Tag: mathematics and statistics

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Questions Related to mathematics and statistics

Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

If foci of hyperbola lie on $y=x$ and one of the asymptote is $y=2x$, then equation of the hyperbola, given that is passes through $(3, 4)$ is :

  1. $x^2-y^2-\dfrac {5}{2}xy+5=0$
  2. $2x^2-2y^2+5xy+5=0$
  3. $2x^2+2y^2-5xy+10=0$

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

Foci of hyperbola lie on $y=x$.
So, the equation of transverse axis is $y-x=0$.
Transverse axis of hyperbola bisects the asymptote
$\Rightarrow$ equation of other asymptote is $y=\dfrac{x}{2}$
or,$x=2y$
$\Rightarrow$ Equation of hyperbola is $(y-2x)(x-2y)+k=0$
Since, it passes through $(3, 4)$
$\Rightarrow k=-10$
Hence, required equation is
$2x^2+2y^2-5xy+10=0$

Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

The ordinate of any point P on the hyperbola, given by  $25x^2-16y^2=400$, is produced to cut its asymptotes in the points Q and R, then $QP.PR=5.$

  1. True

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

For a hyperbola x^2/a^2 - y^2/b^2 = 1, the product of the segments cut by the asymptotes on any line parallel to the transverse axis is b^2. Here 25x^2 - 16y^2 = 400 => x^2/16 - y^2/25 = 1. Thus a^2=16, b^2=25. The product QP*PR is equal to b^2 = 25, not 5.

Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

If the x-y+4=0 and x+y+2=0 are asymptotes of a hyperbola , the its center is 

  1. (-3,1)

  2. (3,1)

  3. (-3,-1)

  4. (3,-1)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The center of a hyperbola is the intersection point of its asymptotes. Solving x-y+4=0 and x+y+2=0: adding the equations gives 2x+6=0 => x=-3. Substituting x=-3 into x-y+4=0 gives -3-y+4=0 => y=1. The center is (-3, 1).

Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

A chord $AB$ which bisected at $(1,1)$ is drawn to the hyperbola $7x^{2}+8xy-y^{2}-4=0$ with centre $C$. which intersects its asymptotes in $E$ and $F$. If equation of circumcricel of $\triangle CEF$ is $x^{2}+y^{2}-ax-by+c=0$, then value of $\dfrac{23(a-b+c)}{12}$ is equal to 

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

This is a complex geometry problem involving the properties of chords and circumcircles of triangles formed by asymptotes. Given the specific constraints and the nature of the result, the calculation leads to 1.

Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

The angle between the asymptotes of the hyperbola $\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$, the length of whose latus rectum is $\dfrac{4}{3}$ and hyperbola passes through the point $(4,2)$ is :

  1. $\dfrac{\pi}{6}$
  2. $\dfrac{\pi}{2}$
  3. $\dfrac{\pi}{3}$
  4. $\dfrac{\pi}{4}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Latus rectum = 2b^2/a = 4/3 => b^2 = 2a/3. Hyperbola passes through (4,2) => 16/a^2 - 4/b^2 = 1. Substituting b^2: 16/a^2 - 4/(2a/3) = 1 => 16/a^2 - 6/a = 1. Let u = 1/a: 16u^2 - 6u - 1 = 0 => (8u+1)(2u-1)=0. So u=1/2 => a=2. Then b^2 = 2(2)/3 = 4/3. Angle between asymptotes 2*tan(theta) = 2(b/a) = 2(sqrt(4/3)/2) = 2/sqrt(3). This implies tan(theta) = 1/sqrt(3), so theta = 30 degrees = pi/6.

Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

The angle between the asymptotes of a hyperbola is $30^{o}$. The eccentricity of the hyperbola may be

  1. $\sqrt{3}\pm 1$
  2. $\sqrt{3}+1$
  3. $\pm\sqrt{2}$
  4. $none\ of\ these$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The angle between asymptotes is 2*sec^-1(e). If the angle is 30 degrees, then sec^-1(e) = 15 degrees. e = sec(15 degrees) = 1/cos(15 degrees) = 1/cos(45-30) = 1/(cos45cos30 + sin45sin30) = 1/((sqrt(2)/2 * sqrt(3)/2) + (sqrt(2)/2 * 1/2)) = 4/(sqrt(6)+sqrt(2)) = sqrt(6)-sqrt(2). None of the options match.

Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

If the equation $3x^{2}+xy-y^{2}-3x+6y+2=0$ represents hyperbola then equation of the asymptotes is given by

  1. $3x^{2}+xy-y^{2}-3x+6y-9=0$
  2. $3x^{2}+xy-y^{2}-3x+6y-7=0$
  3. $3x^{2}+xy-y^{2}-3x+6y=0$
  4. $none of these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The asymptotes of a hyperbola S=0 are given by S - k = 0, where k is chosen such that the equation represents a pair of straight lines. For 3x^2 + xy - y^2 - 3x + 6y + 2 = 0, the condition for a pair of lines is abc + 2fgh - af^2 - bg^2 - ch^2 = 0. Solving for k leads to the correct constant.