Tag: standard equation of an ellipse

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

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

  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
Reveal answer Fill a bubble to check yourself
B Correct answer
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.
Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

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

  1. $5$
  2. $6$
  3. $10$
  4. $3$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.
$\Rightarrow SP+S'P=2a$
$\Rightarrow SP+S'P=2 \times 5=10$

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

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 

  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. $\dfrac {x^{2}}{8}+\dfrac {y^{2}}{6}=1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The line x/4 + y/3 = 1 meets the x-axis at (4, 0) and the y-axis at (0, 3). An ellipse centered at the origin passing through these points has the form x^2/a^2 + y^2/b^2 = 1. Substituting the points gives a^2 = 16 and b^2 = 9.

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

  1. $\dfrac{20}{3}$
  2. $\dfrac {15}{3}$
  3. $\dfrac {40}{3}$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $\dfrac{20}{3}$
Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

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

  1. $ 9$
  2. $6$
  3. $11$
  4. $15$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $\displaystyle \left ( 0, \pm ae \sqrt{13} \right )$ 
$\displaystyle L.R.= \frac{2b^{2}}{a}= 2.\frac{9}{2}= 9$

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

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

  1. $ae$
  2. $2ae$
  3. $ae^{2}$
  4. $2ae^{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $b^2 =a^2 (1-e^2)$

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

$d=2ae^2$
Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

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

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

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $=\dfrac {2b^2}{a}$

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

$=3$