Tag: ellipse

Quizzes (5) Questions (224)

Questions Related to ellipse

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

The length of latus rectum of $\dfrac {x^2}9+\dfrac {y^2}2=1$ is 

  1. $\dfrac 74$
  2. $\dfrac 34$
  3. $\dfrac 43$

  4. None.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The length of latus Rectum of $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$

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

Here $a=3\quad b=\sqrt 2$

$\Rightarrow \dfrac{2b^2}{a}=\dfrac{2(\sqrt 2)^2}{3}=\dfrac{2(2)}{3}=\dfrac 43$

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

An ellipse of semi-axis $a,b,$ slides between two perpendicular lines, then the locus of its foci is, (the two lines being taken  as the axes of coordinates)

  1. $(x^{2}+y^{2})(x^{2}y^{2}+b^{2})=4a^{2}x^{2}y^{2}$
  2. $(x^{2}+y^{2})(x^{2}y^{2}+b^{2})=4b^{2}x^{2}y^{2}$
  3. $(x^{2}-y^{2})(x^{2}y^{2}+b^{2})=4b^{2}x^{2}y^{2}$
  4. $(x^{2}-y^{2})(x^{2}y^{2}+b^{2})=4a^{2}x^{2}y^{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is a classic locus problem involving an ellipse sliding between two perpendicular axes. The locus of the foci is derived using coordinate geometry.

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

If equation $(5x-1)^{2}+(5y-2)^{2}=(\lambda^{2}-2\lambda+1)(3x+4y-1)^{2}$ represents an ellipse, then $\lambda \in$

  1. $(0, 1)$
  2. $(0, 2)$
  3. $(1, 2)$
  4. $(0, 1)\cup (1, 2)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The equation represents an ellipse if the eccentricity e < 1. This condition relates to the coefficients of the quadratic form.

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

The equation $\dfrac{{x}^{2}}{2-r}+\dfrac{{y}^{2}}{r-5}+1=0$ represents an ellipse if

  1. $r>1$
  2. $r>5$
  3. $2 < r< 5$
  4. $r<2$ or $r>5$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For the equation to represent an ellipse, the denominators must be positive and the coefficients must allow for the standard form x^2/A + y^2/B = 1. This requires 2-r > 0 and r-5 < 0, or vice versa, leading to 2 < r < 5.

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

Eccentricity of ellipse $\frac{{{x^2}}}{{{a^2} + 1}} + \frac{{{y^2}}}{{{a^2} + 2}} = 1$ is $\frac{1}{{\sqrt 3 }}$  then length of Latusrectum is 

  1. $\frac{8}{{\sqrt 3 }}$
  2. $\frac{4}{{\sqrt 3 }}$
  3. $2\sqrt 3 $
  4. $\frac{{\sqrt 3 }}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

Eccentricity of ellipse $=\cfrac{1}{\sqrt 3}$
So, $\cfrac{\sqrt{b^2-a^2}}{a}=\cfrac{1}{\sqrt 3}$
Where ellipse 
$\cfrac{x^2}{a^2+1}+\cfrac{y^2}{a^2+2}=1$
So, here
$\cfrac{\sqrt{a^2+2a^2-1}}{a}=\cfrac{1}{\sqrt 3}$
$\implies a^2=2$
So, equation is 
$\cfrac{x^2}{3}+\cfrac{y^2}{4}=1$
Latus rectum $=\cfrac{2b^2}{a}=\cfrac{2\times 4}{\sqrt 3}=\cfrac{8}{\sqrt 3}$

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

The equation $\dfrac { x ^ { 2 } } { 10 - a } + \dfrac { y ^ { 2 } } { 4 - a } = 1$ represents an ellipse if

  1. $a < 4$
  2. $a > 4$
  3. $4 < a < 10$

  4. None of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\dfrac{x^2}{10-a}+\dfrac{y^2}{4-a}=1$

For this equation to represent an ellipse its eccentricity shoule lie between $0$ and $1$.
$\sqrt{1-\dfrac{b^2}{a^2}}< 1$
$0< 1-\dfrac{(4-a)^2}{(10-a)^2} <1$
$0 < (10-a)^2-(4-a)^2 <1$
$0< 84-12a <1$
$0< (7-a)12<1$
$12(7-a)> 0$ and
$12(7-a)<1$
$a< 7$ and $7-a<\dfrac{1}{12}$
$a< 7$ and $a >\dfrac{83}{12}$

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

If the latus rectum of an ellipse $x ^ { 2 } \tan ^ { 2 } \varphi + y ^ { 2 } \sec ^ { 2 } \varphi =$ $1$ is $1 / 2 $ then $\varphi $ is

  1. $\pi / 2$
  2. $\pi / 6$
  3. $\pi / 3$
  4. $5$ $\pi/ 12$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given $x^2 tan^2 \phi + y^2 \, sec^2 \phi = 1$

$\rightarrow \dfrac{x^2}{(1/tan^2 \phi)} + \dfrac{y^2}{(1/sec^2 \phi)} = 1$
$a = \pm \dfrac{1}{tan \phi} , b = \pm \dfrac{1}{sec \phi}$
and $\rightarrow e^2 = 1 - \dfrac{b^2}{a^2}$
$\rightarrow e^2 = 1 - \dfrac{1/sec^2 \phi}{1/tan^2 \phi} = 1 - \dfrac{tan^2 \phi}{sec^2 \phi}$
$\rightarrow e^2 = 1 - sin^2 \phi = cos^2 \phi$
length of latus rectum 
$(LL') = \dfrac{2 b^2}{a} = 2a (1 - e^2)$
$\rightarrow 2a (1 - cos^2 \phi) = 2a. sin^2 \phi = \dfrac{1}{2} $ (Given)
$\therefore 2. \dfrac{cos \phi}{sin \phi} sin^2 \phi = \dfrac{1}{2}$
$\rightarrow 2 cos \phi \, sin \phi = \dfrac{1}{2}$
$\rightarrow sin^2 \phi = \dfrac{1}{2} $
$\rightarrow 2 \phi = \dfrac{\pi}{6} , \dfrac{5 \pi}{6}$
$\therefore \phi = \dfrac{\pi}{12}$    or 
$\phi = \dfrac{ 5 \pi}{12}$

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

vertices of an ellipse are $(0,\pm 10)$ and its eccentricity $e=4/5$ then its equation is 

  1. $90x^2-40y^2=3600$
  2. $80x^2+50y^2=4000$
  3. $36x^2+100y^2=3600$
  4. $100x^2+36y^2=3600$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let the equation of the required ellipse be 

$\dfrac { { x }^{ 2 } }{ { a }^{ 2 } } +\dfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1\longrightarrow \left( 1 \right) $
since the vertices of the ellipse are on $y$-axis, so the coordinate of the vertices are $\left( 0,\pm b \right) $
$\therefore b=10\ Now,\quad { a }^{ 2 }=b^{ 2 }\left( 1-{ e }^{ 2 } \right) \ \Rightarrow { a }^{ 2 }=100\left( 1-\dfrac { 16 }{ 25 }  \right) \ \Rightarrow { a }^{ 2 }=36\ $
substituting the value of ${ a }^{ 2 }$ and ${ b }^{ 2 }$ in equation $(1)$
we get, $\dfrac { { x }^{ 2 } }{ 36 } +\dfrac { { y }^{ 2 } }{ 100 } =1\ \Rightarrow 100{ x }^{ 2 }+36{ y }^{ 2 }=3600\ \Rightarrow 100{ x }^{ 2 }+36{ y }^{ 2 }-3600=0$