Tag: mathematics and statistics

Quizzes (38) Questions (955)

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

A hyperbola, having the transverse axis of length $\displaystyle 2\sin \theta$, is confocal with the ellipse $\displaystyle 3x^{2}+4y^{2}=12$, then its equation is

  1. $\displaystyle x^{2}\text{cosec} ^{2}\theta -y^{2}\sec ^{2}\theta=1$
  2. $\displaystyle x^{2} \sec ^{2}\theta -y^{2}\text{cosec}^{2}\theta=1$
  3. $\displaystyle x^{2} \sin ^{2}\theta -y^{2}\cos ^{2}\theta=1$
  4. $\displaystyle x^{2} \cos ^{2}\theta -y^{2}\sin ^{2}\theta=1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given ellipse may be written as $\cfrac{x^2}{4}+\cfrac{y^2}{3}=1$
$\Rightarrow a^2=4, b^2=3$

$\Rightarrow e= \sqrt{1-\dfrac{3}{4}}=\cfrac{1}{2}$
$\therefore $ Focus of the ellipse $=(\pm ae,0)=(\pm 1, 0)$
Given required hyperbola is confocal to the ellipse
Let $a',b',e'$ are transverse axis, conjugate axis an eccentricity of the hyperbola
$a'e'=1\Rightarrow \sin\theta. e'=1\Rightarrow e'=\cfrac{1}{\sin\theta}$
Using $b'^2=a'^2(e^2-1)\Rightarrow b'^2=1-\sin^2\theta=\cos^2\theta$
Therefore required hyperbola is $\cfrac{x^2}{a'^2}-\cfrac{y^2}{b'^2}=1$
$\Rightarrow \cfrac{x^2}{\sin^2\theta}-\cfrac{y^2}{\cos^2\theta}=1$
$\Rightarrow x^2 \text{cosec}^2\theta-y^2\sec^2\theta=1$

Hence, option 'A' is correct.

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

Consider the hyoerbola ${ 3x^{2} }-{ y }^{ 2 }-{ 24x } + { 4y } { 4 } = 0$

  1. $its centre is \left(4,2\right)$
  2. $its centre is \left(2,4\right)$
  3. $length of latus rectum = 24$
  4. $length of latus rectum = 12$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Completing the square for 3x^2 - y^2 - 24x + 4y + 4 = 0: 3(x^2 - 8x + 16) - (y^2 - 4y + 4) = -4 + 48 - 4, resulting in 3(x-4)^2 - (y-2)^2 = 40. The center is (4, 2).

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

$y=mx+c$ is tangent to hyperbola find $c$ if hyperbola eqn is

  1. $\dfrac{{x}^{2}}{{a}^{2}}-\dfrac{{y}^{2}}{{b}^{2}}=1$$a>b$
  2. $\dfrac{{x}^{2}}{{a}^{2}}-\dfrac{{y}^{2}}{{b}^{2}}=1$$b>a$
  3. $\dfrac{{-x}^{2}}{{a}^{2}}-\dfrac{{y}^{2}}{{b}^{2}}=1$$a>b$
  4. $\dfrac{{-x}^{2}}{{a}^{2}}-\dfrac{{y}^{2}}{{b}^{2}}=1$$b>a$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For line y=mx+c to be tangent to hyperbola x^2/a^2 - y^2/b^2 = 1, we need c^2 = a^2m^2 - b^2. For real c, we need a^2m^2 > b^2. When a > b, we can find m values satisfying this condition. The question is poorly phrased but option A correctly identifies the standard hyperbola condition.

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

Find Directrix, foci and eccentricity of the conics:

$\displaystyle x^{2}+2x-y^{2}+5= 0$

  1. Directrices $\displaystyle y= \pm \sqrt{2}$
  2. foci $\displaystyle \left ( -1,\pm 2\sqrt{2} \right )$
  3. $e= \sqrt{2}$
  4. $e=2$
Reveal answer Fill a bubble to check yourself
A,B,C Correct answer
Explanation

$\displaystyle x^{2}+2x-y^{2}+5= 0$
$\displaystyle x^{2}+2x+1-y^{2}= -4$
$\displaystyle (x+1)^2-y^{2}= -4$
$\displaystyle \frac{y^2}{4}-\frac{(x+1)^2}{4}= 1$
Clearly this equation of rectangular hyperbola with $y-$axis as major axis
eccentricity $e = \sqrt{2}$ directrices $:y = \cfrac{a}{e}=\pm \sqrt{2}$ foci $(-1,\pm ae)=(-1,\pm 2\sqrt{2})$

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

The equation of a hyperbola whose directrix is $2x+y=1$ and focus is at $(1,2)$ with $e=\sqrt{3}$ is :

  1. $7x^2+12xy+2y^2-2x+14y-22=0$
  2. $7x^2+12xy-2y^2-2x+14y-22=0$
  3. $7x^2+12xy-2y^2-2x-14y-22=0$
  4. $7x^2+12xy+2y^2+2x+14y-22=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The definition of a hyperbola is PF = e * PM, where PM is the perpendicular distance to the directrix. Squaring both sides: (x-1)^2 + (y-2)^2 = 3 * ((2x+y-1)/sqrt(2^2+1^2))^2. Expanding this leads to 5(x^2 - 2x + 1 + y^2 - 4y + 4) = 3(4x^2 + y^2 + 1 + 4xy - 4x - 2y). Simplifying results in 7x^2 + 12xy + 2y^2 - 2x + 14y - 22 = 0.

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

Let the eccentricity of the hyperbola $  \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1  $ be reciprocal to that of the ellipse $  x^{2}+4 y^{2}=4 .  $ If thehyperbola passes through a focus of the ellipse, then __________________.

  1. (A) the equation of the hyperbola is $ \frac{x^{2}}{3}-\frac{y^{2}}{2}=1 $
  2. (B) a focus of the hyperbola is $ (2,0) $
  3. (C) the eccentricity of the hyperbola is $ \sqrt{\frac{5}{3}} $
  4. (D) the equation of the hyperbola is $ x^{2}-3 y^{2}=3 $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Ellipse x^2/4 + y^2/1 = 1 has a^2=4, b^2=1. e_e = sqrt(1 - 1/4) = sqrt(3)/2. Hyperbola eccentricity e_h = 2/sqrt(3). Hyperbola x^2/a^2 - y^2/b^2 = 1 passes through focus (ae_e, 0) = (sqrt(3), 0). So 3/a^2 = 1 => a^2 = 3. Then b^2 = a^2(e_h^2 - 1) = 3(4/3 - 1) = 1. Equation: x^2/3 - y^2/1 = 1, or x^2 - 3y^2 = 3.

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

The eccentricity of the hyperbola $\displaystyle \dfrac { \sqrt { 1999 }  }{ 3 } \left( { x }^{ 2 }-{ y }^{ 2 } \right) =1$ is:

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

Equation of hyperbola is $\displaystyle \frac { { x }^{ 2 } }{ 3/\sqrt { 1999 }  } -\frac { { y }^{ 2 } }{ 3/\sqrt { 1999 }  } =1$

Here $\displaystyle { a }^{ 2 }={ b }^{ 2 }=\frac { 3 }{ \sqrt { 1999 }  } $
$\therefore$ Eccentricity $\displaystyle e=\sqrt { 1+\frac { { b }^{ 2 } }{ { a }^{ 2 } }  } =\sqrt { 1+1 } =\sqrt { 2 } $