Tag: hyperbola

Questions Related to hyperbola

Consider
the set of hyperbola $xy = {\text{ }}K,{\text{ K}} \in {\text{R,}}$  let ${e _1}$  be eccentricity
when $K = \sqrt {2017} $  and ${e _2}$ be the
eccentricity when $K = \sqrt {2018} $ , then ${e _1} = {e _2}$   is equal to 

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

  1. -1

  2. 0

  3. 2

  4. 1

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Correct Option: A

The exhaustive interval of $\lambda$ for which the equation $\dfrac{x^2}{(\lambda^2-2\lambda-3)}+\dfrac{y^2}{\lambda^2+2\lambda-8}=1$ represents a hyperbola is 

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

  1. $ \lambda \varepsilon (- \infty, -4) \cup (3, \infty)$

  2. $ \lambda \varepsilon (-4, -1) \cup (2, 3)$

  3. $ \lambda \varepsilon (- \infty, -1) \cup (2, \infty)$

  4. $ \lambda \varepsilon (-4,-1)$

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Correct Option: A

Length of the latus rectum of the hyperbola $xy=c^{2}$, is

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

  1. $2c$

  2. $\sqrt{2}c$

  3. $2\sqrt{2}c$

  4. $4c$

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Correct Option: C

If area of quadrilateral formed by tangents drawn at ends of latus rectum of hyperbola $\dfrac { { x }^{ 2 } }{ { a }^{ 2 } } -\dfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1$ is equal to square of distance between centre and one focus of hyperbola,then ${ e }^{ 3 }$ is (e is eccentricity of hyperbola)

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

  1. $2\sqrt { 2 } $

  2. 2

  3. 3

  4. 8

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Correct Option: A

Eccentricity of a hyperbola is always less than 1.

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

  1. True

  2. False

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Correct Option: B
Explanation:

Standard equation of the hyperbola is $\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}$

The eccentricity of the hyperbola is given by
$e=\sqrt { 1+\dfrac { { b }^{ 2 } }{ { a }^{ 2 } }  }$ which is always greater than $1$.
Thus, the given statement is false.
Hence, option B is correct.

Which of the following equations does not represent a hyperbola?

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

  1. $xy = 4$

  2. $
    \dfrac{1}
    {x^2} + \dfrac{1}
    {y^2} = \dfrac{1}
    {4}
    $

  3. $
    x^2 - xy + y^2 = 4
    $

  4. $
    x^2 - 4xy + 3y^2 = 1
    $

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Correct Option: A

Equation of the latus rectum of the hyperbola $(10x - 5)^{2} + (10y - 2)^{2} = 9(3x + 4y - 7)^{2}$ is

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

  1. $y - 1/5 =-3/4(x - 1/2)$

  2. $x - 1/5 =-3/4(y - 1/2)$

  3. $y + 1/5 =-3/4(x + 1/2)$

  4. $x + 1/5 =-3/4(y + 1/2)$

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Correct Option: A

The equation $\frac{x^2}{1-k}-\frac{y^2}{1+k}=1$, $k<1$ represents 

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

  1. $circle$

  2. $ellipse$

  3. $hyperbola$

  4. $none$

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Correct Option: C
Explanation:
Equating the above equation with the second-degree equation
$A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0$ with $\dfrac{{x}^{2}}{1-k}-\dfrac{{y}^{2}}{1+k}=1$
we get $A=\dfrac{1}{1-k}, B=0, C=\dfrac{1}{1+k},D=0,E=0$ and $F=-1$
$(i)$For the second degree equation to represent a circle , the coefficients must satisfy the discriminant condition ${B}^{2}-4AC=0$ and also $A=C$
$\Rightarrow -4\times \dfrac{1}{1-k}\times \dfrac{1}{1+k}=0$
$\Rightarrow \dfrac{1}{1-{k}^{2}}=0$
This case does not exist
$(ii)$For the second degree equation to represent a ellipse , the coefficients must satisfy the discriminant condition ${B}^{2}-4AC<0$ and also $A\neq C$
$\Rightarrow -4\times \dfrac{1}{1-k}\times \dfrac{1}{1+k}<0$
$\Rightarrow \dfrac{1}{1-{k}^{2}}>0$
$\Rightarrow 1-{k}^{2}<0$
$\Rightarrow -{k}^{2}<-1$
$\Rightarrow {k}^{2}>1$ does not exist since it is given that $k<1$
$(iii)$For the second degree equation to represent a hyperbola, the coefficients must satisfy the discriminant condition ${B}^{2}-4AC>0$ and also $A\neq C$
$\Rightarrow -4\times \dfrac{1}{1-k}\times \dfrac{1}{1+k}>0$
$\Rightarrow \dfrac{1}{1-{k}^{2}}<0$
$\Rightarrow 1-{k}^{2}>0$
$\Rightarrow -{k}^{2}>-1$
$\Rightarrow {k}^{2}<1$ 
$\therefore k<1$
Hence the above equation represents a hyperbola.

The equation $\displaystyle\frac{x^2}{10-\lambda}+\frac{y^2}{6-\lambda}=1$ represents

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

  1. a hyperbola if $\lambda < 6$

  2. an ellipse if $\lambda>6$

  3. a hyperbola if $6 < \lambda < 10$

  4. an ellipse if $0 < \lambda < 6$

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Correct Option: C,D
Explanation:

The general equation of an ellipse is $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ and that of a hyperbola is $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$
Using this, we get that the above equation is an ellipse if $10 - \lambda > 0$ and $6 - \lambda > 0$. The combined solution gives $\lambda < 6.$
For a hyperbola, the coefficient of $x^2$ and $y^2$ must be of opposite sign. Hence, 
 $10 - \lambda > 0$ and $6 - \lambda < 0$ which gives $6 < \lambda < 10$

The point to which the axes are to be translated to eliminate $x$ and $y$ terms in the equation $3x^{2}-4xy-2y^{2}-3x-2y-1=0$ is 

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

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

  2. $(-4,\dfrac{3}{2})$

  3. $ (-2,3)$

  4. $ (2,3)$

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Correct Option: A
Explanation:
Given equation is $3x^{2}-4xy-2y^{2}-3x-2y-1=0$

Let $\left({x} _{1},{y} _{1}\right)$ be a point to which the origin is shifted by translation

Let $\left(X,Y\right)$ be the new coordinates of the point $\left(x,y\right)$

$\therefore\,$ the equations of the transformation are $x=X+{x} _{1},\,y=Y+{y} _{1}$

Now the transformed equation is 
$3{\left(X+{x} _{1}\right)}^{2}-4\left(X+{x} _{1}\right)\left(Y+{y} _{1}\right)-2{\left(Y+{y} _{1}\right)}^{2}-3\left(X+{x} _{1}\right)-2\left(Y+{y} _{1}\right)-1=0$

$\Rightarrow\,3\left({X}^{2}+2X{x} _{1}+{{x} _{1}}^{2}\right)-4\left(XY+X{y} _{1}+Y{x} _{1}+{x} _{1}{y} _{1}\right)-2\left({Y}^{2}+{{y} _{1}}^{2}+2Y{Y} _{1}\right)-3X-{x} _{1}-2Y-2{y} _{1}-1=0$

$\Rightarrow\,3{X}^{2}+3{{x} _{1}}^{2}+6X{x} _{1}-4X{y} _{1}-4{x} _{1}Y-4{x} _{1}{y} _{1}-2{Y}^{2}-2{{y} _{1}}^{2}+4Y{y} _{1}-3X-3{x} _{1}-2Y-2{y} _{1}-1=0$

$\Rightarrow\,\left(3{X}^{2}-4XY-2{Y}^{2}\right)+\left(3{{x} _{1}}^{2}-2{{y} _{1}}^{2}-4{x} _{1}{y} _{1}-3{x} _{1}-2{y} _{1}-1\right)+2X\left(3{x} _{1}-2{y} _{1}-\dfrac{3}{2}\right)+2Y\left(-2{x} _{1}+2{y} _{1}-1\right)=0$

Solving the first degree terms,we have
$3{x} _{1}-2{y} _{1}=\dfrac{3}{2}$

$-2{x} _{1}+2{y} _{1}=1$

Adding the above equations, we get
$3{x} _{1}-2{y} _{1}-2{x} _{1}+2{y} _{1}=\dfrac{3}{2}+1$

$\Rightarrow\,{x} _{1}=\dfrac{5}{2}$

From equation ,$-2{x} _{1}+2{y} _{1}=1$

$\Rightarrow\,2{y} _{1}=1+2{x} _{1}=1+2\times\dfrac{5}{2}=1+5=6$

$\Rightarrow\,{y} _{1}=\dfrac{6}{2}=3$

$\therefore\,\left({x} _{1},{y} _{1}\right)=\left(\dfrac{5}{2},3\right)$


Hence the point is $\left(\dfrac{5}{2},3\right)$