Tag: equations of hyperbola

Questions Related to equations of hyperbola

An ellipse and a hyperbola have the same principle axes. From a point on the ellipse, tangents are drawn to the hyperbola . then  the chord contact of these tangents touches the ellipse.

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

  1. True

  2. False

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Correct Option: A

The eccentricity of the conic represented by$2{x}^{2}+5xy+2{y}^{2}+11x-7y-4=0$ is

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

  1. $\dfrac {\sqrt {10}}{3}$

  2. $\dfrac {\sqrt {10}}{4}$

  3. $\dfrac {5}{4}$

  4. $\dfrac {3}{5}$

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Correct Option: A

The equation $\dfrac{x^{2}}{29 -p} + \dfrac{y^{2}}{4 -p} =1(p\neq4, 29)$ represents - 

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

  1. an ellipse if $p$ is any constant greater than $4$

  2. hyperbola if $p$ is any constant between $4$ and $29$.

  3. a rectanglar hyperbola is $p$ is any constant greater than $29$.

  4. no real curve is $p$ is less than $29$.

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Correct Option: B
Explanation:
Equation of Hyperbola is $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$

if p lies between $4$ and $29$ then coefficient of $y^2$  is negative and coefficient of $x^2$ is positive

Hence, it satisfies the equation of Hyperbola between $4$ and $29$

Which of the following equations in parametric form can represent a hyperbolic profile, where $t$ is a parameter.

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

  1. $x=\dfrac { a }{ 2 } \left( t+\dfrac { 1 }{ t } \right)$ & $y=\dfrac{b}{2}\left( t-\dfrac { 1 }{ t } \right)$

  2. $\dfrac{tx}{a}-\dfrac{y}{b}+t=0$ & $\dfrac{x}{a}-\dfrac{ty}{b}-1=0$

  3. $x={e}^{t}+{e}^{-t}$ & $x={e}^{t}-{e}^{-t}$

  4. ${x}^{2}-6=2\cos{t}$ & ${y}^{2}+2=4{\cos}^{2}\dfrac{t}{2}$

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Correct Option: A
Explanation:
$(a)$ We have $x=\dfrac{a}{2}\left(t+\dfrac{1}{t}\right)$ and $y=\dfrac{b}{2}\left(t-\dfrac{1}{t}\right)$

$\Rightarrow\,\dfrac{2x}{a}=t+\dfrac{1}{t}$ and $\dfrac{2y}{b}=t-\dfrac{1}{t}$

$\Rightarrow\,{\left(\dfrac{2x}{a}\right)}^{2}={\left(t+\dfrac{1}{t}\right)}^{2}$ and ${\left(\dfrac{2y}{b}\right)}^{2}={\left(t-\dfrac{1}{t}\right)}^{2}$

$\Rightarrow\,\dfrac{4{x}^{2}}{{a}^{2}}={t}^{2}+\dfrac{1}{{t}^{2}}+2$ and $\dfrac{4{y}^{2}}{{b}^{2}}={t}^{2}+\dfrac{1}{{t}^{2}}-2$

$\Rightarrow\,\dfrac{4{x}^{2}}{{a}^{2}}-\dfrac{4{y}^{2}}{{b}^{2}}={t}^{2}+\dfrac{1}{{t}^{2}}+2-{t}^{2}-\dfrac{1}{{t}^{2}}+2$

$\Rightarrow\,\dfrac{4{x}^{2}}{{a}^{2}}-\dfrac{4{y}^{2}}{{b}^{2}}=4$

$\Rightarrow\,\dfrac{{x}^{2}}{{a}^{2}}-\dfrac{{y}^{2}}{{b}^{2}}=1$ is in parametric form can represents a hyperbolic profile.

$(b)\dfrac{tx}{a}-\dfrac{y}{b}+t=0$ and $\dfrac{x}{a}-\dfrac{ty}{b}-1=0$

$\Rightarrow\,\dfrac{tx}{a}+t=\dfrac{y}{b}$ and $\dfrac{x}{a}-1=\dfrac{ty}{b}$

$\Rightarrow\,t\left(\dfrac{x}{a}+1\right)=\dfrac{y}{b}$ and $\dfrac{b}{y}\left(\dfrac{x}{a}-1\right)=t$

$\Rightarrow\,t\left(\dfrac{x+a}{a}\right)=\dfrac{y}{b}$ and $t=\dfrac{b}{y}\left(\dfrac{x-a}{a}\right)$

$\Rightarrow\,\dfrac{b}{y}\left(\dfrac{x-a}{a}\right)\left(\dfrac{x+a}{a}\right)=\dfrac{y}{b}$ 

$\Rightarrow\,\dfrac{b}{y}\left(\dfrac{{x}^{2}-{a}^{2}}{{a}^{2}}\right)=\dfrac{y}{b}$ 

$\Rightarrow\,\dfrac{{x}^{2}-{a}^{2}}{{a}^{2}}=\dfrac{{y}^{2}}{{b}^{2}}$ 

$\Rightarrow\,\dfrac{{x}^{2}}{{a}^{2}}-1=\dfrac{{y}^{2}}{{b}^{2}}$ 

$\Rightarrow\,\dfrac{{x}^{2}}{{a}^{2}}-\dfrac{{y}^{2}}{{b}^{2}}=1$ is in parametric form can represents a hyperbolic profile.

$(c)x={e}^{t}+{e}^{-t}$ and $x={e}^{t}-{e}^{-t}$

$\Rightarrow\,{x}^{2}={e}^{2t}+{e}^{-2t}+2{e}^{t}{e}^{-t}$ and $ {x}^{2}={e}^{2t}+{e}^{-2t}-2{e}^{t}{e}^{-t}$

$\Rightarrow\,{x}^{2}={e}^{2t}+{e}^{-2t}+2$ and ${x}^{2}={e}^{2t}+{e}^{-2t}-2$

$\Rightarrow\,{x}^{2}-{x}^{2}={e}^{2t}+{e}^{-2t}+2-{e}^{2t}-{e}^{-2t}+2=4$ is not in parametric form can represents a hyperbolic profile.

$(d){x}^{2}-6=2\cos{t}$ and ${y}^{2}+2=4{\cos}^{2}{\dfrac{t}{2}}$

$\Rightarrow\,{x}^{2}=2\cos{t}+6$ and ${y}^{2}=4{\cos}^{2}{\dfrac{t}{2}}-2$

$\Rightarrow\,{x}^{2}-{y}^{2}=2\cos{t}+6-4{\cos}^{2}{\dfrac{t}{2}}+2$ 

$\Rightarrow\,{x}^{2}-{y}^{2}=2\left(2{\cos}^{2}{\dfrac{t}{2}}-1\right)+8-4{\cos}^{2}{\dfrac{t}{2}}$ 

$\Rightarrow\,{x}^{2}-{y}^{2}=4{\cos}^{2}{\dfrac{t}{2}}-2+8-4{\cos}^{2}{\dfrac{t}{2}}$ 

$\Rightarrow\,{x}^{2}-{y}^{2}=6$ is in parametric form can represents a hyperbolic profile.


The transverse axis of a hyperbola is of length $2a$ and a vertex divides the segment of the axis between the centre and the corresponding focus in the ratio $2:1$. 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. $4x^2-5y^2=4a^2$

  2. $4x^2-5y^2=5a^2$

  3. $5x^2-4y^2=4a^2$

  4. $5x^2-4y^2=5a^2$

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Correct Option: D
Explanation:

We have given  $\cfrac{a}{ae-a}=2\Rightarrow e=\cfrac{3}{2}$
Using $e^2=1+\cfrac{b^2}{a^2}$
$\cfrac{9}{4}=1+\cfrac{b^2}{a^2}\Rightarrow b^2=\cfrac{5}{4}a^2$
Hence required hyperbola is $\cfrac{x^2}{a^2}-\cfrac{y^2}{b^2}=1$
$\cfrac{x^2}{a^2}-\cfrac{y^2}{\frac{5}{4}a^2}=1$
$\Rightarrow 5x^2-4y^2=5a^2$
Hence option 'D' is correct.

The equation of the hyperbola whose directrix is $2x + y = 1$,corresponding focus is $(1, 1)$ and eccentricity $\sqrt { 3 }$, is given by 

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

  1. $7 x ^ { 2 } + 12 x y - 2 y ^ { 2 } - 2 x + 4 y - 7 = 0$

  2. $2 x ^ { 2 } + 12 x y - 7 y ^ { 2 } - 2 x + 14 y - 7 = 0$

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

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

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Correct Option: A
Explanation:
Let$ P(x, y)$ is any point on the hyperbola.
given, focus of parabola is $S(1,1)$.
equation of directrix is $2x + y = 1$
From P draw PM perpendicular to the directrix then $PM = (2x + y – 1)/√(2² + 1²) = (2x + y – 1)/√5$
Also from the definition of the hyperbola, we have
$SP/PM = e ⇒ SP = ePM$
$⇒ √{(x–1)² + (y–1)²} = √3{(2x + y – 1)/√5}$
$⇒ (x – 1)² + (y – 1)² = 3 (2x + y – 1)²/5$
$⇒ 5[(x² – 2x + 1) + (y² –2y + 1)] = 3(4x² + y² + 1 + 4xy – 4x – 2y)$
$⇒5x² - 10x + 5 + 5y² - 10y + 5 = 12x² + 3y² + 3 + 12xy - 12x - 6y $
$⇒7x² + 2y² + 12xy - 2x + 4y - 7 = 0$
hence, equation of hyperbola is $7x² - 2y² + 12xy - 2x + 4y - 7 = 0$

The equation of the hyperbola whose foci are $(8,3)$ and $(0,3)$ and eccentricity$=\cfrac { 4 }{ 3 } $ is

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

  1. $ 7{\left(x-4 \right ) }^{2} -9{\left(y-3 \right) }^{2}=63$

  2. ${ 7x }^{ 2 }-{ 9y }^{ 2 }=63$

  3. $ 9{ \left( x-4 \right) }^{ 2 }-9{ \left( y-3 \right) }^{ 2 }=63$

  4. $7{ \left( x+4 \right) }^{ 2 }-9{ \left( y+3 \right) }^{ 2 }=63$

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Correct Option: A
Explanation:

The centre of the hyperbola is the mid-point of the line joining the two
foci. So, the coordinates of the centre are $\left( \cfrac { 8+0 }{ 2 },\cfrac { 3+3 }{ 2 }  \right) \quad $ i.e $(4,3)$
Let $2a,2b$ be the length of the transverse and conjugate axes and let $e$ be the
eccentricity. Then, the equation of the hyperbola is
$\cfrac { {\left( x-4 \right)  }^{ 2 } }{ { a }^{ 2 } } -\cfrac { { \left( y-3 \right)  }^{ 2 } }{ { b }^{ 2 } } =1\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (i)$
Distance between the two foci$ = 2ae$
$\Rightarrow \sqrt { { \left( 8-0 \right)  }^{ 2 }+{ \left( 3-3 \right)  }^{ 2 } } =2ae$
$\Rightarrow ae=4\Rightarrow a=3$
$\therefore\quad { b }^{ 2 }={ a }^{ 2 }\left( { e }^{ 2 }-1 \right) \Rightarrow {b }^{ 2 }=9\left( \cfrac { 16 }{ 9 } -1 \right) =7$
Substituting the value of $a$ and $b$ in $(i)$, we find that the equation of the hyperbola is
$\cfrac{ { \left( x-4 \right)  }^{ 2 } }{ 9 } +\cfrac { { \left( y-3 \right)  }^{ 2 } }{ 7 } =1\quad or\quad 7{ \left( x-4 \right)  }^{ 2 }-9{ \left( y-3 \right)  }^{ 2 }=63$
Hence, option 'A' is correct.

The equation of the hyperbola whose directrix is $x + 2y = 1$, focus is $(2, 1)$ and eccentricity $2$ is

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

  1. $x^2 + 16 xy - 11y^2 - 12 x + 6y + 21 = 0$

  2. $x^2 - 16 xy - 11y^2 - 12 x + 6y + 21 = 0$

  3. $x^2 - 4 xy - y^2 - 12 x + 6y + 21 = 0$

  4. none of these

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Correct Option: B
Explanation:

Using Definition of hyperbola
$PS^2=e^2\cdot PM^2$
$(x-2)^2+(y-1)^2=2^2\left(\cfrac{x+2y-1}{\sqrt{5}}\right)^2$
$5(x^2+y^2-4x-2y+5)=4(x^2+4y^2+1+4xy-2x-4y)$
$\Rightarrow x^2 - 16 xy - 11y^2 - 12 x + 6y + 21 = 0$
Hence, option 'B' is correct.

If ${ e } _{ 1 }$ is the eccentricity of the ellipse $\cfrac { { x }^{ 2 } }{ 16 } +\cfrac { { y }^{ 2 } }{ 25 } =1$ and ${ e } _{ 2 }$ is the eccentricity of the hyperbola passing through the foci of the ellipse and ${ e } _{ 1 }.{ e } _{ 2 }=1$, 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. $\cfrac { { x }^{ 2 } }{ 9 } -\cfrac { { y }^{ 2 } }{ 16 } =1$

  2. $\cfrac { { x }^{ 2 } }{ 16 } -\cfrac { { y }^{ 2 } }{ 9 } =-1$

  3. $\cfrac { { x }^{ 2 } }{ 9 } -\cfrac { { y }^{ 2 } }{ 25 } =1$

  4. none of these

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Correct Option: B
Explanation:

We have ${ e } _{ 1 }=\sqrt { 1-\cfrac { 16 }{ 25 }  } =\cfrac { 3 }{ 5 } $
$\because \quad { e } _{ 1 }{ e } _{ 2 }=1\Rightarrow { e } _{ 2 }=\cfrac { 5 }{ 3 } $
Clearly y-axis is transverse axis of the ellipse.
Thus, coordinates of foci of the ellipse are $(0,\pm b e _1)$ or $\left( 0,\pm 3 \right) $.
Let the equation of hyperbola is, $\dfrac{y^2}{b^2}-\cfrac{x^2}{a^2}=1$ ..... $(1)$
Since, hyperbola passes through foci of the ellipse
$\Rightarrow b^2=9$ and also $a^2=b^2(e^2-1)=9(25/9-1)=16$
Therefore, the required hyperbola is, $\cfrac{x^2}{16}-\cfrac{y^2}{9}=-1$
Hence, option 'B' is correct.

Equation of the hyperbola whose vertices are at ($\pm3, 0$) and focii at ($\pm5, 0$) is

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

  1. $16x^2 - 9y^2 = 144$

  2. $9x^2 - 16y^2 = 144$

  3. $25x^2 - 9y^2 = 255$

  4. $9x^2 - 25y^2 = 81$

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Correct Option: A
Explanation:

Given $a=3$ and $ae=5\Rightarrow e=\cfrac{5}{3}$
Using $e^2=1+\cfrac{b^2}{a^2}$
$\cfrac{25}{9}=1+\cfrac{b^2}{9}\Rightarrow b^2=16$
Therefore required hyperbola is, $\cfrac{x^2}{9}-\cfrac{y^2}{16}=1$
$\Rightarrow 16x^2-9y^2=144$
Hence, option 'A' is correct.