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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) 
$\Rightarrow \ C:x^2 +3xy+2y^2+2x+3y+2=0$