Tag: mathematics and statistics

Quizzes (38) Questions (955)

Questions Related to mathematics and statistics

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, $\dfrac { { x }^{ 2 } }{ 36 } +\dfrac { { y }^{ 2 } }{ 100 } =1\ \Rightarrow 100{ x }^{ 2 }+36{ y }^{ 2 }=3600\ \Rightarrow 100{ x }^{ 2 }+36{ y }^{ 2 }-3600=0$

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

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

  1. $y=\pm \sqrt{5}$
  2. $y=- \sqrt{5}$
  3. $y=1\pm \sqrt{5}$
  4. $y=-1\pm \sqrt{5}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$9x^2+4y^2-18x-8y-23=0$
$=(3x-3)^2+(2y-2)^2-13-23=0$
$\Rightarrow \ 9(x-1)^2+4(y-1)^2=36$
$\Rightarrow \ \dfrac {(x-1)^2}{4}+\dfrac {(y-1)^2}{9}=1$
Shifting origin to $(1,1)\Rightarrow x-1=x,\ y-1=y$
$\Rightarrow \ \dfrac {x^2}{4}+\dfrac {y^2}{9}=1$
$a=2,= b=3,\ e^2=1+\dfrac {a^2}{b^2}=1+\dfrac {4}{9}=\dfrac {5}{9}$
$\Rightarrow \ e=\pm \sqrt {\dfrac {5}{9}}+\dfrac {\sqrt 5}{3}$
$\Rightarrow \ $ Focus $=(0,\ \pm be)=(0,\ \pm \sqrt 5)$
$\Rightarrow \ $ latus ractum $\Rightarrow \ y=\pm \sqrt {5}$
Shifting back, $y=y-1$
$\Rightarrow \ y-1=\pm \sqrt {5}$
$\Rightarrow \ y=1\pm \sqrt 5\  \Rightarrow \ (C) $


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

If there is exactly one tangent at a distance of $4$ units from one of the locus of $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{a^{2}-16}=1, a>4$, then length of latus rectum is :-

  1. $16$
  2. $\dfrac{8}{3}$
  3. $12$
  4. $15$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equation represents a hyperbola for a > 4. The condition of exactly one tangent at a distance of 4 units implies the distance from the center to the tangent is 4, which corresponds to the distance to the asymptotes or a specific property of the hyperbola. Solving for the latus rectum length 2(a^2-16)/a leads to 16.

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. $r>2$
  2. $r\in \left(2,\:\dfrac{7}{2}\right)\cup \left(\dfrac{7}{2},5\right)$
  3. $r>5$
  4. $r<2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Equating the equation of the ellipse with the second-degree equation
$A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0$ with $\dfrac{{x}^{2}}{2-r}+\dfrac{{y}^{2}}{r-5}+1=0$
we get $A=\dfrac{1}{2-r}, B=0, C=\dfrac{1}{r-5},D=0,E=0$ and $F=1$
For the second degree equation to represent an ellipse, the coefficients must satisfy the discriminant condition ${B}^{2}-4AC<0$ and also $A\neq C$
$\Rightarrow {\left(0\right)}^{2}-4\left(\dfrac{1}{2-r}\right)\left(\dfrac{1}{r-5}\right)<0$
$\Rightarrow -4\left(\dfrac{1}{2-r}\right)\left(\dfrac{1}{r-5}\right)<0$
$\Rightarrow \left(\dfrac{1}{2-r}\right)\left(\dfrac{1}{r-5}\right)>0$
$\Rightarrow \left(2-r\right)\left(r-5\right)<0$
$\Rightarrow \left(r-2\right)\left(r-5\right)>0$
$\Rightarrow r>2$