Tag: hyperbola

Questions Related to hyperbola

Find the locus of the point of intersection of the lines $\sqrt{3}x-y-4\sqrt{3} \lambda=0$ and $\sqrt{3}\lambda x+\lambda y-4\sqrt{3}=0$ for different values of $\lambda$.

mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

  1. $4x^2-y^2=48$

  2. $x^2-4y^2=48$

  3. $3x^2-y^2=48$

  4. $y^2-3x^2=48$

Show answer
Correct Option: C
Explanation:

Let $(h,k)$ be the point of intersection of the given lines. Then,


$\sqrt{3}h-k-4\sqrt{3} \lambda=0$ and $\sqrt{3} \lambda h+\lambda k-4\sqrt{3}=0$


$\sqrt{3}h-k=4\sqrt{3}\lambda$ and $\lambda(\sqrt{3}h+k)=4\sqrt{3}$

$(\sqrt{3}h-k)\lambda(\sqrt{3}h+k)=(4\sqrt{3}\lambda)(4\sqrt{3})$

$3h^2-k^2=48$

Hence, the locus of (h,k) is $3x^2-y^2=48$.

The AFC Curve passes through the Origin statement is -

mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

  1. True

  2. False

  3. Partially True

  4. Nothing can be said

Show answer
Correct Option: B
Explanation:

AFC curve is downward sloping because fixed costs are distributed over large volume when quantity produced increases.

So, it doesn't pass through origin.

$Center\quad of\quad the\quad hyperbola\quad { x }^{ 2 }+4{ y }^{ 2 }+6xy+8x-2y+7=0\quad is\quad $

mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

  1. $(1,1)$

  2. $(0,2)$

  3. $(2,0)$

  4. $None\quad of\quad these$

Show answer
Correct Option: D
Explanation:
Given Hyperbola: $x^{2}+4y^{2}+6xy+8x-2y+7=0$
Centre: Point of intersection of asymptotes of hyperbola.
Now finding Asymptotes of given equation, taking equation of asymptote $y=mx+c$ by replacing $x\rightarrow 1, y\rightarrow m$ in $\phi _n(m)$
When $n=2$, $\phi _{2}(m)=1+4m^{2}+6m$
$\phi _{1}(m)=8-2m$
$\phi _{0}(m)=7$
$\phi _{2}^{1}(m)=8m+6$
$\phi _{1}(m)=8-2m$

Putting $\phi _{2}(m)=0$
$\Rightarrow 1+6m+4m^{2}=0$
$\Rightarrow  m=\cfrac{-6\pm \sqrt {36-16}}{2(4)}$
$\Rightarrow m=\cfrac{-6\pm\sqrt 20}{2(4)}$
$\Rightarrow m=\cfrac{-3\pm \sqrt 5}{4}$
So, m$=\cfrac{-3+\sqrt 5}{4}, \cfrac{-3-\sqrt 5}{4}$
Value of $c$, when $m$ is different
$c= \cfrac{-\phi _{1}(m)}{\phi _{2}^{'}(m)}=\cfrac{8-2m}{8m+6}$

For $m=\cfrac{-3+\sqrt 5}{4}, c=\cfrac{8-2(\cfrac{-3+\sqrt 5}{4})}{8\cfrac{-3+\sqrt 5}{4})+6}=\cfrac{3-\sqrt 5+16}{-6+2\sqrt 5+6}=\cfrac{19\sqrt 5-5}{10}$

For $m=\cfrac{-3-\sqrt 5}{4}, c=\cfrac{8-2(\cfrac{-3-\sqrt 5}{4})}{8\cfrac{-3-\sqrt 5}{4})+6}=\cfrac{19+15}{-2 \sqrt 5}=\cfrac{-(19\sqrt 5+5)}{10}$
Equation of Asymptotes : $y=\cfrac{-3+\sqrt 5}{4}x +\cfrac{19\sqrt 5-5}{10}$ & $y=\cfrac{-3-\sqrt 5}{4}x-(\cfrac{5+19\sqrt 5}{10})$
On solving them for x & y, putting LHS-RHS
$\Rightarrow 0=\cfrac{\sqrt 5}{2}x+\cfrac{19\sqrt 5}{5} \Rightarrow x=\cfrac{-38}{5}$
Now putting values of x in Asymptotes equation, we get
$y=\cfrac{-3-\sqrt 5(-19)}{10}+\cfrac{+5+19\sqrt 5}{10}=\cfrac{+52}{10}$
Centre$(\cfrac{-38}{5}, \cfrac{+52}{10})$.

Circles are drawn on chords of the rectangular hyperbola $xy=4$ parallel to the line $y=x$ as diameters.All such circles pass through two fixed points whose coordinates are 

mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

  1. $\left(2,2\right)$

  2. $\left(2,-2\right)$

  3. $\left(-2,2\right)$

  4. $\left(-2,-2\right)$
Show answer
Correct Option: A,D
Explanation:
Given:Rectangular hyperbola $xy=4={c}^{2}$
$\Rightarrow\,{c}^{2}=4$
$\Rightarrow\,c=2$
Let $P$ and $Q$ be the end points on the Rectangular hyperbola where $P\left(2{t} _{1},\dfrac{2}{{t} _{1}}\right)$ and $Q\left(2{t} _{2},\dfrac{2}{{t} _{2}}\right)$
Using the end points of diameter the equation of the circle

$C:\left(x-2{t} _{1}\right)\left(x-2{t} _{2}\right)+\left(y-\dfrac{2}{{t} _{1}}\right)\left(y-\dfrac{2}{{t} _{2}}\right)=1$         ..........$(1)$

Now,Slope of the $PQ=\dfrac{\dfrac{2}{{t} _{2}}-\dfrac{2}{{t} _{1}}}{2{t} _{2}-2{t} _{1}}$

$=\dfrac{2\left(\dfrac{1}{{t} _{2}}-\dfrac{1}{{t} _{1}}\right)}{2\left({t} _{2}-{t} _{1}\right)}$

$=\dfrac{\dfrac{{t} _{1}-{t} _{2}}{{t} _{1}{t} _{2}}}{\left({t} _{2}-{t} _{1}\right)}$

$=\dfrac{\dfrac{{t} _{1}-{t} _{2}}{{t} _{1}{t} _{2}}}{\left({t} _{2}-{t} _{1}\right)}$

$=\dfrac{-1}{{t} _{1}{t} _{2}}$

Hence $PQ$ is the diameter for circle and it is parallel to the line $y=x$

Slope of $PQ=$Slope of the line $y=x$

$\Rightarrow\,\dfrac{-1}{{t} _{1}{t} _{2}}=1$

$\Rightarrow\,{t} _{1}{t} _{2}=-1$

$(1)\Rightarrow\,\left(x-2{t} _{1}\right)\left(x-2{t} _{2}\right)+\left(y-\dfrac{2}{{t} _{1}}\right)\left(y-\dfrac{2}{{t} _{2}}\right)=1$ 

$\Rightarrow\,x\left(x-2{t} _{2}\right)-2{t} _{1}\left(x-2{t} _{2}\right)+y\left(y-\dfrac{2}{{t} _{2}}\right)-\dfrac{2}{{t} _{1}}\left(y-\dfrac{2}{{t} _{2}}\right)=1$
 
$\Rightarrow\,{x}^{2}-2x{t} _{2}-2x{t} _{1}+4{t} _{1}{t} _{2}+{y}^{2}-\dfrac{2y}{{t} _{2}}-\dfrac{2y}{{t} _{1}}+\dfrac{4}{{t} _{1}{t} _{2}}=1$

$\Rightarrow\,{x}^{2}-2x{t} _{2}-2x{t} _{1}+4\times -1+{y}^{2}-\dfrac{2y}{{t} _{2}}-\dfrac{2y}{{t} _{1}}+\dfrac{4}{\times -1}=1$ using ${t} _{1}{t} _{2}=-1$

$\Rightarrow\,{x}^{2}+{y}^{2}-2x\left({t} _{2}+{t} _{1}\right)-4-2y\left(\dfrac{1}{{t} _{2}}+\dfrac{1}{{t} _{1}}\right)-4=1$

$\Rightarrow\,{x}^{2}+{y}^{2}-8-2x\left({t} _{2}+{t} _{1}\right)-2y\left(\dfrac{{t} _{1}+{t} _{2}}{{t} _{1}{t} _{2}}\right)=1$

$\Rightarrow\,{x}^{2}+{y}^{2}-8-2x\left({t} _{2}+{t} _{1}\right)-2y\left(\dfrac{{t} _{1}+{t} _{2}}{-1}\right)=1$ using ${t} _{1}{t} _{2}=-1$

$\Rightarrow\,{x}^{2}+{y}^{2}-8-2x\left({t} _{2}+{t} _{1}\right)+2y\left({t} _{1}+{t} _{2}\right)=1$ 

$\Rightarrow\,{x}^{2}+{y}^{2}-8+\left(2y-2x\right)\left({t} _{2}+{t} _{1}\right)=1$ is of the form $C+\lambda\,L$ 

where $C={x}^{2}+{y}^{2}-8=0$ is the equation of a circle.
and $L=2y-2x=0$ is the equation of a line.
$\Rightarrow\,y-x=0$ or $x=y$

Substituting $x=y$ in the equation ${x}^{2}+{y}^{2}-8=0$ we get
$\Rightarrow\,2{x}^{2}-8=0$

$\Rightarrow\,2\left({x}^{2}-4\right)=0$

$\Rightarrow\,\left(x-2\right)\left(x+2\right)=0$

$\therefore\,x=2,-2$

$\Rightarrow\,y=2,-2$ since $x=y$

Hence the coordinates of the fixed points are $\left(2,2\right)$ and $\left(-2,-2\right)$ 

Centre of the hyperbola ${x^2} + 4{y^2} + 6xy + 8x - 2y + 7 = 0$ is 

mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

  1. $(1,1)$

  2. $(0,2)$

  3. $(2,0)$

  4. None of these

Show answer
Correct Option: D
Explanation:

Consider equation of a hyperbola as $F=ax^2+2by+cy^2+2dx+2ey+f=0$

The centre of this hyperbola can be found by applying the concepts of partial differentiation
We first find $\dfrac{\delta F}{\delta x} $ and $\dfrac{\delta F}{\delta y}$
We then solve $\dfrac{\delta F}{\delta x} =0$ and $\dfrac{\delta F}{\delta y}=0 $ to find $x,y$ which is the centre of the hyperbola .

Given that,
$F=x^2+4y^2+6xy+8x-2y+7$

$\Rightarrow \dfrac{\delta F}{\delta x}=2x+0+6y+8+0+0$

$\Rightarrow \dfrac{\delta F}{\delta x}=2x+6y+8$         ...$(1)$

$\Rightarrow \dfrac{\delta F}{\delta y}=0+8y+6x+0-2+0$

$\Rightarrow \dfrac{\delta F}{\delta y}=6x+8y-2$      ....$(2)$
 
$(1) \rightarrow 2x+6y+8=0$                    

$(2) \rightarrow 6x+8y-2=0$

Solving $(1), (2)$ we get,

$\Rightarrow x=\dfrac{19}{5}, y=\dfrac{-13}{5}$

Therefore the centre of hyperbola is $(\dfrac{19}{5},\dfrac{-13}{5})$



The eccentricity of the hyperbola whose latus-return is $8$ and length of the conjugate axis is equal to half the distance between the foci, is

mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

  1. $\dfrac43$

  2. $\dfrac4{\surd 3}$

  3. $\dfrac2{\surd 3}$

  4. $None\ of\ these$

Show answer
Correct Option: C
Explanation:

Given that the length of the latus rectum is $8$ and length of the conjugate axis is equal to half the distance between the foci.


$\Rightarrow \dfrac{2b^2}{a}=8$ and $2b=\dfrac{1}{2}(2ae)$

$\therefore \dfrac{2}{a}\left(\dfrac{ae}{2}\right)^2=8$

$\Rightarrow ae^2=16$ ...(1)

We have $\dfrac{2b^2}{a}=8$

$\Rightarrow b^2=4a$

$\Rightarrow a^2(e^2-1)=4a$

$\Rightarrow ae^2-a=4$

$\Rightarrow 16-a=4$ (by (1))

$\Rightarrow a=12$

Substitute $a=12$ in (1)

$\Rightarrow 12e^2=16$

$\Rightarrow e^2=\dfrac{4}{3}$

$\therefore e=\dfrac{2}{\sqrt{3}}$

From any point on the hyperbola $\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ tangents are drawn to the hyperbola $\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = 2$. The area cut-off by the chord of contact on the asymptotes is equal to

mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

  1. $\displaystyle \frac{ab}{2}$

  2. ab

  3. 2 ab

  4. 4 ab

Show answer
Correct Option: D
Explanation:

Let $P\left( { x } _{ 1 },{ y } _{ 1 } \right) $ be a point on the hyperbola $\cfrac { { x }^{ 2 } }{ { a }^{ 2 } } -\cfrac { { y }^{ 2 }
}{ { b }^{ 2 } } =1$. Then,
$\cfrac { { { x } _{ 1 } }^{ 2 } }{ { a }^{ 2 } } -\cfrac { { { y } _{ 1 } }^{ 2 } }{ { b }^{ 2 } } =1$
The chord of contact of tangents from $P$ to the hyperbola $\cfrac { { x

}^{ 2 } }{ { a }^{ 2 } } -\cfrac { { y }^{ 2 } }{ { b }^{ 2 } } =2$ is
$\cfrac

{ { x }{ x } _{ 1 } }{ { a }^{ 2 } } -\cfrac { { y }{ y } _{ 1 } }{ { b

}^{ 2 } } =2\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (i)$

The equations of the asymptotes are
$\cfrac { x }{ a } -\cfrac { y }{ b } =0$ and $\cfrac { x }{ a } +\cfrac { y }{ b } =0$

The points of intersection of $(i)$ with the two asymptotes are given by
${

x } _{ 1 }=\cfrac { 2a }{ \cfrac { { x } _{ 1 } }{ a } -\cfrac { { \quad y

} _{ 1 } }{ b }  } \quad ,{ \quad y } _{ 1 }=\cfrac { 2b }{ \cfrac { { x

} _{ 1 } }{ a } -\cfrac { { \quad y } _{ 1 } }{ b }  } $
${ x } _{ 2

}=\cfrac { 2a }{ \cfrac { { x } _{ 1 } }{ a } -\cfrac { { \quad y } _{ 1 }

}{ b }  } \quad ,{ \quad y } _{ 2 }=\cfrac {- 2b }{ \cfrac { { x } _{ 1 }

}{ a } -\cfrac { { \quad y } _{ 1 } }{ b }  } $
$\quad \therefore $ Area of the triangle
$\cfrac

{ 1 }{ 2 } \left( { x } _{ 1 }{ y } _{ 2 }-{ x } _{ 2 }{ y } _{ 1 } \right)

=\cfrac { 1 }{ 2 } \left( \cfrac { 4ab\times 2 }{ \cfrac { { { x } _{ 1 }

}^{ 2 } }{ { a }^{ 2 } } +\cfrac { { { y } _{ 1 } }^{ 2 } }{ { b }^{ 2 }

}  }  \right) =4ab$

Let $a, b$ be non-zero real numbers. The equation $\displaystyle \left ( ax^{2}+by^{2}+c \right )\left ( x^{2}-5xy+6y^{2} \right )$ represents

mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

  1. four straight lines, when $\displaystyle c=0$ and $a, b$ are of the same sign

  2. two straight lines and a circle, when $\displaystyle a=b$ and $c$ is of sign opposite to that of $a$

  3. two straight lines and a hyperbola, when $a$ and $b$ are of the same sign and $c$ is of sign opposite to that of $a$

  4. a circle and an ellipse, when $a$ and $b$ are of the same sign

Show answer
Correct Option: B
Explanation:

From given expression, $\displaystyle { x }^{ 2 }-5xy+6{ y }^{ 2 }=0$ are pair of straight lines $\displaystyle y=\frac { x }{ 2 } $ and $ y=\dfrac { x }{ 3 } $
Now, $\displaystyle a{ x }^{ 2 }+b{ y }^{ 2 }+c=0$ will be cirlce of radius $\displaystyle \sqrt { -\frac { c }{ a }  } $.

If $\displaystyle a=b$ and $\displaystyle c$ is of opposite sign of $\displaystyle a  $ and $  b$, then $\displaystyle { x }^{ 2 }+{ y }^{ 2 }=-\frac { c }{ a } $

If a hyperbola passes through the foci of the ellipse $\displaystyle \frac {x^2}{25} + \frac {y^2}{16} = 1$ and its traverse and conjugate axis coincide with major and minor axes of the ellipse, and product of the eccentricities is 1, then:

mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

  1. Equations of the hyperbola is $\displaystyle \frac {x^2}{9} - \frac {y^2}{16} = 1$

  2. Equations of the hyperbola is $\displaystyle \frac {x^2}{9} - \frac {y^2}{25} = 1$

  3. Focus of the hyperbola is $\displaystyle (5, 0)$

  4. Focus of the hyperbola is $\displaystyle (5 \sqrt 3, 0)$

Show answer
Correct Option: A,C
Explanation:

Given ellipse is, $\displaystyle \frac {x^2}{25} + \frac {y^2}{16} = 1$
Eccentricity of the ellipse is, $\displaystyle e _e  =\sqrt{1-\frac{b^2}{a^2}}=\frac{3}{5}$
So the foci of the ellipse is, $(\pm ae _e,0)=(\pm 3 , 0)$
Let eccentricity of the required  hyperbola is $e _h$ and semi major and minor axes are $a$ and $b$, so the equation of hyperbola is, $\displaystyle \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
Given hyperbola passes trough $(\pm 3,0)\Rightarrow \displaystyle \frac{9}{a^2}-\frac{0}{b^2}=1\Rightarrow a^2 = 9$
Also given that $\displaystyle  e _e\times e _h = 1$ $\Rightarrow e _h=\cfrac{5}{3}$ $\Rightarrow$ $b^2 =a^2(e _h^2-1)=16 $
Hence required hyperbola is, $\displaystyle \frac{x^2}{9}-\frac{y^2}{16}=1$
And foci of the hyperbola is, $(\pm 5,0)$

The equation ${x}^{2}+9=2{y}^{2}$ is an example of which of the following curves?

mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

  1. hyperbola

  2. circle

  3. ellipse

  4. parabola

  5. line

Show answer
Correct Option: A
Explanation:

Given, ${x}^{2}+9=2{y}^{2}$ 

$\Rightarrow 2{y}^{2}-{x}^{2}=9$
$\Rightarrow \dfrac { { y }^{ 2 } }{ 9/2 } -\dfrac { { x }^{ 2 } }{ 9 } =1$
It is in the form of $\dfrac { { y }^{ 2 } }{ { a }^{ 2 } } -\dfrac { { x }^{ 2 } }{ { b }^{ 2 } } =1$ which is the equation of hyperbola.
Therefore, the given equation is a equation of hyperbola.