Tag: equations of hyperbola

Questions Related to equations of hyperbola

The equation of the conic with focus at $(1, -1)$, directrix along $x - y + 1= 0$ and with eccentricity $\sqrt{2}$ is

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

  1. $x^2 - y^2 = 1$

  2. $xy = 1$

  3. $2xy - 4x + 4y + 1 = 0$

  4. $2xy + 4x - 4y - 1 = 0$

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Correct Option: C
Explanation:

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

The tangent of a point $P$ on the hyperbola $\dfrac {x^{2}}{a^{2}}-\dfrac {y^{2}}{b^{2}}=1$ passes through the point $(0,\ -b)$ and the normal at $P$ pases through the point $(2a\sqrt {2},\ 0)$. Then the eccentricity of the hyperbola is   

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

  1. $2$

  2. $\sqrt {2}$

  3. $3$

  4. $\sqrt {3}$

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Correct Option: A

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

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

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

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

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

  4. none of above

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Correct Option: A

Eccentricity of the hyperbola satisfying the differential equation $2xy\dfrac{dy}{dx}=x^2+y^2$ and passing through $(2,1)$ is

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

  1. $\sqrt2$

  2. $2\sqrt2$

  3. $3\sqrt2$

  4. $5\sqrt2$

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Correct Option: A

Find the equation to the hyperbola of given transverse xis (2a) whose vertex bisects the distance between the centre and the focus

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

  1. $3x^2-2y^2=12a^2$

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

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

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

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Correct Option: A

The ecentricity of the hyperbola passing through the origin and whose asymptotes are given by straight lines $y=3x-1$ and $x+3y=3$, is

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

  1. $\sqrt{2}$

  2. $3$

  3. $2\sqrt{2}$

  4. $\dfrac{3}{\sqrt{2}}$

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Correct Option: A

A hyperbola passes through the points $(3, 2)$ and $(-17, 12)$ and has its centre at origin and transverse axis is along $x-axis$. The length of its transverse axis is:

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

  1. $2$

  2. $4$

  3. $6$

  4. $None\ of\ these$

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Correct Option: A

If a hyperbola passes through the focii of the ellipse$\dfrac { { x }^{ 2 } }{ 25 } +\dfrac { { y }^{ 2 } }{ 16 } =1.$ Its transverse and conjugate axes coincide respectively with the major and minor axes of the ellipse and if the product of eccentricities hyperbola and ellipse is 1, then

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

  1. the equation of hyperbola is $\dfrac { x^{ 2 } }{ 9 } -\dfrac { { y }^{ 2 } }{ 16 } =1\ \quad \quad $

  2. the equation of hyperbola is $\dfrac { x^{ 2 } }{ 9 } -\dfrac { { y }^{ 2 } }{ 25 } =1\ \quad \quad $

  3. focus of hyperbola is (5,0)

  4. focus of hyperbola is $\left( 5\sqrt { 3, } 0 \right) $

Show answer
Correct Option: A
Explanation:
Formula,

$e^2=1-\dfrac{b^2}{a^2}$

$=1-\dfrac{16}{25}$

$\therefore e=\dfrac{3}{5}$

$e _2 \times e =1$

$\Rightarrow e _2=\dfrac{5}{3}$

Equation,

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

Given,

$\Rightarrow (3,0)$

$\dfrac{3^2}{a^2}=1$

$\Rightarrow a^2=9$

we have,

$e _2^2=1+\dfrac{b^2}{a^2}$

$\dfrac{25}{9}=1+\dfrac{b^2}{9}$

$\Rightarrow b^2=16$

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

Hence the required equation.

The hyperbola $\displaystyle \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}}=1$ passes through the point $\displaystyle \left ( 2, : 3 \right )$ and has the eccentricity $2$. Then the transverse axis of the hyperbola has the length

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

  1. $1$

  2. $3$

  3. $2$

  4. $4$

Show answer
Correct Option: C
Explanation:

Given hyperbola is,  $\displaystyle \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}}=1$
It passes through $(2,3)$
$\cfrac{4}{a^{2}} - \cfrac{9}{b^{2}}=1 ..(1)$
Also eccentricity is $2$,
$\Rightarrow e^2=1+\cfrac{b^2}{a^2}=4\Rightarrow \cfrac{b^2}{a^2}=3   ..(2)$
Solving (1) and (2) we get $a=1, b=\sqrt{3}$
Hence, length of transverse axis is $=2a=2$

If in a hyperbola the eccentricity is $\displaystyle \sqrt{3}$, and the distance between the foci is $9$ then the equation of the hyperbola in the standard form is

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

  1. $\displaystyle \dfrac{x^{2}}{\left ( \dfrac{\sqrt{3}}{2} \right )^{2}} - \dfrac{y^{2}}{\left ( \sqrt{\dfrac{3}{2}} \right )^{2}} = 1$

  2. $\displaystyle \dfrac{x^{2}}{\left ( \dfrac{3 \sqrt{3}}{2} \right )^{2}} - \dfrac{y^{2}}{\left ( \dfrac{3\sqrt{3}}{\sqrt{2}} \right )^{2}} = 1$

  3. $\displaystyle \dfrac{x^{2}}{\left ( \dfrac{3\sqrt{3}}{\sqrt{2}} \right )^{2}} - \dfrac{y^{2}}{\left ( \dfrac{3\sqrt{2}}{2} \right )^{2}} = 1$

  4. none of these

Show answer
Correct Option: B
Explanation:

Given eccentricity of the hyperbola $e=\sqrt{3}$
and distance between focii is 9. $\Rightarrow 2ae=9\Rightarrow a=\cfrac{3\sqrt{3}}{2}$
also $b^2=a^2(e^2-1)=\cfrac{27}{4}(3-1)=\cfrac{27}{2}\Rightarrow b=\cfrac{3\sqrt{3}}{\sqrt{2}}$
Hence equation of required hyperbola is,
$\cfrac{x^2}{\left(\cfrac{3\sqrt{3}}{2}\right )^2}-\cfrac{y^2}{\left (\cfrac{3\sqrt{3}}{\sqrt{2}}\right )^2}=1$
Hence, option 'B' is correct.