Tag: mathematics and statistics

Questions Related to mathematics and statistics

If e is the eccentricity of $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ and $'\Theta '$ be the angle between its asymptotes, then $cos(\Theta /2)$ is equal to,

  1. 1/2e

  2. 1/e

  3. $1/e^{2}$

  4. none of these


Correct Option: A

The equation of the line passing through the centre of a rectangle hyperbola is $x-y-1=0$. If one of its asymptotes is $3x-4x-6=0$, the equation of the other asymptote is $

  1. $4x+3y+17=0$

  2. $4x-3y+8=0$

  3. $3x-2y+15=0$

  4. $None of these$


Correct Option: A

if the product of the perpendicular distances from any point on the hyperbola$\frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\quad of\quad eccentrincity\quad e=\sqrt { 3 } $ on its asymptotes is equal to 6 then the length of the transverse axis of the hyperbola is;

  1. 3

  2. 6

  3. 8

  4. 12


Correct Option: A

if the product of the perpendicular distances from any point on the hyperbola $\frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1$ of eccentrincity $e=\sqrt { 3 } $ on the asymptotes is equal to 6 then the length of transverse axis of the hyperbola is

  1. 3

  2. 6

  3. 8

  4. 12


Correct Option: A

If $e$ is the eccentricity of $\dfrac {x^{2}}{a^{2}}-\dfrac {y^{2}}{b^{2}}=1$ and '$\theta $' be the angle between its asymptotes then $\cos (\theta /2)$ is equal to.

  1. $1/ 2e$

  2. $1/ e$

  3. $2/e^{2}$

  4. $none\ of\ these$


Correct Option: B

The asymptotes of the hyperbola $xy-3x+4y+2=0$

  1. $x=-4$

  2. $x=4$

  3. $y=-3$

  4. $y=3$


Correct Option: A,D
Explanation:

Since the equation of a hyperbola and its asymptotes differ in constant terms only. Therefore, the equations of asymptotes of the given hyperbola are given by $xy-3x+4y+k=0$

where $k$ is a constant to be determined  by the condition that $abc+2fgh-{ af }^{ 2 }-{ bg }^{ 2 }-{ ch }^{ 2 }=0$
i.e., $\displaystyle 0+2\times 2\times \left( \frac { -3 }{ 2 }  \right) \times \frac { 1 }{ 2 } -0-0-k\times { \left( \frac { 1 }{ 2 }  \right)  }^{ 2 }=0\Rightarrow k=-12$
$\because $ Asymptotes of the given hyperbola are $xy-3x+4y-12=0$ or $(x+4)(y-3)=0$
i.e., $x=-4$ and $y=3.$

If $x + 2 = 0$ and $y = 1$ are the equation of asymptotes of rectangular hyperbola passing through (1,0).Then which of the following is(are) not the equation(s) of hyperbola :

  1. $xy + 2y -1 = 0$

  2. $xy - 2y + 1 = 0$

  3. $xy - 2y - 1 = 0$

  4. $xy-x+2y+1=0$


Correct Option: A,B,C
Explanation:

Equation of hyperbola is of the form $(x+2)(y-1)=k$
Since, it passes through $(1,0)$
Therefore, $(1+2)(0-1)=k$
$\Rightarrow k=-3$
Therefore, equation of hyperbola is $xy-x+2y+1=0$

If ax + by + c = 0 and $\displaystyle \varphi \chi $ + my + n = 0 are asymptotes of a hyperbola, then: 

  1. $\displaystyle am\neq b\varphi $

  2. $\displaystyle \frac{am+b\varphi }{a\varphi +bm}\neq 0$

  3. $\displaystyle a\varphi \neq bm$

  4. none of these


Correct Option: A
Explanation:

Asymptotes of Hyperbola are Intersecting each other 


So, These line will be anti-parallel or intersecting

Condition for intersecting lines is $\dfrac{a}{b}\neq\dfrac{\varphi}{m}\Rightarrow am \neq b\varphi$

If $\theta$ is the angle between the asymptotes of the hyperbola $\displaystyle \frac{x^2}{a^2}\, -\, \displaystyle \frac{y^2}{b^2}\, =\, 1$ with eccentricity $e$, then $\sec \displaystyle  \frac{\theta}{2}$can be

  1. $e$

  2. $\dfrac{e}2$

  3. $\dfrac{e}3$

  4. $\displaystyle \frac{e}{\sqrt{e^2\, -\, 1}}$


Correct Option: A,D
Explanation:

$\tan\, \displaystyle \frac{\theta}{2}\, =\, \displaystyle \frac{b}{a}\,

\Rightarrow\, e^2\, -\, 1\, =\, \tan^2\, \displaystyle

\frac{\theta}{2}\, \Rightarrow\, \sec \displaystyle \frac{\theta}{2}\,

=\, e$
or $e^2\, -\, 1\, =\, \cot^2\, \displaystyle \frac{\theta}{2}\, \Rightarrow\, co\sec\, \displaystyle \frac{\theta}{2}\, =\, e$
$\Rightarrow\, \sec\, \displaystyle \frac{\theta}{2}\, =\, \displaystyle \frac{e}{\sqrt{e^2\, -\, 1}}$.

The asymptotes of a hyperbola are parallel to lines $2x + 3y = 0$ and $3x + 2y = 0.$ The hyperbola has its centre at $(1, 2)$ and it passes through $(5, 3).$ Find its equation.

  1. $(2x\, +\, 3y\, -\, 8) (3y\, +\, 2y\, -\, 7)\, =\, 154$

  2. $(2x\, +\, 3y\, -\, 7) (3y\, +\, 2y\, -\, 8)\, =\, 154$

  3. $(2x\, +\, 3y\, -\, 7) (3y\, +\, 2y\, -\, 8)\, =\, 127$

  4. $(2x\, +\, 3y\, -\, 8) (3y\, +\, 2y\, -\, 7)\, =\, 127$


Correct Option: A
Explanation:

let the equation of asymtotes be $2x+3y=a$ and $3x+2y=b$
both asymtotes intersect at centre $(1,2)$
Therefore, $a=2+3(2)=8$ and $b=3+2(2)=7$
now, the equation of hyperbola is of the form $(2x+3y-8)(3x+2y-7)=k$
It passes through $(5,3)$
Therefore, $(2(5)+3(3)-8)(3(5)+2(3)-7)=k$
Thus $k=154$