Tag: hyperbola

Questions Related to hyperbola

The angle between the asymptotes of the hyperbola ${27x}^{2}-{9y}^{2}=24$ is 

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

  1. ${30}^{o}$

  2. ${120}^{o}$

  3. ${60}^{o}$

  4. ${90}^{o}$

Show answer
Correct Option: B
Explanation:
$27 x ^ { 2 } - 9 y ^ { 2 } = 24$
$\Rightarrow \quad \dfrac { x ^ { 2 } } { \left( \dfrac { 24 } { 27 } \right) } - \dfrac { y ^ { 2 } } { \left( \dfrac { 24 } { 9 } \right) } = 1$
$\Rightarrow \dfrac { x ^ { 2 } } { ( \dfrac 89 ) } - \dfrac { y ^ { 2 } } { \dfrac 8 3 } = 1$
$\Rightarrow a ^ { 2 } = \dfrac 8  9 \quad , \quad b ^ { 2 } = \dfrac 8 3$
$\Rightarrow \quad \dfrac { b ^ { 2 } } { a ^ { 2 } } = \dfrac { 8 } { 3 } \times \dfrac { 9 } { 8 } = 3$
$\Rightarrow \quad \dfrac { b } { a } = \sqrt { 3 }$

The angle between the asymptotes 
$\begin{aligned} & = 2 \tan ^ { - 1 } \left( \frac { b } { a } \right) \ = & 2 \tan ^ { - 1 } \left( \sqrt 3 \right) \ = & 2 \cdot 60 \ = & 120 \end{aligned}$

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

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

  1. $x = - 4,y=3$

  2. $x = 4,y=3$

  3. $x =2, y =- 3$

  4. $x =2, y = 3$

Show answer
Correct Option: A
Explanation:

Given : Hyperbola,
$xy-3x+4y+2=0$---------------1
For Asymptotes,
Let the Asymptote's Equation be $y=mx+c$
And then finding $\phi _{n}(m)$ by replacing $y\rightarrow m$ and $x\rightarrow 1$
As $n=2$,
$\phi _{2}(m)=m$
putting $\phi _{2}(m)=0$, we get $m=0$
By taking $m=0$, we will get only one asymptote parallel to X-axis, so let's find them with putting the co-efficients of higher terms to zero.
For Asymptote parallel to X-axis, we put co-efficient of highest degree of x to zero that is here 1, so co-efficient of x$=0$
$\Rightarrow (y-3)=0$------------2(from Equation 1)
For Asymptote parallel to Y-axis, we put co-efficient of highest degree of y to zero which is 1 here, co-efficient of y$=0    (from Equation 1)
$\Rightarrow x+4=0$------------3
The Equation 2 & 3 are asymptotes to Equation 1.
$x+4=0$ & $y-3=0$

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

Show answer
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

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

Show answer
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. $\sqrt 3$

Show answer
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. $\cos ec^2\lambda$

Show answer
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. $18$

Show answer
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 $