Tag: hyperbola

Quizzes (10) Questions (269)

Questions Related to hyperbola

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.

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

The equation of the line passing through the centre of a rectangle hyperbola is $x-y-1=0$. If one of its asymptotes is $3x-4x-6=0$, the equation of the other asymptote is $

  1. $4x+3y+17=0$
  2. $4x-3y+8=0$
  3. $3x-2y+15=0$
  4. $None of these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a rectangular hyperbola, the asymptotes are perpendicular. One asymptote is 3x - 4y - 6 = 0 (note: this appears to be a typo in the question, should be 3x - 4y - 6 = 0, not 3x - 4x - 6 = 0). The center lies on x - y - 1 = 0. The other asymptote must be perpendicular to the first and pass through the center. A line perpendicular to 3x - 4y - 6 = 0 has equation 4x + 3y + k = 0. Finding the intersection of x - y - 1 = 0 with 3x - 4y - 6 = 0 gives the center, and substituting this in 4x + 3y + k = 0 gives k = 17.

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

If $e$ is the eccentricity of $\dfrac {x^{2}}{a^{2}}-\dfrac {y^{2}}{b^{2}}=1$ and '$\theta $' be the angle between its asymptotes then $\cos (\theta /2)$ is equal to.

  1. $1/ 2e$
  2. $1/ e$
  3. $2/e^{2}$
  4. $none\ of\ these$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The angle between asymptotes is 2*sec^-1(e). Thus theta/2 = sec^-1(e), which means sec(theta/2) = e. Therefore, cos(theta/2) = 1/e.

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

The asymptotes of the hyperbola $xy-3x+4y+2=0$

  1. $x=-4$
  2. $x=4$
  3. $y=-3$
  4. $y=3$
Reveal answer Fill a bubble to check yourself
A,D Correct answer
Explanation

Since the equation of a hyperbola and its asymptotes differ in constant terms only. Therefore, the equations of asymptotes of the given hyperbola are given by $xy-3x+4y+k=0$

where $k$ is a constant to be determined  by the condition that $abc+2fgh-{ af }^{ 2 }-{ bg }^{ 2 }-{ ch }^{ 2 }=0$
i.e., $\displaystyle 0+2\times 2\times \left( \frac { -3 }{ 2 }  \right) \times \frac { 1 }{ 2 } -0-0-k\times { \left( \frac { 1 }{ 2 }  \right)  }^{ 2 }=0\Rightarrow k=-12$
$\because $ Asymptotes of the given hyperbola are $xy-3x+4y-12=0$ or $(x+4)(y-3)=0$
i.e., $x=-4$ and $y=3.$