Tag: hyperbola

Questions Related to hyperbola

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

conjugate hyperbola hyperbola conic section maths

  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. $2x^{2}-3y^{2}-4x+6y-8=0$

Show answer
Correct Option: A

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

conjugate hyperbola hyperbola conic section maths

  1. $-2$

  2. $4$

  3. $0$

  4. $1$

Show answer
Correct Option: C
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, $c=0$.

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

conjugate hyperbola hyperbola conic section maths

  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. $a^{2}l^{2}-b^{2}m^{2}=3$

Show answer
Correct Option: A
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
$a^{2}l^{2}-b^{2}m^{2}=0$

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

conjugate hyperbola hyperbola conic section maths

  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

Show answer
Correct Option: A
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 $

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

conjugate hyperbola hyperbola conic section maths

  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. $x^2+3xy+2y^2+2x+3y+4=0$

Show answer
Correct Option: B
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) 
$\Rightarrow \ C:x^2 +3xy+2y^2+2x+3y+2=0$

The equation of the curve which is such that the protion of the axis of x cut off between the origin and tangent at any point is proportional to the ordinate of that point is _______________.

normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

  1. $\log x = b y ^ { 2 } + a$

  2. $x = y ( a + b \log y )$

  3. $x = y ( b - a \log y )$

  4. None of these

Show answer
Correct Option: C

The hyperbola $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$, normals are drawn to curve $\left( {{{\left( {\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}}} \right)}^2} - 1} \right)\left( {\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}}} \right) = 0$.
Find the sum;  of abscissa of foot of all such normals.

normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

  1. $\frac{{6{a^2}h}}{{\left( {{a^2} + {b^2}} \right)}}$

  2. $\frac{{8{a^2}h}}{{\left( {{a^2} + {b^2}} \right)}}$

  3. $\frac{{6a{h^2}}}{{\left( {{a^2} + {b^2}} \right)}}$

  4. $\frac{{8a{h^2}}}{{\left( {{a^2} + {b^2}} \right)}}$

Show answer
Correct Option: B

If the straight line $(a - 2) x - by + 4 = 0$ is normal to the hyperbola $xy = 1$ then which of the followings does not hold?

normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

  1. $a > 1, b > 0$

  2. $a > 1, b < 0$

  3. $a < 1, b < 0$

  4. $a < 1, b > 0$

Show answer
Correct Option: A,C
Explanation:

Every normal to $xy = 1$ must have positive slope as $\dfrac {-dx}{dy} = x^{2}$. So $\dfrac {a - 1}{b} > 0$.

The normal to the hyperbola $4x^2-9y^2=36$ meets the axes in $M$ and $N$ and the lines $MP$, $NP$ are drawn right angles at the axes. The locus of $P$ is the hyperbola 

normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

  1. $9x^2-4y^2=169$

  2. $4x^2-9y^2=169$

  3. $3x^2-4y^2=169$

  4. $None\ of\ these$

Show answer
Correct Option: D
Explanation:

$\dfrac {x^2}9-\dfrac {y^2}4=1$.Let $P(x _1, y _1)$ be the point on hyperbola.

Eqn of the normal is$\dfrac {a^2x}{x _1}-\dfrac {b^2y}{y _1}=a^2b^2\M=\left( \dfrac { { a }^{ 2 }-{ b }^{ 2 } }{ { a }^{ 2 } }  \right) { x } _{ 1 }=x\N=\left( \dfrac { { a }^{ 2 }-{ b }^{ 2 } }{ { a }^{ 2 } }  \right) { y } _{ 1 }=y\P=(x, y)$$x _1=\dfrac {a^2(x)}{a^2-b^2}$ $(x _1, y _1)$ lies at hyperbola.$y _1=\dfrac {b^2(y)}{a^2-b^2}$ $(x _1, y _1)$ lies at hyperbola.Now, $a^2=9, b^2=4$Therefore, $x _1=\dfrac {9x}5, y _1=\dfrac {4y}5$$\dfrac {x _1^2}9-\dfrac {y _1^2}4=1\\left(\dfrac {9x}5\right)^2\dfrac {x _1^2}9-\left(\dfrac {4y}5\right)^2\dfrac {y _1^2}4=1\\implies 9x^2-4y^2=25$

A normal to the hyperbola, $4x^2-9y^2=36$ meets the co-ordinate axes x and y at A and B, respectively. If the parallelogram $OABP$($O$ being the origin) is formed, then the locus of $P$ is?

normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

  1. $4x^{2}+9y^{2}=121$

  2. $9x^{2}+4y^{2}=169$

  3. $4x^{2}-9y^{2}=121$

  4. $9x^{2}-4y^{2}=169$

Show answer
Correct Option: D