Tag: standard equation of ellipse

Questions Related to standard equation of 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. $A, C, B, D$


Correct Option: C
Explanation:

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

The focal chord of a conic perpendicular to axis is 

  1. Tangent

  2. Vertex

  3. Focal distance

  4. Latus rectum


Correct Option: D
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 $latus\ rectum$.

The locus of a planet orbiting around the sun is: 

  1. A circle

  2. A straight line

  3. A semicircle

  4. An ellipse


Correct Option: D
Explanation:

It is a fact & proof of it can be seen from higher education physics books

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


Correct Option: B
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

Equation of the ellipse in its standard form is $\displaystyle \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

  1. True

  2. False

  3. Nither

  4. Either


Correct Option: B
Explanation:

Equation of ellipse in standard form is 

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

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. ${x^2} + ay - {b^2}$


Correct Option: A

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. $x=1 \pm \sqrt {5}$


Correct Option: C
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 } $

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. $4$


Correct Option: B
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)
$16-9=b^2\Rightarrow b^2=7$

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


Correct Option: A
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$\Rightarrow \dfrac{x^2}{\dfrac{1}{4}}+\dfrac{y^2}{\dfrac{5}{4}}=1$

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. $\dfrac{(x-1)^2}{45}+\dfrac{(y-2)^2}{16}=1$


Correct Option: C