Tag: conic section

Quizzes (8) Questions (30)

Questions Related to conic section

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
Reveal answer Fill a bubble to check yourself
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. $10$
Reveal answer Fill a bubble to check yourself
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. $\sqrt{3}$
Reveal answer Fill a bubble to check yourself
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, $p=1$

Multiple choice conjugate hyperbola hyperbola conic section maths

The equation to the conjugate hyperbola of $2x^{2}-3y^{2}-4x+6y-15=0$ is 

  1. $2x^{2}-3y^{2}-4x+6y+13=0$
  2. $2x^{2}-3y^{2}-4x+6y-1=0$
  3. $2x^{2}-3y^{2}-4x+6y+15=0$
  4. $2x^{2}-3y^{2}-4x+6y-8=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The conjugate hyperbola of ax^2 + by^2 + 2gx + 2fy + c = 0 is ax^2 + by^2 + 2gx + 2fy + c' = 0, where the center (h, k) is the same. For 2x^2 - 3y^2 - 4x + 6y - 15 = 0, center is (1, -1). The constant term changes such that the new equation is 2x^2 - 3y^2 - 4x + 6y + k = 0. The conjugate hyperbola has the same asymptotes, so the constant term is adjusted.

Multiple choice conjugate hyperbola hyperbola conic section maths

If the hyperbolas, $ x^2+3xy+2y^2+2x+3y+2=0 $ and $ x^2+3xy+2y^2+2x+3y+c=0 $ are conjugate of each other, the value of $c$ is equal to

  1. $-2$
  2. $4$
  3. $0$
  4. $1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The given hyperbola is $x^2+3xy+2y^2+2x+3y+2=0$         ...(1)

We already know that the equation of the asymptote of a hyperbola differs from the hyperbola by a constant.
$\therefore$ Let $x^2+3xy+2y^2+2x+3y+k=0$            ...(2)
be the equation of the asymptotes of the given hyperbola.
Hence, equation (2) must represent a pair of straight lines, the condition for which is
$abc+2fgh-af^2-bg^2-ch^2 = 0$
$\Rightarrow (1)(2)(k)+2(\cfrac 32)(1)(\cfrac 32)-1(\cfrac 32)^2-2(1)^2-k(\cfrac 32)^2=0$
$\Rightarrow 2k+\cfrac 92-\cfrac 94-2-\cfrac 94 k=0$
$\Rightarrow k=1$
Therefore, the asymptotes are given by $x^2+3xy+2y^2+2x+3y+1=0$ .

The equation of conjugate hyperbola is $2A-H=0$
where, $A$ is the equation of asymptotes
            $H$ is the equation of given hyperbola
$2(x^2+3xy+2y^2+2x+3y+1) -$$(x^2+3xy+2y^2+2x+3y+2)=0$
$\therefore x^2+3xy+2y^2+2x+3y=0$

Hence, $c=0$.

Multiple choice conjugate hyperbola hyperbola conic section maths

If the line $lx+my+n=0$ meets the hyperbola $\displaystyle \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ at the extermities of a pair of conjugate diameters, then

  1. $a^{2}l^{2}-b^{2}m^{2}=0$
  2. $a^{2}l^{2}-b^{2}m^{2}=1$
  3. $a^{2}l^{2}-b^{2}m^{2}=2$
  4. $a^{2}l^{2}-b^{2}m^{2}=3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let $ \theta$ and $\phi$ be the essentric angles of the conjugate diameters, 
Then,
$\theta +\phi =\dfrac{\pi}{2}$
$(a\sec \theta,b\tan \theta)   , (a\sec \phi, b \tan\phi)$
$\Rightarrow (a \sec \theta,b \tan \theta)   , (a\ \text{cosec }\theta ,b \cot \theta)$

$(y-b\tan \theta)=\dfrac{b\cot \theta-b\tan\theta}{a\ \text{cosec }\theta-a\sec\theta}(x-a\sec\theta)$

$\Rightarrow ya-ab\tan \theta=xb(\sin\theta+\cos \theta)-ab(\tan\theta+1)$
$ya=xb(\sin\theta+\cos\theta) - ab$
$ ya - xb (\sin\theta+\cot\theta)+ab=0$
$lx+my+n=0$

Comparing both, we get

$\dfrac{a}{m}=\dfrac{-b(\sin\theta+\cos\theta)}{l}=\dfrac{ab}{n}$

On elliminating $\theta $, we can get
$a^{2}l^{2}-b^{2}m^{2}=0$

Multiple choice conjugate hyperbola hyperbola conic section maths

Find the equation to the hyperbola,conjugate to the hyperbola $ 2x^2+3xy-2y^2-5x+5y+2=0 $.

  1. $ 2x^2+3xy-2y^2-5x+5y-8=0 $
  2. $ x^2+3xy-y^2-5x+5y-8=0 $
  3. $ x^2+3xy-y^2-5x+5y+8=0 $

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

Let Asymptotes :
$ 2x^2+3xy-2y^2-5x+5y+\lambda=0 $.
$ \therefore abc+2fgh-af^2-bg^2-ch^2=0 $
$ \lambda=-5 $
Equation Hyperbola + Conjugate Hyperbola$=$2 (Asympototes)
$ \therefore $ Conjugate Hyperbola$=$2 (Asymptotes) $- $Hyperbola
Therefore, equation of conjugate hyperbola is $  2x^2+3xy-2y^2-5x+5y-8=0 $

Multiple choice conjugate hyperbola hyperbola conic section maths

The equation of a hyperbola, conjugate to the hyperbola $x^2+3xy+2y^2+2x+3y=0$ is?

  1. $x^2+3xy+2y^2+2x+3y+1=0$
  2. $x^2+3xy+2y^2+2x+3y+2=0$
  3. $x^2+3xy+2y^2+2x+3y+3=0$
  4. $x^2+3xy+2y^2+2x+3y+4=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$H:x^2+3xy +2y^2+2x+3y=0$
Let the pair of asympt at is be
$A:x^2 +3xy+2y^2+2x+3y+k=0$
$\therefore \ $ It represent a apir of straight lines, satisfying the condition :
$abc+2fgh-af^2 -bg^2-ch^2=0$
$\Rightarrow \ 1\times 2\times k+2\left (\dfrac {3}{2}\right) \times 1\times \left (\dfrac {3}{2}\right) -1 \left (\dfrac 32\right)^2 -2(1)^2 -k\left (\dfrac 32 \right)^2 =0$
$\Rightarrow \ 2k+\dfrac {9}{4}-2-\dfrac {9}{4}k=0\ \Rightarrow \ k=1$
$\Rightarrow \ A^2 .x^2 +3xy+2y^2 +2x+3y+1=0$
As $H+C=2A\ \Rightarrow \ C$ (conjugate $=2A-H$ hypergate) 
$\Rightarrow \ C:x^2 +3xy+2y^2+2x+3y+2=0$