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 arrangement of the following conics in the descending order of their lengths of semi latus rectum is
A) $ 6= r (1 + 3\cos \theta )$
B) $10= r (1 + 3\cos \theta )$
C) $8= r (1 + 3\cos \theta )$
D) $12= r (1 + 3\cos \theta )$

  1. $D, A, B, C$
  2. $B, C, D, A$
  3. $D, B, C, A$
  4. $A, C, B, D$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Comparing given equation with standard equation $r(1+e\cos\theta)=l$ where $l$ is semi latus rectum
Hence order is $D,B,C,A$

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

The focal chord of a conic perpendicular to axis is 

  1. Tangent

  2. Vertex

  3. Focal distance

  4. Latus rectum
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

A perpendicular from a point on the conic to the axis is called an ordinate, and if produced to meet the conic again it is called a double ordinate. The double ordinate through the focus is called the $latus\ rectum$.

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

The sum of the focal distances of a point on the ellipse $\cfrac { { x }^{ 2 } }{ 4 } +\cfrac { { y }^{ 2 } }{ 9 } =1$ is:

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

The sum of focal distances from a point of ellipse is 2 times the major axis.
For the given ellipse , length of semi major axis i.e. $b$ is $3$.
So required length $=2\times 3=6$
Option B is true

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

The focus of extremities of the latus rectum of the family of the ellipse  ${b^2}{x^2} + {a^2}{y^2} = {a^2}{b^2}{\text{ is }}\left( {b \in R} \right)$ 

  1. ${x^2} - ay = {a^3}$
  2. ${x^2} - ay - {e^2}$
  3. ${x^2} \pm ay = {a^2}$
  4. ${x^2} + ay - {b^2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The extremities of the latus rectum for the ellipse x^2/a^2 + y^2/b^2 = 1 are (ae, b^2/a) and (ae, -b^2/a). With b^2 = a^2(1-e^2), the coordinates are (ae, a(1-e^2)). Eliminating e leads to the locus x^2 = ay.

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

The equation of the latusrecta of the ellipse $9x^{2}+4^{2}-18x-8y-23=0$ are 

  1. $y=\pm \sqrt {5}$
  2. $x=\pm \sqrt {5}$
  3. $y=1 \pm \sqrt {5}$
  4. $x=1 \pm \sqrt {5}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

equation of ellipse is ${ 9x }^{ 2 }+{ 4y }^{ 2 }-18x-8y-23=0$

$\Rightarrow (9x^{ 2 }-18x+9)+(4y^{ 2 }-8y+4)-23-9-4=0$
$ \Rightarrow 9(x-1)^{ 2 }+4(y-1)^{ 2 }=36$
$ \Rightarrow \cfrac { (x-1)^{ 2 } }{ 4 } +\cfrac { (y-1)^{ 2 } }{ 9 } =1$
So, equation of latus recta is $(y-1)=\pm be$
$y=1\pm \sqrt { b^{ 2 }-a^{ 2 } } \Rightarrow y=1\pm \sqrt { 5 } $

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

The foci of the ellipse $\dfrac{x^{2}}{16} + \dfrac{y^{2}}{b^{2}} =1$ and the hyperbola $\dfrac{x^{2}}{144} - \dfrac{y^{2}}{81} =\dfrac{1}{25}$ coincide, then the value of $b^{2}$ is:

  1. $5$
  2. $7$
  3. $9$
  4. $4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The foci of the ellipse are also the foci of an hyperbola,
then we have, for the ellipse,

$a^2 -c^2 = b^2$
so
$16 -c^2 = b^2...............(1) $
 
Equation of Hyperbola can also be written as $\dfrac{x^2}{\dfrac{144}{25}}-\dfrac{y^2}{\dfrac{81}{25}}=1$

For the hyperbola, which must have its transverse axis on the x-axis, the equation
$c^2 - a^2 = b^2\Rightarrow c^2-\dfrac{144}{25}=\dfrac{81}{25}\Rightarrow c^2=\dfrac{225}{25}=9$

Putting this value in equation (1)
$16-9=b^2\Rightarrow b^2=7$

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

If foci are points $(0,1)(0,-1)$ and minor axis is of length $1$, then equation of ellipse is

  1. $\dfrac { { x }^{ 2 } }{ 1/4 } +\dfrac { { y }^{ 2 } }{ 5/4 } =1$
  2. $\dfrac { { x }^{ 2 } }{ 5/4 } +\dfrac { { y }^{ 2 } }{ 1/4 } =1$
  3. $\dfrac { { x }^{ 2 } }{ 3/4 } +\dfrac { { y }^{ 2 } }{ 1/4 } =1$
  4. $\dfrac { { x }^{ 2 } }{ 1/4 } +\dfrac { { y }^{ 2 } }{ 3/4 } =1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given that focii are $(0,1), (0,-1)$


Axis lies along $y-axis$

Distance between focii$\rightarrow 2be=2$

$\Rightarrow be=1$

Given that $2a=1$

$\Rightarrow a=\dfrac{1}{2}$

We know that $\Rightarrow e^2=1-\dfrac{a^2}{b^2}$

$\Rightarrow b^2e^2=b^2-a^2$

Substituting above obtained values in this expression we get,

$\Rightarrow 1=b^2-(\dfrac{1}{2})^2$

$\Rightarrow 1=b^2-\dfrac{1}{4}$

$\Rightarrow b^2=\dfrac{5}{4}$

Thus equation of ellipse$\Rightarrow \dfrac{x^2}{\dfrac{1}{4}}+\dfrac{y^2}{\dfrac{5}{4}}=1$

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

The equation of the ellipse with its focus at $(6, 2)$, centre at $(1, 2)$ and which passes through the point $(4, 6)$ is?

  1. $\dfrac{(x-1)^2}{25}+\dfrac{(y-2)^2}{16}=1$
  2. $\dfrac{(x-1)^2}{25}+\dfrac{(y-2)^2}{20}=1$
  3. $\dfrac{(x-1)^2}{45}+\dfrac{(y-2)^2}{20}=1$
  4. $\dfrac{(x-1)^2}{45}+\dfrac{(y-2)^2}{16}=1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Center (1, 2), Focus (6, 2). Distance ae = 5. Ellipse passes through (4, 6). Using the definition of an ellipse (sum of distances to foci = 2a), we can solve for a and b.