Tag: hyperbola

Questions Related to hyperbola

Find the equation to the hyperbola, whose eccentricity is $\displaystyle \frac{5}{4}$, focus is $(a, 0)$ and whose directrix is $4x - 3y = a$.

mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

  1. $7y^2\, +\, 24xy\, -\, 12ax\, -\, 3ay\, +\, 15a^2\, =\, 0$

  2. $7y^2\, +\, 24xy\, +\, 12ax\, +\, 3ay\, +\, 15a^2\, =\, 0$

  3. $7y^2\, +\, 24xy\, +\, 24ax\, +\, 6ay\, +\, 15a^2\, =\, 0$

  4. $7y^2\, +\, 24xy\, -\, 24ax\, -\, 6ay\, +\, 15a^2\, =\, 0$

Show answer
Correct Option: D
Explanation:

Using Hyperbola definition, $PS^2=e^2.PM^2$
$\displaystyle (x-a)^2+(y-0)^2=\frac{25}{16}\left| \frac{4x-3y-a}{\sqrt{3^2+4^2}} \right|^2$
$\Rightarrow 16(x^2+y^2-2ax+a^2)=16x^2+9y^2+a^2-24xy+6ya-8ax$
$\Rightarrow 7y^2\, +\, 24xy\, -\, 24ax\, -\, 6ay\, +\, 15a^2\, =\, 0$

If the centre, vertex and focus of a hyperbola be $(0,0), (4, 0)$ and $(6,0)$ respectively, then the equation of the hyperbola is

mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

  1. $4x^2\, -\, 5y^2\, =\, 8$

  2. $4x^2\, -\, 5y^2\, =\, 80$

  3. $5x^2\, -\, 4y^2\, =\, 80$

  4. $5x^2\, -\, 4y^2\, =\, 8$

Show answer
Correct Option: C
Explanation:

Given $a=4, ae=6\Rightarrow e=\cfrac{3}{2}\Rightarrow \cfrac{b^2}{a^2}=e^2-1=\cfrac{9}{4}-1=\cfrac{5}{4}\Rightarrow b^2=20$
Therefore, required hyperbola is, $\cfrac{x^2}{16}-\cfrac{y^2}{20}=1$
$\Rightarrow 5x^2-4y^2=80$
Hence, option 'C' is correct.

The foci of the ellipse $\displaystyle \frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1$ and the hyperbola $\displaystyle \frac { { x }^{ 2 } }{ 144 } -\frac { { y }^{ 2 } }{ 81 } =\frac { 1 }{ 25 } $ coincide. The value of ${ b }^{ 2 }$ is

mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

  1. $9$

  2. $1$

  3. $5$

  4. $7$

Show answer
Correct Option: D
Explanation:

The equation of hyperbola is $\displaystyle \frac { { x }^{ 2 } }{ 144 } -\frac { { y }^{ 2 } }{ 81 } =\frac { 1 }{ 25 } $


Here, $\displaystyle a=\sqrt { \frac { 144 }{ 25 }  } ,b=\sqrt { \frac { 81 }{ 25 }  } ,e=\sqrt { 1+\frac { 81 }{ 144 }  } =\frac { 15 }{ 12 } =\frac { 5 }{ 4 } $
$\therefore$ Foci $=\left( \pm 3,0 \right) $
Also, focus of ellipse $\displaystyle =\left( 3,0 \right) \Rightarrow e=\frac { 3 }{ 4 } $
$\displaystyle \therefore { b }^{ 2 }=16\left( 1-\frac { 9 }{ 16 }  \right) =7$

The hyperbola $\dfrac{x^2}{a^2}\, -\, \dfrac{y^2}{b^2}\, =\, 1\, (a,\, b\, >\, 0)$ passes through the point of intersection of the lines $7x + 13y - 87 = 0$ & $5x - 8y + 7 = 0$ and the latus rectum is $\dfrac{32 \sqrt{2}}5$. The values of $a$ and $b$ are:

mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

  1. $\displaystyle a =\frac{5}{\sqrt{2}},\, b=3$.

  2. $\displaystyle a =\frac{5}{\sqrt{2}},\, b=4$.

  3. $\displaystyle a =\frac{7}{\sqrt{2}},\, b=3$.

  4. None of these

Show answer
Correct Option: B
Explanation:

Point of intersection of lines
$7x + 13y - 87 = 0$ & $5x - 8y + 7 = 0$ is $(5, 4)$.
Also this point lies on the given hyperbola
$\therefore \displaystyle \frac{25}{a^2}\, -\, \frac{16}{b^2}\, =\, 1$ ......(1)
Also latus rectum $\displaystyle LR\, =\, \frac{2b^2}{a}\, =\, \frac{32 \sqrt{2}}{5}$
$\displaystyle \Rightarrow\, b^2\, =\, \frac{16 \sqrt{2}a}{5}$ .....(ii)
From (i) & (ii) $\displaystyle a^2\, =\, \frac{25}{2},\, b^2\, =\, 16$.

For the hyperbola $16x^2\, -\, 9y^2\, +\, 32x\, +\, 36y\,-\, 164\, =\, 0$, find $2(a+b)$.

mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

  1. $8$

  2. $6$

  3. $14$

  4. $12$

Show answer
Correct Option: C
Explanation:

Given hyperbola can be written as


$16x^2+32x+16-9y^2+36y-36=144$

$\displaystyle\, \frac{(x\, +\, 1)^2}{9}\, -\, \frac{(y\, -\, 2)^2}{16}\, =\, 1$

$\Rightarrow a =3, b = 4$

Length of major axis $=2\times 4=8$ and length of minor axis $=2\times 3 = 6$. 

Hence sum is $2(a+b)=2(3+4)=14.$

A hyperbola having the transverse axis of length $\sqrt{2}$ is confocal with $3x^2 + 4y^2 = 12$, then its equation is:

mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

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

  2. $2x^2+2y^2=1$

  3. $x^2+y^2=2$

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

Show answer
Correct Option: A
Explanation:

Given ellipse may be written as, $\displaystyle \frac {x^2}{4}+\frac {y^2}{3}=1$
$\displaystyle \Rightarrow a^2 = 4, b^2 = 3\therefore e = \sqrt{1-\frac{3}{4}}=\frac{1}{2}$
Thus foci of ellipse $(\pm 1, 0)$
Hence foci of the hyperbola is $(\pm 1, 0)$
And semi-major axis $a =\cfrac{1}{\sqrt{2}}$
$\Rightarrow \pm 1=\sqrt {a^2+b^2}\Rightarrow b^2=\frac {1}{2}$
or $b^2=1-a^2=1-\frac {1}{2}=\frac {1}{2}$
The required equation of hyperbola is,
$\displaystyle \frac {x^2}{1/2}-\frac {y^2}{1/2}=1\Rightarrow 2x^2-2y^2=1$

Find the equation to the hyperbola, the distance between whose foci is $16$ and whose eccentricity is $\sqrt{2}$.

mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

  1. $x^2\, -\, y^2\, =\, 32$

  2. $x^2\, -\, y^2\, =\, 18$

  3. $x^2\, -\, y^2\, =\, 64$

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

Show answer
Correct Option: A
Explanation:

Given eccentricity of the hyperbola is $\sqrt{2}$
Hence hyperbola is rectangular $a=b$
Also given distance between focii is 16. $\Rightarrow 2ae = 16\Rightarrow a =4\sqrt{2}$
Hence required hyperbola is given by, $x^2-y^2=a^2=32$

A parabola is drawn with its vertex at $(0,-3)$, the axis of symmetry along the conjugate axis of the hyperbola $\displaystyle \frac { { x }^{ 2 } }{ 49 } -\frac { { y }^{ 2 } }{ 9 } =1$ and passing through the two foci of the hyperbola. The coordinates of the focus of the parabola are :

mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

  1. $\displaystyle \left( 0,\frac { 11 }{ 6 }  \right) $

  2. $\displaystyle \left( 0,-\frac { 11 }{ 6 }  \right) $

  3. $\displaystyle \left( 0,\frac { 11 }{ 12 }  \right) $

  4. $\displaystyle \left( 0,-\frac { 11 }{ 12 }  \right) $

Show answer
Correct Option: A
Explanation:

Equation of hyperbola is $\displaystyle \frac { { x }^{ 2 } }{ 49 } -\frac { { y }^{ 2 } }{ 9 } =1$

Its conjugate axis is y-axis
Also, $\displaystyle e=\sqrt { 1+\frac { { b }^{ 2 } }{ { a }^{ 2 } }  } =\sqrt { 1+\frac { 9 }{ 49 }  } =\frac { \sqrt { 58 }  }{ 7 } $
$\therefore$ Foci of hyperbola is $\left( \pm ae,0 \right) \Rightarrow \left( \pm \sqrt { 58 } ,0 \right) $
Now equation of parabola with vertex at $(0,-3)$ and axis along y-axis is ${ x }^{ 2 }=l\left( y+3 \right) $
It passes through $\left( \pm \sqrt { 58 } ,0 \right) $
$\displaystyle \therefore 58=l\left( 0+3 \right) \Rightarrow l=\frac { 58 }{ 3 } $
$\therefore$ parabola is $\displaystyle { x }^{ 2 }=\frac { 58 }{ 3 } \left( y+3 \right) $
Its focus is $\displaystyle \left( 0,-3+\frac { 58 }{ 4.3 }  \right) \equiv \left( 0,\frac { 11 }{ 6 }  \right) $

Which of the following is true for the hyperbola $9x^2\, -\, 16y^2\, -\, 18x\, +\, 32y\, -\, 151\, =\, 0$?

mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

  1. The length of the transverse axes is $4$

  2. Length of latus rectum is $9$

  3. Equation of directrix is $x\, =\, \displaystyle \frac{21}{5}$ and $x\, =\, - \displaystyle \frac{11}{5}$

  4. None of these

Show answer
Correct Option: C
Explanation:

$9x^2\, -\, 16y^2\, -\, 18x\, +\, 32y\, -\, 151\, =\, 0$
$9(x^2-2x)-16(y^2-2y)=151$
$9(x^2-2x+1)-16(y^2-2y+1)=151-7=144$
$9(x-1)^2-16(y-1)^2=144$
$\cfrac{(x-1)^2}{16}-\cfrac{(y-1)^2}{9}=1$
$\Rightarrow a^2=16, b^2=9$
$\therefore e=\sqrt{1+\dfrac{b^2}{a^2}}=\cfrac{5}{4}$

$\therefore$ The length of the transverse axis is $=2a=8$
Length of latus rectum is $=2\cfrac{b^2}{a}=\cfrac{9}{2}$
Equation of directrix is, $x=1\pm \cfrac{a}{e}=1 \pm \cfrac{16}{5}=\cfrac{21}{5}$ or $-\cfrac{11}{5}$
Hence, option 'C' is correct.

An ellipse intersects the hyperbola $\displaystyle 2x^{2}-2y^{2}=1$ orthogonally at point $P$. The eccentricity of the ellipse is reciprocal to that of the hyperbola. If the axes of the ellipse are along the co-ordinate axes and product of focal distances of $P$ is $x$ then $2x$ is:

mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

  1. $1$

  2. $2$

  3. $3$

  4. $5$

Show answer
Correct Option: C
Explanation:

As ellipse and hyperbola intersect orthogonally 
$\displaystyle \Rightarrow $ their foci are coincident
Now we have $\displaystyle SP+S,P=2a\Rightarrow $ length of major axis ellipse ..........(1)
And $\displaystyle \left | SP-S,P \right |=2a'\Rightarrow $ length of transverse axis of hyperbola ....... (2)
$\displaystyle (1)^{2}-(2)^{2}\Rightarrow 4PS'PS=4(a^{2}-a'^{2})=4\left [ \left ( \frac{a'e}{e} \right )^{2} -a'^{2}\right ]=4\left ( 2-\frac{1}{2} \right )=6$
$\displaystyle \Rightarrow 2x = 3$