Tag: mathematics and statistics

Questions Related to mathematics and statistics

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

  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


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

  1. $\sqrt2$

  2. $2\sqrt2$

  3. $3\sqrt2$

  4. $5\sqrt2$


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

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

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

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

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


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

  1. $\sqrt{2}$

  2. $3$

  3. $2\sqrt{2}$

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


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:

  1. $2$

  2. $4$

  3. $6$

  4. $None\ of\ these$


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

  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) $


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

  1. $1$

  2. $3$

  3. $2$

  4. $4$


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

  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


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.

If any point on a hyperbola has the coordinates $\displaystyle \left ( 5 \tan \phi , : 4 \sec \phi \right )$ then the ecentricity of the hyperbola is

  1. $\displaystyle \frac{5}{4}$

  2. $\displaystyle \frac{\sqrt{41}}{5}$

  3. $\displaystyle \frac{25}{16}$

  4. $\displaystyle \frac{\sqrt{41}}{4}$


Correct Option: D
Explanation:

We have   $x=5\tan\phi, y=4\sec\phi$
Eliminating $\phi$ we get
$\cfrac{y^2}{16}-\cfrac{x^2}{25}=\sec^2\phi-\tan^2\phi=1$
This is a hyperbola. Therefore, its eccentricity $= \sqrt{1+\cfrac{25}{16}}=\cfrac{\sqrt{41}}{4}$

If the eccentricity of the hyperbola $\displaystyle \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ is $e$ then the eccentricity of the hyperbola $\displaystyle \frac{y^{2}}{b^{2}} - \frac{x^{2}}{a^{2}} = 1$ is :

  1. $e$

  2. $\displaystyle \frac{e}{\sqrt{e^{2} - 1}}$

  3. $\displaystyle e \sqrt{e^{2} - 1}$

  4. $\displaystyle e^{2} - e$


Correct Option: B
Explanation:

For hyperbola $\displaystyle \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$
eccentricity $\Rightarrow e=\sqrt{1+\cfrac{b^2}{a^2}}\Rightarrow \cfrac{b^2}{a^2}=e^2-1$
For hyperbola $\displaystyle \frac{x^{2}}{b^{2}} - \frac{y^{2}}{a^{2}} = 1$
Required eccentricity $ e'=\sqrt{1+\cfrac{a^2}{b^2}}=\sqrt{1+\cfrac{1}{e^2-1}}=\displaystyle \cfrac{e}{\sqrt{e^{2} - 1}}$