Tag: parametric equation of the hyperbola

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Questions Related to parametric equation of the hyperbola

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

The centre of the hyperbola $\dfrac {x^{2} + 4x + 4}{25} - \dfrac {y^{2} - 6x + 9}{16} = 1$ is: 

  1. $(-4, -9)$
  2. $(-2, 3)$
  3. $(2, -3)$
  4. $(5, 4)$
  5. $(25, 16)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
  • the equation of hyperbola is $\dfrac { { x }^{ 2 }+4x+4 }{ 25 } -\dfrac { { y }^{ 2 }-6x+9 }{ 16 } =1$
  • $\dfrac { { (x+2) }^{ 2 } }{ 25 } -\dfrac { { (y-3) }^{ 2 } }{ 16 } =1$
  • Therefore the center is $(-2,3)$
Multiple choice mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

For hyperbola  $-\dfrac{(x-1)^2}{3}+\dfrac{(y+2)^2}{16}=1$ distance between directrices is ?

  1. $\dfrac{2}{\sqrt{19}}$
  2. $\dfrac{3}{\sqrt{19}}$
  3. $\dfrac{4}{\sqrt{19}}$
  4. $\dfrac{32}{\sqrt{19}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Comparing the equation of given hyperbola with the standard equation
$\dfrac{(y-k)^2}{a^2}-\dfrac{(x-h)^2}{b^2}=1$
$h=1,k=-2,a^2=16,b^2=3$

$e=\sqrt{1+\dfrac{b^2}{a^2}}=\dfrac{\sqrt{19}}{4}$

Distance between the directrices $=\dfrac{2a}{e}=\dfrac{32}{\sqrt{19}}$

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

For hyperbola  $-\dfrac{(x-1)^2}{3}+\dfrac{(y+2)^2}{16}=1$ vertices are

  1. $(\pm\sqrt3,0)$
  2. $(\pm\sqrt3+1,-2)$
  3. $(\pm1,-2)$
  4. $(0,0)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given, hyperbola is conjugate hyperbola of $\dfrac { { (x-1) }^{ 2 } }{ 3 } -\dfrac { { (y+2) }^{ 2 } }{ 16 } =-1$

So the vertices of given hyperbola are
${ (x-1) }^{ 2 }=3,{ (y+2) }^{ 2 }=0\ \Rightarrow x-1=\pm \sqrt { 3 } ,y+2=0\ \Rightarrow x=1\pm \sqrt { 3 } ,y=-2\ \Rightarrow \left( 1\pm \sqrt { 3 } ,-2 \right) $
So, option B is correct.

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

Find the equation to the hyperbola, referred to its axes as axes of coordinates, whose transverse axis is $7$ and which passes through the point $\left( 3,-2 \right) $.

  1. $65y^2-16x^2=196$
  2. $65y^2-14x^2=196$
  3. $85y^2-16x^2=196$
  4. $85y^2-16x^2=147$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

General equation of hyperbola is $\dfrac{y^2}{b^2}-\dfrac{x^2}{a^2}=1$

Length of transverse axis is $2a$.
So, $2a=7$
$\Rightarrow a=\dfrac{7}{2}$
Equation becomes,
$\dfrac{y^2}{b^2}-\dfrac{4x^2}{49}=1$
It passes through $(3,-2)$, so it should satisfy the parabola,
$\dfrac{4}{b^2}-\dfrac{36}{49}=1$
On solving, we get 
$85y^2-16x^2=196$

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

Equation of the hyperbola with vertices at $(\pm 5, 0)$ and foci at $(\pm 7, 0)$ is

  1. $24x^2-25y^2=600$
  2. $25x^2-24y^2=600$
  3. $\displaystyle \frac{x^2}{25}-\frac{y^2}{24}=1$
  4. $\displaystyle \frac{x^2}{24}-\frac{y^2}{25}=1$
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

$Vertices(\pm 5,0)\quad Foci(\pm 7,0)$ 

$a=\pm 5$ and $ae=\pm 7$ 
And $e=\dfrac { 7 }{ 5 } $ 
We know $e=\sqrt { 1+\frac { { b }^{ 2 } }{ { a }^{ 2 } }  }$ 
On squaring both sides we get:
${ e }^{ 2 }=\quad 1+\frac { { b }^{ 2 } }{ { a }^{ 2 } }$ 
Or $\dfrac { 49 }{ 25 } =1+\frac { { b }^{ 2 } }{ 25 }$ 
Or $\dfrac { { b }^{ 2 } }{ 25 } =\frac { 24 }{ 25 } $
${ b }^{ 2 }=24$ 
The equation of hyperbola is 
$\dfrac { { x }^{ 2 } }{ 25 } -\dfrac { y^{ 2 } }{ 24 } =1$ 
$24{ x }^{ 2 }-25y^{ 2 }=600$

Hence, Option [A] and [C] are correct.

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

The equation of a hyperbola is given in its standard form as $16x^2-9y^2=144$.Equations of directrices is

  1. $5x \pm 16=0$
  2. $5y \pm 16=0$
  3. $5x \pm 12=0$
  4. $5y \pm 12=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$Given\quad :\quad 16{ x }^{ 2 }−9{ y }^{ 2 }=144\quad \quad \quad \ Or,\quad \frac { { x }^{ 2 } }{ 9 } -\frac { { y }^{ 2 } }{ 16 } =1\ We\quad know,\ be=\sqrt { { a }^{ 2 }+{ b }^{ 2 } } \quad \quad \quad \quad \quad \because (b>a)\ Or,\quad be=\sqrt { 9+16 } \ Or,\quad be=\pm 5\ Or,\quad \frac { b }{ e } \quad =\frac { 16 }{ \pm 5 } \ We\quad know\quad equation\quad of\quad directrix\quad is\quad y=\frac { b }{ e } \ \therefore \quad 5y\pm 16=0$


Option [B]

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

Equation of the transverse and conjugate axis of a hyperbola are respectively $x+2y-3=0$, $2x-y+4=0$ and their respectively lengths are $\sqrt {2}$ and $\cfrac { 2 }{ \sqrt { 3 }  } $ then answer the following 
Equation of one of the directrix is

  1. $2x-y+4+\sqrt { \cfrac { 3 }{ 2 } } =0\quad $
  2. $x+2y+4-\sqrt { \cfrac { 2 }{ 3 } } =0$
  3. $2x-y=\sqrt { \cfrac { 3 }{ 2 } } $
  4. $2x-y+4+\sqrt { \cfrac { 3 }{ 2 } } =\sqrt { 3 } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The directrix of a hyperbola is given by the line parallel to the conjugate axis at a distance a/e from the center. Given the axes and lengths, one can determine the specific equation for the directrix.

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

The equation of a hyperbola is given in its standard form as $16x^2-9y^2=144$.Coordinates of foci is

  1. $(0, \pm 1)$
  2. $(0, \pm 1, 0)$
  3. $(\pm 5, 0)$
  4. $(0, \pm 5)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$Given\quad :\quad 16{ x }^{ 2 }−9{ y }^{ 2 }=144\quad \quad \quad \ Or,\quad \frac { { x }^{ 2 } }{ 9 } -\frac { { y }^{ 2 } }{ 16 } =1\ We\quad know,\ be=\sqrt { { a }^{ 2 }+{ b }^{ 2 } } \quad \quad \quad \quad \quad \because (b>a)\ Or,\quad be=\sqrt { 9+16 } \ Or,\quad be=\pm 5\ \therefore \quad Focii\quad is\quad (0,\pm 5)\quad $


Option [D]

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

Hyperbola $\dfrac{{x}^{2}}{{a}^{2}}-\dfrac{{y}^{2}}{3}=1$ of eccentricity $e$ is confocal with the ellipse $\dfrac{{x}^{2}}{8}+\dfrac{{y}^{2}}{4}=1$. Let $A$, $B$, $C$ & $D$ are points of intersection of hyperbola & ellipse, then-

  1. $e=\dfrac{5}{2}$
  2. $e=2$
  3. $A$, $B$, $C$, $D$ are concyclic points
  4. Number of common tangents of hyperbola & ellipse is $2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

The foci of hyperbola $9x^2-16y^2+18x+32y=151$ are 

  1. $(-4,1),(6,1)$
  2. $(-11,2),(-6,1)$
  3. $(4,1),(-6,1)$
  4. $(2,1),(1,-6)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$9{x}^{2}−16{y}^{2}+18x+32y=151$
$\left(9{x}^{2}+18x\right)-\left(16{y}^{2}-32y\right)=151$
$\Rightarrow \left({\left(3x\right)}^{2}+2\times 3x\times 3+{3}^{2}-{3}^{2}\right)-\left({\left(4y\right)}^{2}-2\times 4y\times 4+{4}^{2}-{4}^{2}\right)=151$ by completing the square method
$\Rightarrow {\left(3x+3\right)}^{2}-9-{\left(4y-4\right)}^{2}+16=151$
$\Rightarrow {\left(3x+3\right)}^{2}-{\left(4y-4\right)}^{2}=151-7$
$\Rightarrow {\left(3x+3\right)}^{2}-{\left(4y-4\right)}^{2}=144$
$\Rightarrow 9{\left(x+1\right)}^{2}-16{\left(y-1\right)}^{2}=144$
$\Rightarrow \dfrac{9{\left(x+1\right)}^{2}}{144}-\dfrac{16{\left(y-1\right)}^{2}}{144}=1$ by dividing both sides by $144$
$\Rightarrow \dfrac{{\left(x+1\right)}^{2}}{16}-\dfrac{{\left(y-1\right)}^{2}}{9}=1$ is the equation of the horizontal hyperbola.
center$=\left(-1,1\right)$
We have $a=4$ and $b=3$
${c}^{2}={a}^{2}+{b}^{2}={4}^{2}+{3}^{2}=16+9=25$
$\therefore c=\sqrt{25}=\pm 5$
Foci$=\left(-1\pm 5, 1\right)$
$\therefore$Foci$=\left(-1+5,1\right)$ and $\left(-1-5,1\right)$
Hence Foci$=\left(4,1\right)$ and $\left(-6,1\right)$