Tag: mathematics and statistics

Questions Related to mathematics and statistics

The number of parabolas that can be drawn if two ends of the latus rectum are given 

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

  1. 1

  2. 2

  3. 4

  4. 3

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Correct Option: B

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

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

  1. $r>1$

  2. $r>5$

  3. $2 < r< 5$

  4. $r<2$ or $r>5$

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Correct Option: C

The locus of the mid points of the portion of the tangents to the ellipse intercepted between the axes

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

  1. $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=4$

  2. $\dfrac{a^{2}}{x^{2}}+\frac{b^{2}}{y^{2}}=4$

  3. $\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=4$

  4. none of these

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Correct Option: A

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 

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

  1. $\frac{8}{{\sqrt 3 }}$

  2. $\frac{4}{{\sqrt 3 }}$

  3. $2\sqrt 3 $

  4. $\frac{{\sqrt 3 }}{2}$

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Correct Option: A
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 $=\cfrac{2b^2}{a}=\cfrac{2\times 4}{\sqrt 3}=\cfrac{8}{\sqrt 3}$

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

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

  1. $a < 4$

  2. $a > 4$

  3. $4 < a < 10$

  4. None of these

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Correct Option: D
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 $a >\dfrac{83}{12}$

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

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

  1. $\pi / 2$

  2. $\pi / 6$

  3. $\pi / 3$

  4. $5$ $\pi/ 12$

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Correct Option: D
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 
$\phi = \dfrac{ 5 \pi}{12}$

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

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

  1. $90x^2-40y^2=3600$

  2. $80x^2+50y^2=4000$

  3. $36x^2+100y^2=3600$

  4. $100x^2+36y^2=3600$

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

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

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

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

  2. $y=- \sqrt{5}$

  3. $y=1\pm \sqrt{5}$

  4. $y=-1\pm \sqrt{5}$

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Correct Option: C
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) $


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 :-

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

  1. $16$

  2. $\dfrac{8}{3}$

  3. $12$

  4. $15$

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Correct Option: A

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

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

  1. $r>2$

  2. $r\in \left(2,:\dfrac{7}{2}\right)\cup \left(\dfrac{7}{2},5\right)$

  3. $r>5$

  4. $r<2$

Show answer
Correct Option: A
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 $A\neq C$
$\Rightarrow {\left(0\right)}^{2}-4\left(\dfrac{1}{2-r}\right)\left(\dfrac{1}{r-5}\right)<0$
$\Rightarrow -4\left(\dfrac{1}{2-r}\right)\left(\dfrac{1}{r-5}\right)<0$
$\Rightarrow \left(\dfrac{1}{2-r}\right)\left(\dfrac{1}{r-5}\right)>0$
$\Rightarrow \left(2-r\right)\left(r-5\right)<0$
$\Rightarrow \left(r-2\right)\left(r-5\right)>0$
$\Rightarrow r>2$