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

The vertices and the foci of a hyperbola are the points $\displaystyle \left ( \pm 5, 0 \right )$ and $\displaystyle \left ( \pm 7, 0 \right )$.Which of the following holds true?

  1. $\displaystyle a^{2}\neq b^{2}$
  2. $a^2=b^2$
  3. $\dfrac{a^2}{b^2}=2$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\displaystyle a= 5, ae = 7\Rightarrow a^{2}e^{2}= 49$
or $\displaystyle a^{2}\left ( 1+\frac{b^{2}}{a^{2}} \right )= 49$ 

Or 
$\displaystyle a^{2}+b^{2}=49$ 
$\displaystyle b^{2}= 24$ 
Since, $\displaystyle a^{2}\neq b^{2},$ hence hyperbola is not rectangular.

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

An ellipse intersects the hyperbola $2x^{2}-2y^{2}=1$ orthogonally. The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinates axes, then

  1. equation of ellipse is $x^{2}+2y^{2}=2$
  2. the foci of ellipse are $\left ( \pm 1, 0 \right )$
  3. equation of ellipse is $x^{2}+2y^{2}=4$
  4. the foci of ellipse are $\left ( \pm \sqrt{2}, 0 \right )$
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation

Eccentricity of the hyperbola is $\sqrt{2}$ as it is a rectangular hyperbola. 

So eccentricity $e$ of the ellipse is $\dfrac1{\sqrt{2}}$
Let the equation of the ellipse be $\displaystyle \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ where $b^{2}=a^{2}\left ( 1-e^{2} \right )=\dfrac{a^{2}}2\Rightarrow a^{2}=2b^{2}$
So equation of the ellipse is $x^{2}+2y^{2}=a^{2}$
Let $\left ( x _{1}, y _{1} \right )$ be a point of intersection of the ellipse and the hyperbola
Then $2x _{1}^{2}-2y _{1}^{2}=1$ and $x _{1}^{2}+2y _{1}^{2}=a^{2}$          (1)
Equations of the tangents at $\left ( x _{1}, y _{1} \right )$ to the two conics are
   $2xx _{1}-2yy _{1}=1$ and $xx _{1}+2yy _{1}=a^{2}$
Since the two conics intersect orthogonally
$\displaystyle \left ( \frac{x _{1}}{y _{1}} \right )\left ( -\frac{x _{1}}{2y _{1}} \right )=-1\Rightarrow x _{1}^{2}=2y _{1}^{2}$
And from (1) we get $x _{1}^{2}=1$, $a^{2}=2$.
Hence the equation of the ellipse is $x^{2}+2y^{2}=2$ and its focus is
   $\displaystyle \left ( \pm ae, 0 \right )=\left ( \pm \sqrt{2}\times \frac{1}{\sqrt{2}}, 0 \right )=\left ( \pm 1, 0 \right )$

Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

The equation of the ellipse whose equation of directrix is $3x+4y-5=0$, coordinates of the focus are $(1,2)$ and the eccentricity is $\dfrac{1}{2}$ is $91x^2+84y^2-24xy-170x-360y+475=0$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let $P(x,y)$ be any point on the ellipse and PM be the perpendicular from P upon the directrix $3x+4y-5=0$.

Then by the definition,
$\dfrac{SP}{PM}=e$

$SP=e.PM$
$\sqrt{(x-1)^2+(y-2)^2}=\dfrac{1}{2}|\dfrac{3x+4y-5}{\sqrt{3^2+4^2}}|$

$(x-1)^2+(y-2)^2=\dfrac{1}{4}. \dfrac{(3x+4y-5)^2}{25}$

$100(x^2+y^2-2x-4y+5)=9x^2+16y^2+24xy-30x-40y+25$
$91x^2+84y^2-24xy-170x-360y+475=0$ is the equation of the ellipse.

Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

The equation of the ellipse whose foci are $(\pm5,0)$ and of the directrix is $5x=36$, is

  1. $\dfrac{x^2}{36}+\dfrac{y^2}{11}=1$
  2. $\dfrac{x^2}{6}+\dfrac{y^2}{\sqrt{11}}=1$
  3. $\dfrac{x^2}{6}+\dfrac{y^2}{11}=1$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given foci $(\pm 5,0)$ and directrix $x=\cfrac{36}{5}$

Then $ae=5$ (focus coordinates ($\pm ae,0)]$....(1)
$\cfrac{a}{e}=\cfrac{36}{5}$ (directrix equation $x=\cfrac{a}{e}$]....(2)
From (1) and (2) ${a}^{2}=36\Rightarrow$ $a=6$
$e=\cfrac{5}{6}\Rightarrow $ $\sqrt { 1-\cfrac { { b }^{ 2 } }{ { a }^{ 2 } }  } =\cfrac { 5 }{ 6 } $
$1-\cfrac { { b }^{ 2 } }{ 36 } =\cfrac{25}{36}$
$b=\sqrt 11$
required equation $\cfrac{{x}^{2}}{36}+\cfrac{{y}^{2}}{11}=1$

Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

If the eccentricity of the ellipse $\dfrac{x^2}{a^2 + 1} + \dfrac{y^2}{a^2 + 2 } = 1$ is $\dfrac{1}{\sqrt{6}}$, then the length of latusrectum is

  1. $\dfrac{5}{\sqrt{6}}$
  2. $\dfrac{10}{\sqrt{6}}$
  3. $\dfrac{8}{\sqrt{6}}$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given the ellipse equation x^2/(a^2+1) + y^2/(a^2+2) = 1, we identify the semi-axes. Since a^2+2 > a^2+1, the ellipse is vertical. The eccentricity e = 1/sqrt(6). Using e^2 = 1 - (a^2+1)/(a^2+2) = 1/(a^2+2), we find 1/6 = 1/(a^2+2), so a^2+2 = 6, a^2 = 4. The semi-axes are b^2 = 5 and a^2 = 6. Latus rectum = 2 * (minor^2) / major = 2 * 5 / sqrt(6) = 10/sqrt(6).

Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

If focus of the parabola is $(3,0)$ and length of latus rectum is $8$, then its vertex is

  1. $(2,0)$
  2. $(1,0)$
  3. $(0,0)$
  4. $(-1,0)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given, focus $=(3,0)$ and Length of latus rectum $= 8$

$\Rightarrow 4a=8$ $\Rightarrow a=2$

$\Rightarrow$ Vertex = $(3-a,0)$ $=(1,0)$

$\therefore $ Option B is correct
Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

If $(0,0)$ be the vertex and $3x-4y+2=0$ be the directrix of a parabola, then the length of its latus rectum is

  1. $4/5$
  2. $2/5$
  3. $8/5$
  4. $1/5$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Distance of vertex from directrix = $\dfrac{\left | 3(0)-4(0)+2 \right |}{\sqrt{3^{2}+4^{2}}}= \dfrac{2}{5}=a$

Length of latus rectum = $4a= \dfrac{8}{5}$

$\therefore $ Option C is correct
Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

Which of the following can be the equation of an ellipse?

  1. $x^{2} + y^{2} = 5$
  2. $\dfrac {x^{2}}{9} + \dfrac {x^{2}}{9} = 1$
  3. $2x^{2} + 3y^{2} = 5$
  4. $2x + 2y = 5$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

An ellipse equation in standard form is x^2/a^2 + y^2/b^2 = 1. Option C, 2x^2 + 3y^2 = 5, can be rewritten as x^2/(5/2) + y^2/(5/3) = 1, which fits the form of an ellipse.