Tag: hyperbola

Questions Related to hyperbola

The focal length of the hyperbola $x^2-3y^2-4x-6y-11=0$, is?

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

  1. $4$

  2. $6$

  3. $8$

  4. $10$

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Correct Option: A

Consider the hyperbola $3{x^2} - {y^2} - 24x + 4y - 4 = 0$

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

  1. Its centre is (4, 2)

  2. its centre is (2, 4)

  3. Length of latus rectum= 24

  4. length of latus rectum=12

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Correct Option: B

Find Directrix, foci and eccentricity of the conics:

$\displaystyle x^{2}+2x-y^{2}+5= 0$

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

  1. Directrices $\displaystyle y= \pm \sqrt{2}$

  2. foci $\displaystyle \left ( -1,\pm 2\sqrt{2} \right )$ 

  3. $e= \sqrt{2}$

  4. $e=2$

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Correct Option: A,B,C
Explanation:

$\displaystyle x^{2}+2x-y^{2}+5= 0$
$\displaystyle x^{2}+2x+1-y^{2}= -4$
$\displaystyle (x+1)^2-y^{2}= -4$
$\displaystyle \frac{y^2}{4}-\frac{(x+1)^2}{4}= 1$
Clearly this equation of rectangular hyperbola with $y-$axis as major axis
eccentricity $e = \sqrt{2}$ directrices $:y = \cfrac{a}{e}=\pm \sqrt{2}$ foci $(-1,\pm ae)=(-1,\pm 2\sqrt{2})$

The equation of a hyperbola whose directrix is $2x+y=1$ and focus is at $(1,2)$ with $e=\sqrt{3}$ is :

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

  1. $7x^2+12xy+2y^2-2x+14y-22=0$

  2. $7x^2+12xy-2y^2-2x+14y-22=0$

  3. $7x^2+12xy-2y^2-2x-14y-22=0$

  4. $7x^2+12xy+2y^2+2x+14y-22=0$

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Correct Option: A

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

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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 $=\cfrac{2b^2}{a}=\cfrac{8}{3}$