Tag: parametric equation of the hyperbola

Questions Related to parametric equation of the hyperbola

Let the eccentricity of the hyperbola $  \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1  $ be reciprocal to that of the ellipse $  x^{2}+4 y^{2}=4 .  $ If thehyperbola passes through a focus of the ellipse, then __________________.

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

  1. (A) the equation of the hyperbola is $ \frac{x^{2}}{3}-\frac{y^{2}}{2}=1 $

  2. (B) a focus of the hyperbola is $ (2,0) $

  3. (C) the eccentricity of the hyperbola is $ \sqrt{\frac{5}{3}} $

  4. (D) the equation of the hyperbola is $ x^{2}-3 y^{2}=3 $

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Correct Option: D

The eccentricity of the hyperbola $\displaystyle \dfrac { \sqrt { 1999 }  }{ 3 } \left( { x }^{ 2 }-{ y }^{ 2 } \right) =1$ is:

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

  1. $\sqrt { 2 } $

  2. $2$

  3. $2\sqrt { 2 } $

  4. $\sqrt { 3 } $

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Correct Option: A
Explanation:

Equation of hyperbola is $\displaystyle \frac { { x }^{ 2 } }{ 3/\sqrt { 1999 }  } -\frac { { y }^{ 2 } }{ 3/\sqrt { 1999 }  } =1$

Here $\displaystyle { a }^{ 2 }={ b }^{ 2 }=\frac { 3 }{ \sqrt { 1999 }  } $
$\therefore$ Eccentricity $\displaystyle e=\sqrt { 1+\frac { { b }^{ 2 } }{ { a }^{ 2 } }  } =\sqrt { 1+1 } =\sqrt { 2 } $

The equation of the hyperbola whose foci are $(6,5), (-4, 5)$ and eccentricity $\dfrac54$ is:

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

  1. $\displaystyle \frac{(x\, -\, 1)^2}{16}\, -\, \frac{(y\, -\, 5)^2}{9}\, =\, 1$

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

  3. $\displaystyle \frac{(x\, -\, 1)^2}{16}\, -\, \frac{(y\, -\, 5)^2}{9}\, =\, -1$

  4. $\displaystyle \frac{(x\, -\, 1)^2}{4}\, -\, \frac{(y\, -\, 5)^2}{9}\, =\, 1$

Show answer
Correct Option: A
Explanation:

Centre of the ellipse $=$ mid point of foci $=(1,5)$

Distance between foci $=\sqrt{(6-(-4))^2+(5-5^2)}$ $= 10$

$2ae=6-(-4)=10\Rightarrow a=5/e=4$

$\Rightarrow b^2 = a^2(e^2-1) = 9$

Hence required hyperbola is $\cfrac{(x-1)^2}{16}-\cfrac{(y-5)^2}{9}=1$

The eccentricity of the hyperbola $4x^2\, -\, 9y^2\, -\, 8x\, =\, 32$ is

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

  1. $\displaystyle \frac{\sqrt{5}}{3}$

  2. $\displaystyle \frac{\sqrt{13}}{3}$

  3. $\displaystyle \frac{4}{3}$

  4. $\displaystyle \frac{3}{2}$

Show answer
Correct Option: B
Explanation:

$4x^2\, -\, 9y^2\, -\, 8x\, =\, 32$
$\Rightarrow 4(x^2-2x)-9y^2=32$
$\Rightarrow 4(x^2-2x+1)-9y^2=32+4=36$
$\Rightarrow \cfrac{(x-1)^2}{9}-\cfrac{y^2}{4}=1$
$\Rightarrow a^2=9, b^2=4$
$\therefore e=\sqrt{1+\cfrac{b^2}{a^2}}=\cfrac{\sqrt{13}}{3} $
Hence, option 'B' is correct.

The vertices of a hyperbola are at $(0, 0)$ and $(10,0)$ and one of its focus is at $(18,0)$. The possible equation of the hyperbola is

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

  1. $\displaystyle \frac{x^2}{25}\, -\, \frac{y^2}{144}\, =\, 1$

  2. $\displaystyle \frac{(x\, -\, 5)^2}{25}\, -\, \frac{y^2}{144}\, =\, 1$

  3. $\displaystyle \frac{x^2}{25}\, -\, \frac{(y\, -\, 5)^2}{144}\, =\, 1$

  4. $\displaystyle \frac{(x\, -\, 5)^2}{25}\, -\, \frac{(y\, -\, 5)^2}{144}\, =\, 1$

Show answer
Correct Option: B
Explanation:

Centre of hyperbola is $(5, 0)$, so equation is
$\displaystyle \frac{(x\, -\, 5)^2}{a^2}\, -\, \frac{y^2}{b^2}\, =\, 1$
$a\, =\, 5,\, ae\, -\, a\, =\, 8\, \Rightarrow\, e\, =\, \displaystyle \frac{13}{5}$

$b^2\, =a^2(e^2-1)=\, 144$
So required equation is,  $\displaystyle \frac{(x\, -\, 5)^2}{25}\, -\, \frac{y^2}{144}\, =\, 1$
Hence, option 'B' is correct.

In the hyperbola $4x^2\, -\, 9y^2\, =\, 36$, find lengths of the axes, the co-ordinates of the foci, the eccentricity, and the latus rectum.

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

  1. $6, 4;\, (\pm\, \sqrt{13},\, 0);\, \dfrac{\sqrt{13}}3;\, \dfrac8 3$

  2. $9, 4;\, (\pm\, \sqrt{8},\, 0);\, \dfrac{\sqrt{8}}3;\, \dfrac83$

  3. $9, 4;\, (\pm\, \sqrt{13},\, 0);\, \dfrac{\sqrt{13}}3;\, \dfrac83$

  4. $6, 4;\, (\pm\, \sqrt{8},\, 0);\, \dfrac{\sqrt{8}}3;\, \dfrac83$

Show answer
Correct Option: A
Explanation:

Given hyperbolas may be written as,  $\displaystyle \frac{x^2}{9}-\frac{y^2}{4}=1$
$\Rightarrow a^2 =9, b^2=4$
$\therefore$ Eccentricity is, $\displaystyle e= \sqrt{1+\frac{b^2}{a^2}}=\frac{\sqrt{13}}{3}$
Thus, length of axes are $2a$ and $2b \Rightarrow 6 $ and $4$
Focus $\equiv (\pm ae, 0) =(\pm \sqrt{13},0)$
And length of latus rectum $=\cfrac{2b^2}{a}=\cfrac{8}{3}$

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) $\displaystyle a^2\, =\, \frac{25}{2},\, b^2\, =\, 16$.