Tag: asymptote

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Questions Related to asymptote

Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

The curve ${ y }^{ 2 }\left( x-2 \right) ={ x }^{ 2 }\left( 1+x \right) $ has:

  1. An asymtote parallel to $x$-axis
  2. An asymtote parallel to $y$-axis

  3. Asymtotes parallel to both axes

  4. No asymptote
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

To find the asymtote, we need to get one variable in terms of other variable.

By observation, we see that it is very easy to get $y$ in terms of $x$. 
So, that's exactly what we will do:
$y^{ 2 }(x-2)=x^{ 2 }(1+x)$
$\Rightarrow  y^{ 2 }=\dfrac { x^{ 2 }(1+x) }{ x-2 }$ 
By definition, asymtote can be found when for a finite value of one co-ordinate, other tends to $\infty$ or $-\infty$.
So, we see when $x=2$,  $y= \infty$.
So, $x=2$ is an asymptote which is parallel to $y$-axis.

Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

If e is the eccentricity of the hyperbola and $\theta$ is angle between the asymptotes, then $\dfrac{cos\theta}{2}$ = 

  1. $\dfrac{(1-e)}{e}$
  2. $\dfrac{1}{e}-1$
  3. $\dfrac{1}{e}$

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

Let $\dfrac {x^{2}}{a^{2}}-\dfrac {y^{2}}{b^{2}}=1$  be the hyperbola


It has asymptotes $y=\pm \dfrac {b} {a} x$
Angle between the asymptotes $= 2tan^-1 (\dfrac{b}{a})=\theta$
$\Rightarrow \tan \dfrac {\theta} {2}=\pm \dfrac {b} {a} $

$\Rightarrow sec^{2}\dfrac {\theta} {2}=1+\tan ^{2}\dfrac {\theta} {2}=1+\dfrac {b^{2}}{a^{2}}$

$\Rightarrow \sec ^{2}\dfrac {\theta} {2}=\sqrt {1+\dfrac{b^{2}}{a^{2}}} $

$\Rightarrow \sec ^{2}\dfrac {\theta} {2} =e^{2}$

$\Rightarrow \cos \dfrac {\theta} {2}=\dfrac {1}{e}$

Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

Through any P of the hyperbola $\frac{x^2}{a^2}- \frac{y^2}{b^2} =1 $ a line $PQR$ is drawn with a fixed gradient $m$, meeting the asymptotes in $Q\ &\ R$. Then the product,$ (QP) (PR) =\frac{a^2b^2(1+m^2)}{b^2- a^2m^2}$.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The product of the segments cut by the asymptotes on a line with gradient m is given by the formula (a^2b^2(1+m^2)) / (b^2 - a^2m^2). This is a standard result for hyperbolas.

Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

The asymptotes of the hyperbola $6{x^2} + 13xy + 6{y^2} - 7x - 8y - 26 = 0$ are 

  1. $2x + 3y - 1 = 0$,$3x + 2y + 2 = 0$
  2. $2x + 3y = 1,3x + 2y = 2$
  3. $3x + 3y = 0,3x + 2y = 0$
  4. $2x + 3y = 3,3x + 2y = 4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The asymptotes of a hyperbola S=0 are given by S - k = 0, where k is a constant such that the equation represents a pair of lines. For 6x^2 + 13xy + 6y^2 - 7x - 8y - 26 = 0, the homogeneous part factors as (2x+3y)(3x+2y). The asymptotes are of the form (2x+3y+c1)(3x+2y+c2)=0. Expanding and matching coefficients with the original equation (with a constant adjustment) yields the lines 2x+3y=1 and 3x+2y=2.

Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

From a point $P (1, 2)$ two tangents are drawn to a hyperbola $H$ in which one tangent is drawn to each arm of the hyperbola. If the equations of asymptotes of hyperbola $H$ are $\sqrt 3x-y+5=0$ and $\sqrt 3x+y-1=0$, then eccentricity of $H$ is :

  1. $2$
  2. $\dfrac {2}{\sqrt 3}$
  3. $\sqrt 2$
  4. $\sqrt 3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Since ${c} _{1}{c} _{2}\left({a} _{1}{a} _{2}+{b} _{1}{b} _{2}\right)<0$

$\therefore$ origin lies in acute angle. 

$P\left(1,2\right)$ lies in obtuse angle

Slope of asymptotes${m} _{1}=\sqrt{3},\,{m} _{2}=-\sqrt{3}$

$\tan{\theta}=\left|\dfrac{{m} _{1}-{m} _{2}}{1+{m} _{1}{m} _{2}}\right|$

$=\left|\dfrac{\sqrt{3}-\left(-\sqrt{3}\right)}{1+\sqrt{3}\times-\sqrt{3}}\right|$

$=\left|\dfrac{2\sqrt{3}}{1-3}\right|$

$=\left|\dfrac{2\sqrt{3}}{-2}\right|$

$\Rightarrow\,\tan{\theta}=\sqrt{3}$

Acute angle between the asymptotes is $\dfrac{\pi}{3}$

Hence eccentricity $e=\sec{\dfrac{\theta}{2}}=\sec{\dfrac{\pi}{6}}=\dfrac{2}{\sqrt{3}}$
Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

The asymptotes of the hyperbola $\dfrac {x^2}{a^2}-\dfrac {y^2}{b^2}=1$ form with any tangent to the hyperbola a triangle whose area is $a^2 \tan\lambda$ in magnitude, then its eccentricity is :

  1. $\sec \lambda$
  2. $\cos ec \lambda$
  3. $\sec^2\lambda$
  4. $\cos ec^2\lambda$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Any tangent to hyperbola forms a triangle with the asymptotes which has constant area $ab$.

$\Rightarrow ab=a^2 \tan\lambda$

$\displaystyle \Rightarrow \frac {b}{a}=\tan \lambda$

$\displaystyle e=\sqrt{1+\frac{b^2}{a^2}} $

$\Rightarrow e = \sqrt{1+\tan^2{\lambda}} =\sec{\lambda}$

Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

If $S=0$ be the equation of the hyperbola $x^2+4xy+3y^2-4x+2y+1=0$, then the value of $k$ for which $S+k=0$ represents its asymptotes is :

  1. $20$
  2. $-16$
  3. $-22$
  4. $18$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$S+k=x^2+4xy+3y^2-4x+2y+1+k =0$
For equation $S+k=0$ to represent a pair of lines,
$\triangle =0$
$\begin{vmatrix} 1& 2 & -2\ 2 & 3 & 1\ -2 & 1 & 1+k\end{vmatrix}=0$
$\Rightarrow 3(1+k)-1-2(2+2k+2)-2(2+6)=0$
$\Rightarrow k=-22$

Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

One of the asymptotes (with negative slope) of a hyperbola passes through (2, 0) whose transverse axis is given by x - 3y + 2 = 0 then equation of hyperbola if it is given that the line y = 7x - 11 can intersect the hyperbola at only one point (2, 3) is given by

  1. $\displaystyle 7x^{2}+xy-y^{2}+10x-4y-3=0$
  2. $\displaystyle 7x^{2}-xy-y^{2}-10x-5y+2=0$
  3. $\displaystyle 7x^{2}+xy-y^{2}-19x-5y+28=0$
  4. $\displaystyle 7x^{2}+6xy-y^{2}-20x-4y-3=0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

As $y=7x-11$ intersects the hyperbola at only one point 


$ \displaystyle \Rightarrow $ it is parallel to one of the asymptotes

$ \displaystyle \Rightarrow $ Equation of one asymoptote can be taken as $7x-y+k=0$ clearly mirror image of $(2,0)$ about transverse axis $x-3y=2 $lies on other asymplote 

$ \displaystyle \Rightarrow \left ( \frac{6}{5},\frac{12}{5} \right )$ lies on $7x-y+k=0$

$ \displaystyle \Rightarrow k=-6$

$ \displaystyle \Rightarrow $other asymptote is $7x-y-6=0$

$ \displaystyle \Rightarrow $ centre is $(1,1)$

$ \displaystyle \Rightarrow $Asymptote through $(2,0)$ is $x+y=2$

Equation of hyperbola is $(7x-y-6)(x+y-2)-(7* 2-3-6)(2+3-2)=0$

$ \displaystyle \Rightarrow 7x^{2}+6xy-y^{2}-20x-4y-3=0 $

Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

The second-degree curve and pair of asymptotes differ by a constant. Let the second-degree curve $S = 0$ represent the hyperbola then respective pair of asymptote is given by.$\displaystyle S+\lambda =0\left ( \lambda \in R \right )$ which represent a pair of straight lines so $\lambda$  can be determined. The equation of asymptotes is $\displaystyle A=s+\lambda =0$ if equation of conjugate hyperbola of the curve $S =0$ be represents by $S _{1}$, then $A$ is arithmetic mean of the curves $S _{1}$, & $ S $.

A hyperbola passing through origin has $\displaystyle 2x-y+3=0$ and $\displaystyle x-2y+2=0$ as its asymptotes, then equation of its transverse and conjugate axes are:

  1. $\displaystyle x-y+2=0$ and $\displaystyle 3x-3y+5=0$
  2. $\displaystyle x+y+2=0$ and $\displaystyle 3x-3y+5=0$
  3. $\displaystyle x-y+1=0$ and $\displaystyle 3x-3y+5=0$
  4. $\displaystyle2 x-2y+1=0$ and $\displaystyle 3x-3y+5=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The transverse axis of hyperbola is the bisector of the angle between the asymptotes containing the origin and the conjugate axis is the other bisector.


 And equation of bisector of angle of the asymptotes are given by

$\displaystyle \frac{2x-y+3}{\sqrt{5}}=\pm \frac{x-2y+2}{\sqrt{5}}$

$\displaystyle \Rightarrow  2x-y+3 =\pm \left ( x-2y+2 \right )$

$\displaystyle \Rightarrow  2x-y+3 =x-2y+2$

and $\displaystyle  2x-y+3 =x-2y+2= -\left ( x-2y+2 \right )$

$\displaystyle  \Rightarrow x+y+1=0 \ and \ 3x-3y+5=0$

Hence, option 'C' is correct.

Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

The asymptotes of a hyperbola have equations $y-1=\dfrac{3}{4}(x+3).$ If a focus of the hyperbola has coordinates $(7,1)$, the equation of the hyperbola is

  1. $\dfrac{(x+3)^2}{16}-\dfrac{(y-1)^2}{9} = 1$
  2. $\dfrac{(y-1)^2}{9}-\dfrac{(x+3)^2}{16} = 1$
  3. $\dfrac{(x+3)^2}{64}-\dfrac{(y-1)^2}{36} = 1$
  4. $\dfrac{(y-1)^2}{36}-\dfrac{(x+3)^2}{64} = 1$
  5. $\dfrac{(x+3)^2}{4}-\dfrac{(y-1)^2}{3} = 1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Equation of asymptotes are 

$y-1=\dfrac { 3 }{ 4 } (x+3  )$    ......(i)

$ y-1=-\dfrac { 3 }{ 4 } (x+3)$     .....(ii)

Centre of the hyperbola is point of intersection of asymptotes.

Therefore, by solving (i) and (ii), we get centre as $C(-3,1)$.

Slope of asymptotes $=\dfrac { b }{ a } $

$\Rightarrow \dfrac { b }{ a } =\pm \dfrac { 3 }{ 4 }$      ......(i)

Focus is $(7,1)$.

Focus for hyperbola of form $\dfrac { { (x-h) }^{ 2 } }{ { a }^{ 2 } } -\dfrac { { (y-k) }^{ 2 } }{ { b }^{ 2 } } =1$ is $(h+ae,k)$

$\Rightarrow 7=-3+ae\\ \Rightarrow ae=10\\ \Rightarrow a\dfrac { \sqrt { { a }^{ 2 }+{ b }^{ 2 } }  }{ a } =10\\ \Rightarrow \sqrt { { a }^{ 2 }+{ b }^{ 2 } } =10$

Substituting $b$ from (i), we get

$\Rightarrow \sqrt { { a }^{ 2 }+{ \left( \pm a \dfrac { 3 }{ 4 }  \right)  }^{ 2 } } =10\\ \Rightarrow \dfrac { 5a }{ 4 } =10\\ \Rightarrow a=8\\ \Rightarrow b=\pm \dfrac { 3 }{ 4 } a=\pm 6$

So, the equation of hyperbola is

$\dfrac { { (x+3) }^{ 2 } }{ { 8 }^{ 2 } } -\dfrac { { (y-1) }^{ 2 } }{ { 6 }^{ 2 } } =1$

$\dfrac { { (x+3) }^{ 2 } }{ 64 } -\dfrac { { (y-1) }^{ 2 } }{ 36 } =1$

So, option C is correct.