{2}}{3}=1$

  4. $\dfrac {x^{2}}{8}+\dfrac {y^{2}}{6}=1$

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Correct Option: A

The equation $\sqrt{(x-3)^{2}+(y-1)^{2}}+\sqrt{(x-3)^{2}+(y-1)^{2}}=6$ represents : 

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

  1. an ellipse

  2. a pair of straight lines

  3. a circle

  4. the line segment joining the point $(-3,1)$ to the point $(3,1)$

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Correct Option: A

If a chord of $y^{ 2 } = 4ax$ makes an angle $\alpha ,\alpha \epsilon \left( 0,\pi /4 \right)$ with the positive direction of $X-axis$, then the minimum length of this focal chord is 

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

  1. $2 \sqrt{ 2 } a units$

  2. $4 \sqrt{ 2 } a units$

  3. $8a units$

  4. $16 a units$

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Correct Option: A

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

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

$=3$