Tag: asymptote

Questions Related to asymptote

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

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

  1. An asymtote parallel to $x$-axis

  2. An asymtote parallel to $y$-axis

  3. Asymtotes parallel to both axes

  4. No asymptote

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Correct Option: B
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.

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

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

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

  2. $\dfrac{1}{e}-1$

  3. $\dfrac{1}{e}$

  4. None of these

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Correct Option: C
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}}} $

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

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

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}$.

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

  1. True

  2. False

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Correct Option: A

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

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

  1. $2x + 3y - 1 = 0$,$3x + 2y + 2 = 0$

  2. $2x + 3y = 1,3x + 2y = 2$

  3. $3x + 3y = 0,3x + 2y = 0$

  4. $2x + 3y = 3,3x + 2y = 4$

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Correct Option: B

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 :

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

  1. $2$

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

  3. $\sqrt 2$

  4. $\sqrt 3$

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Correct Option: B
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 $e=\sec{\dfrac{\theta}{2}}=\sec{\dfrac{\pi}{6}}=\dfrac{2}{\sqrt{3}}$

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 :

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

  1. $\sec \lambda$

  2. $\cos ec \lambda$

  3. $\sec^2\lambda$

  4. $\cos ec^2\lambda$

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Correct Option: A
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}} $

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

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 :

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

  1. $20$

  2. $-16$

  3. $-22$

  4. $18$

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Correct Option: C
Explanation:

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

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

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

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

Show answer
Correct Option: D
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 $

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:

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

  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 $\displaystyle 3x-3y+5=0$

Show answer
Correct Option: C
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.

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

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

  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. $\dfrac{(x+3)^2}{4}-\dfrac{(y-1)^2}{3} = 1$

Show answer
Correct Option: C
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.