Tag: hyperbola

Questions Related to hyperbola

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

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

  1. a straight line or hyperbola

  2. a straight line or parabola

  3. a parabola or hyperbola

  4. circles

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Correct Option: C

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

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

  1. $\ \sqrt { 1+k } $

  2. $\ \sqrt { 1-k } $

  3. $\ \sqrt {2 } $

  4. $\2 \sqrt {2 } $

Show answer
Correct Option: C
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 }  } } $

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

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

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

  1. $(5\sqrt3,0)$

  2. $(5,0)$

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

  4. none of these

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Correct Option: A

The foci of the hyperbola $xy=4$ are 

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

  1. $(2\surd{2},2\surd{2})$

  2. $(-2\surd{2},-2\surd{2})$

  3. $(-2\surd{2},2\surd{2})$

  4. $None of these$

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Correct Option: A

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 

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

  1. $(3,\infty)$

  2. $(1,3/2)$

  3. $(2,3)$

  4. $(-3,-2)$

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Correct Option: A

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

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

  1. $\dfrac{13}{6}$

  2. $\dfrac{32}{3}$

  3. $\dfrac{8}{3}$

  4. $\dfrac{4}{3}$

Show answer
Correct Option: B
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.

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

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

  1. (3,4)

  2. (3,5)

  3. (5,3)

  4. none of these

Show answer
Correct Option: B
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 $(3,5)$



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

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

  1. $(2,3)$

  2. $(-2,-3)$

  3. $(-2, 3)$

  4. none of these

Show answer
Correct Option: A
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 $(h,k)=(2,3)$

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:

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

  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. $\dfrac{x^2}{1}-\dfrac{y^2}{24}=1$

Show answer
Correct Option: D
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 $\dfrac{x^2}{1}-\dfrac{y^2}{24}=1$ 

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

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

  1. Ellipses

  2. Parabolas

  3. Hyperbolas

  4. None of these

Show answer
Correct Option: C
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.