Tag: conjugate hyperbola

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Questions Related to conjugate hyperbola

Multiple choice conjugate hyperbola hyperbola conic section maths

The equation of the conjugate axis of the hyperbola $\frac{{{{\left( {y - 2} \right)}^2}}}{9} - \frac{{{{\left( {x + 3} \right)}^2}}}{{16}} = 1$ is

  1. $y=2$
  2. $y=6$
  3. $y=8$
  4. $y=3$
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A Correct answer
Multiple choice conjugate hyperbola hyperbola conic section maths

If variable has its interceptson the coordinates axes $e$ and $e'$ where $e/2$ and $e'/2$ are the eccentricities of hyperbola and conjugate hyperbola, Then the line always touches the circle $x^{2}+y^{2}=r^{2}$, where $r=$ 

  1. $1$
  2. $2$
  3. $3$
  4. $Cannot\ be\ decided$
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A Correct answer
Explanation

If the line x/e + y/e' = 1 touches x^2 + y^2 = r^2, the perpendicular distance from the origin to the line must equal r. Using the formula for a line touching a circle, r = 1/sqrt(1/e^2 + 1/e'^2). Given the relationship between eccentricities of conjugate hyperbolas 1/e^2 + 1/e'^2 = 1, we find r = 1.

Multiple choice conjugate hyperbola hyperbola conic section maths

Let $e$ be the eccentricity of a hyperbola $\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$, and $f(e)$ be the eccentricity of hyperbola $-\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$, then $\displaystyle \int _{ 1 }^{ 3 } \underbrace { fff.....f\left( e \right)  } _{ n\quad times } de$ is equal to

  1. $2$, if $n$ is even
  2. $4$, if $n$ is even
  3. $2\sqrt{2}$, if $n$ is odd
  4. $4\sqrt{2}$, if $n$ odd
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Explanation

The eccentricity of a hyperbola and its conjugate are related by 1/e^2 + 1/e'^2 = 1. Applying the function f(e) repeatedly toggles between the two eccentricities. The integral of the constant values over the interval [1, 3] yields the result.

Multiple choice conjugate hyperbola hyperbola conic section maths

$e _{1}$ and $e _{2}$ are respectively the eccentricities of a hyperbola and its conjugate then  $\dfrac{1}{e^{2} _{1}}$+$\dfrac{1}{e^{2} _{2}}$=1.

  1. True

  2. False
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A Correct answer
Explanation
Let $ \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$     $.......(1)$
And 
$\dfrac{y^2}{b^2}-\dfrac{x^2}{a^2}=1$     $.........(2)$ are two hyperbola conjugate to each other.

Also let, $e _1$ and $e _2$ are the eccentricities of $(1)$ and $(2)$ respectively.
Then, 
$e _1^2=1+\dfrac{b^2}{a^2}$ and $e _2^2=1+\dfrac{a^2}{b^2}$

Therefore,
$\Rightarrow \dfrac{1}{e _1^2}+\dfrac{1}{e _2^2}$

$\Rightarrow \dfrac{1}{1+\dfrac{b^2}{a^2}}+\dfrac{1}{1+\dfrac{a^2}{b^2}}$

$\Rightarrow \dfrac{a^2}{a^2+b^2}+\dfrac{b^2}{a^2+b^2}$

$\Rightarrow \dfrac{a^2+b^2}{a^2+b^2}$

$\Rightarrow 1$

Hence, proved.

Multiple choice conjugate hyperbola hyperbola conic section maths

The area of quadrilateral formed by focil hyperbola $\dfrac{x^2}{4}-\dfrac{y^2}{3}=1$ & its conjugate hyperbola is

  1. $14$
  2. $24$
  3. $12$
  4. $10$
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A Correct answer
Explanation

For x^2/4 - y^2/3 = 1, a=2, b=sqrt(3). Foci are (+-ae, 0). e = sqrt(1 + 3/4) = sqrt(7)/2. Foci are (+-sqrt(7), 0). For the conjugate hyperbola y^2/3 - x^2/4 = 1, foci are (0, +-be') where e' = sqrt(1 + 4/3) = sqrt(7)/sqrt(3). Foci are (0, +-sqrt(7)). The quadrilateral vertices are (+-sqrt(7), 0) and (0, +-sqrt(7)). Area = 1/2 * d1 * d2 = 1/2 * (2*sqrt(7)) * (2*sqrt(7)) = 14.

Multiple choice conjugate hyperbola hyperbola conic section maths

The eccentricity of the hyperbola length of whose conjugate axis is equal to half of the distance betweet the foci is 

  1. $\dfrac{4}{\sqrt{3}}$
  2. $\dfrac{4}{3}$
  3. $\dfrac{2}{\sqrt{3}}$
  4. $\sqrt{3}$
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A Correct answer
Multiple choice conjugate hyperbola hyperbola conic section maths

Assertion(A): lf the lines $3x+y+p=0$ and $2x+5y-3=0$ are conjugate with respect to $3x^{2}-2y^{2}=6$ then $\mathrm{p}=1$

Reason(R): lf the lines $l _{1}x+m _{1}y+n _{1}=0$ and $l _{2}x+m _{2}y+n _{2}=0$ are conjugate with respect to the hyperbola $\mathrm{S}=0$ is $a^{2}l _{1}l _{2}+b^{2}m _{1}m _{2}=n _{1}n _{2}$


  1. Both A and R are true and R is the correct

    explanation of A.

  2. Both A and R are true but R is not correct

    explanation of A.

  3. A is true but R is false

  4. A is false but R is true
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

 lf the lines $l _{1}x+m _{1}y+n _{1}=0$ and $l _{2}x+m _{2}y+n _{2}=0$ are conjugate with re- spect to the hyperbola $\mathrm{S}=0$ is $a^{2}l _{1}l _{2}-b^{2}m _{1}m _{2}=n _{1}n _{2}$
then $2(3)(2)-3(1)(5)=-3p$
therefore, $p=1$