Tag: mathematics and statistics

Questions Related to mathematics and statistics

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

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

  1. $(\pm\sqrt3,0)$

  2. $(\pm\sqrt3+1,-2)$

  3. $(\pm1,-2)$

  4. $(0,0)$

Show answer
Correct Option: B
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.

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) $.

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

  1. $65y^2-16x^2=196$

  2. $65y^2-14x^2=196$

  3. $85y^2-16x^2=196$

  4. $85y^2-16x^2=147$

Show answer
Correct Option: C
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 
$85y^2-16x^2=196$

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

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

  1. $24x^2-25y^2=600$

  2. $25x^2-24y^2=600$

  3. $\displaystyle \frac{x^2}{25}-\frac{y^2}{24}=1$

  4. $\displaystyle \frac{x^2}{24}-\frac{y^2}{25}=1$

Show answer
Correct Option: A,C
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.

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

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

  1. $5x \pm 16=0$

  2. $5y \pm 16=0$

  3. $5x \pm 12=0$

  4. $5y \pm 12=0$

Show answer
Correct Option: B
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]

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

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

  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 } $

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Correct Option: A

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

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

  1. $(0, \pm 1)$

  2. $(0, \pm 1, 0)$

  3. $(\pm 5, 0)$

  4. $(0, \pm 5)$

Show answer
Correct Option: D
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]

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-

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

  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 $2$

Show answer
Correct Option: A

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

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

  1. $(-4,1),(6,1)$

  2. $(-11,2),(-6,1)$

  3. $(4,1),(-6,1)$

  4. $(2,1),(1,-6)$

Show answer
Correct Option: C
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 $\left(-6,1\right)$

If foci of $\dfrac {x^2}{a^2}-\dfrac {y^2}{b^2}=1$ coincide with the foci of $\dfrac {x^2}{25}+\dfrac {y^2}{9}=1$ and eccentricity of the hyperbola is 2, then :

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

  1. $a^2+b^2=16$

  2. there is no director circle to the hyperbola

  3. centre of the director circle is $(0, 0)$

  4. length of latus rectum of the hyperbola $=12$

Show answer
Correct Option: A,B,D
Explanation:

$\displaystyle \frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 9 } =1$
which is an ellipse.
For ellipse, $a=5$ , $b=3$ 
$\Rightarrow \displaystyle e=\sqrt {\frac {25-9}{25}}=\frac {4}{5}$
$\therefore ae=4$
Hence, the foci are $(-4, 0)$ and $(4, 0)$.
For the hyperbola,
$ae=4, e=2$
$\Rightarrow a=2$
$b^2=4(4-1)=12$
$\Rightarrow b=\sqrt {12}$
$\Rightarrow a^2+b^2=16$
Since $b^2>a^2$, so there is no director circle to the hyperbola.
Length of latus rectum $\displaystyle=\frac{2b^2}{a}=12$

The centre of the conic section $14x^2-4xy+11y^2-44x-58y+71=0$ is

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

  1. $(2, 3)$

  2. $(2, -3)$

  3. $(-2, 3)$

  4. $(-2, -3)$

Show answer
Correct Option: A
Explanation:

Let $S\equiv 14x^2-4xy+11y^2-44x-58y+71$
To know centre of this hyperbola, $\cfrac{\partial S}{\partial x}=0$ and $\cfrac{\partial S}{\partial y}=0 $
$\Rightarrow \cfrac{\partial S}{\partial x}=28x-4y-44=0\Rightarrow 7x-y-11=0  ..(1)$
and $\cfrac{\partial S}{\partial y}=-4x+22y-58=0\Rightarrow 2x-11y+29=0  ..(2)$
Solving $(1)$ and $(2)$ we get required centre $(2,3)$
Hence, option 'A' is correct.