Tag: standard equation of hyperbola

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Questions Related to standard equation of hyperbola

Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

General solution of the equation $ y=x\dfrac{dy}{dx}+\dfrac {dx}{dy}$ represents _____________.

  1. a straight line or hyperbola

  2. a straight line or parabola

  3. a parabola or hyperbola

  4. circles
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The given differential equation can be rearranged into a form that describes a family of curves, which are hyperbolas or parabolas depending on the integration constants.

Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

Eccentricity of hyperbola$ \dfrac { { x }^{ 2 } }{ k } -\dfrac { { y }^{ 2 } }{ k } =1$

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

$Standard\, hyperbola\, : $


$\dfrac { { { x^{ 2 } } } }{ k } -\dfrac { { { y^{ 2 } } } }{ k } =1$

$Now, \ \dfrac { { { { \left( { x-h } \right)  }^{ 2 } } } }{ { { a^{ 2 } } } } -\dfrac { { { { \left( { y-k } \right)  }^{ 2 } } } }{ { { b^{ 2 } } } } =1$

$Therefore\, Hyperbola\, properties\, are \ (h,k)=\left( { 0,0 } \right) ,\, \, a=\sqrt { k } ,\, b=\sqrt { k }  $

$=\dfrac { { \sqrt { { { \left( \sqrt k \right)  }^{ 2 } }+{ { \left( { \sqrt { k }  } \right)  }^{ 2 } } }  } }{ { \sqrt { k }  } } $

$=\sqrt { 2 }  \ Hence,\, the\, option\, C\, is\, the\, correct\, answer$

Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

A hyperbola passes through the focus of the ellipse $\dfrac{x^2}{25}+\dfrac{y^2}{16}=1,$ and its transverses and conjugate axes coincide with the major and minor axes of the ellipse. If the product of the eccentricites of the two curve is $1$, then the focus of the hyperbola is

  1. $(5\sqrt3,0)$
  2. $(5,0)$
  3. $\left(\dfrac{5}{3},0\right)$

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

The focus of the ellipse x^2/25 + y^2/16 = 1 is (3, 0). If the hyperbola passes through (3, 0) and shares axes, its equation is x^2/a^2 - y^2/b^2 = 1. Using the eccentricity product condition, one can solve for the focus.

Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

If eccentricity of the hyperbola $\dfrac {x^{2}}{\cos^{2}\theta}-\dfrac {y^{2}}{\sin^{2}\theta}=1$ is more then $2$ when $\theta\ \in \ \left(0,\dfrac {\pi}{2}\right)$. Find the possible values of length of latus rectum 

  1. $(3,\infty)$
  2. $(1,3/2)$
  3. $(2,3)$
  4. $(-3,-2)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The eccentricity e = sqrt(1 + tan^2(theta)) = sec(theta). If e > 2, then sec(theta) > 2, so cos(theta) < 1/2. The latus rectum length is 2*sin^2(theta)/cos(theta). Solving this interval gives (3, infinity).

Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

The latus rectum of the hyperbola $16{x^2} - 9{y^2} = 144$ is-

  1. $\dfrac{13}{6}$
  2. $\dfrac{32}{3}$
  3. $\dfrac{8}{3}$
  4. $\dfrac{4}{3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We have,


$16{x^2} - 9{y^2} = 144$

$\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$

Here, $a=3,  b=4$

We know that the latus rectum 

$=\dfrac{2b^2}{a}$

Therefore,

$=\dfrac{2\times 16}{3}$

$=\dfrac{32}{3}$

Hence, this is the answer.

Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

The Vertex of the parabola $y^{2} - 10y + x + 22=0$ is.

  1. (3,4)

  2. (3,5)

  3. (5,3)

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given,

$y^2-10y+x+22=0$

$\Rightarrow x=-y^2+10y-22$

$x=-\left(y-5\right)^2+3$

$x-3=-\left(y-5\right)^2$

$-\left(x-3\right)=\left(y-5\right)^2$

$4\left(-\frac{1}{4}\right)\left(x-3\right)=\left(y-5\right)^2$

$\left(h,\:k\right)=\left(3,\:5\right),\:p=-\frac{1}{4}$

Vertex of parabola $(3,5)$



Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

The centre of the hyperbola 9x$^2$ - 36 x - 16y$^2$ + 96y - 252 = 0 is

  1. $(2,3)$
  2. $(-2,-3)$
  3. $(-2, 3)$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given,

$9x^2-36x-16y^2+96y-252=0$

$9x^2-36x-16y^2+96y=252$

$9\left(x^2-4x\right)-16\left(y^2-6y\right)=252$

$\left(x^2-4x\right)-\dfrac{16}{9}\left(y^2-6y\right)=28$

$\dfrac{1}{16}\left(x^2-4x\right)-\dfrac{1}{9}\left(y^2-6y\right)=\dfrac{7}{4}$

$\dfrac{1}{16}\left(x-2\right)^2-\dfrac{1}{9}\left(y-3\right)^2=\dfrac{7}{4}+\dfrac{1}{16}\left(4\right)-\dfrac{1}{9}\left(9\right)$

$\dfrac{\left(x-2\right)^2}{16}-\dfrac{\left(y-3\right)^2}{9}=1$

$\dfrac{\left(x-2\right)^2}{4^2}-\dfrac{\left(y-3\right)^2}{3^2}=1$

Center $(h,k)=(2,3)$
Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

Find the locus of a point which moves so that the difference of its distances from the points, $(5, 0)$ and $(-5, 0)$ is $2$ is:

  1. $\dfrac{x^2}{1}+\dfrac{y^2}{24}=1$
  2. $\dfrac{x^2}{24}+\dfrac{y^2}{1}=1$
  3. $\dfrac{x^2}{24}-\dfrac{y^2}{2}=1$
  4. $\dfrac{x^2}{1}-\dfrac{y^2}{24}=1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The locus is nothing but hyperbola.
Difference of distance of a point from foci $=2a$ 

Given distance is $2 \Rightarrow a=1$
Distance between foci $=2ae=2\sqrt{a^2+b^2}=\sqrt{(5+5)^2}$
                                                 $\Rightarrow a^2+b^2 =25$
                                                  $\Rightarrow b^2=24$
Therefore, locus is $\dfrac{x^2}{1}-\dfrac{y^2}{24}=1$ 

Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

If $e$ and $e'$ be the eccentricities of two conics $S$ and $S'$ such that $\displaystyle e^{2}+(e')^{2}= 3,$  then both $S$ and $S'$ are

  1. Ellipses

  2. Parabolas

  3. Hyperbolas

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

For a parabola the eccentricity is $1$

$\therefore e^2 + e'^2 = 1 + 1 = 2$
For an ellipse the eccerntricity is less than $1$
$\therefore$ for a hyperbola the eccentricity is greater than $1$
So, the conics can be hyperbolas
Hence, hyperbola correct.