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Questions Related to mathematics and statistics

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

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
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the definition of a hyperbola (distance from focus = e * distance from directrix), the equation is (x-1)^2 + (y-2)^2 = 3 * (2x+y-1)^2 / (2^2 + 1^2). Expanding this yields the provided equation.

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

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. $5\sqrt2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Solving the differential equation 2xy dy/dx = x^2 + y^2 leads to the hyperbola x^2 - y^2 = c. Passing through (2, 1) gives 4 - 1 = c, so c = 3. The hyperbola is x^2 - y^2 = 3. For a rectangular hyperbola, e = sqrt(2).

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 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. $3x^2-y^2=2a^2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given transverse axis is 2a, so the center is at the origin and vertex is at (a, 0). The focus is at (ae, 0). Since the vertex bisects the distance between the center and the focus, a = ae / 2, which means e = 2. Using b^2 = a^2(e^2 - 1) gives b^2 = 3a^2, yielding the hyperbola equation 3x^2 - y^2 = 3a^2.

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

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. $\dfrac{3}{\sqrt{2}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The asymptotes are 3x - y - 1 = 0 and x + 3y - 3 = 0. The angle between them is 90 degrees because the product of their slopes is -1 (3 * -1/3 = -1). A hyperbola with perpendicular asymptotes is a rectangular hyperbola, which has an eccentricity of sqrt(2).

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

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) $
Reveal answer Fill a bubble to check yourself
A Correct answer
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.
Multiple choice mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

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. $4$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $=2a=2$

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

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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. $\displaystyle \frac{\sqrt{41}}{4}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $= \sqrt{1+\cfrac{25}{16}}=\cfrac{\sqrt{41}}{4}$

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

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. $\displaystyle e^{2} - e$
Reveal answer Fill a bubble to check yourself
B Correct answer
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}}$