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Asymptote - class-XI

Description: asymptote
Number of Questions: 52
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Tags: two dimensional analytical geometry-ii maths conic sections mathematics and statistics hyperbola
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Asymptotes of the hyperbola $xy=4x+3y$ are 

  1. x=3, y=4

  2. x=4, y=3

  3. x=2, y=6

  4. x=6, y=2


Correct Option: A
Explanation:

correct answer is A.

Given,

$xy=4x+3y$
$\Rightarrow (x-y)=3(y-4)+12$
$\Rightarrow (x-4)(y-4)=12$
$\therefore$ joint equation of asymptotes is $(x-3)(y-4)=0$
hence,
$x-3=0$
$x=3$
and,
$y-4=0$

$y=4$

The angle between the asymptotes to the hyperbola $\dfrac { { x }^{ 2 } }{ 16 } -\dfrac { { y }^{ 2 } }{ 9 } =1$ is

  1. $\pi -2\tan ^{ -1 }{ \left( \dfrac { 3 }{ 4 } \right) } $

  2. $\pi -2\tan ^{ -1 }{ \left( \dfrac { 4 }{ 3 } \right) } $

  3. $2\tan ^{ -1 }{ \left( \dfrac { 3 }{ 4 } \right) } $

  4. $2\tan ^{ -1 }{ \left( \dfrac { 4 }{ 3 } \right) } $


Correct Option: C
Explanation:

Fact: Angle between asymptotes of hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ is $2\tan^{-1}\left(\dfrac{b}{a}\right)$


Hence angle between the asymptotes to the hyperbola $\dfrac{x^2}{16}-\dfrac{y^2}{9}=1$ is $2\tan^{-1}\left(\dfrac{3}{4}\right)$

The asymptote of the hyperbole $\dfrac {x^{2}}{a^{2}-y^{2}b^{2}}=1$ from with any tangent to the hyperbola a triangle whose area is $a^{2}tan\lambda$ in magnitude then its eccentricity is ?

  1. $Sec\lambda$

  2. $csc\lambda$

  3. $sec^{2}\lambda$

  4. $csc^{2}\lambda$


Correct Option: C

Differential equation of all hyperbolas which pass through the origin, and have their asymptotes parallel to the coordinate axes is?

  1. $xy\dfrac{d^2y}{dx^2}-2x\left(\dfrac{dy}{dx}\right)^2+2y=0$

  2. $xy\dfrac{d^2y}{dx^2}-2\left(\dfrac{dy}{dx}\right)^2+2y\left(\dfrac{dy}{dx}\right)=0$

  3. $xy\left(\dfrac{d^2y}{dx^2}\right)-2x\left(\dfrac{dy}{dx}\right)^2+2y\dfrac{dy}{dx}=0$

  4. $xy\dfrac{d^2y}{dx^2}+2x\left(\dfrac{dy}{dx}\right)^2+y\left(\dfrac{dy}{dx}\right)=0$


Correct Option: A

Area of triangle formed by the tangent at one vertex and asymptotes of the hyperbola xy=2

  1. 2sq. units

  2. 3 units

  3. 1 sq. unit

  4. none of these


Correct Option: A

The product of perpendiculars drawn from any point of a hyperbola with principal axes $2a$ and $2b$ upon its asymptotes is equal to:

  1. $\frac{a^2b^2}{a^2+b^2}$

  2. $\frac{a^2 +b^2}{a^2b^2}$

  3. $\frac{ab}{a^2+b^2}$

  4. $\frac{ab(a+b)}{\sqrt a+\sqrt b}$


Correct Option: A

The angle between the asymptotes of the hyperbola $24x^2 - 8y^2 = 27$ is 

  1. $90^o$

  2. $60^o$

  3. $120^o$

  4. $45^o$


Correct Option: C
Explanation:
$24x^{2}-8y^{2}=27$

divide the above equation with 27 
$\displaystyle \frac{n^{2}}{\dfrac{27}{24}}=\frac{y^{2}}{\dfrac{27}{8}}=1$

$\displaystyle \frac{n^{2}}{\dfrac{9}{8}}-\frac{y^{2}}{\dfrac{27}{8}}=1$

$\displaystyle a^{3}=\frac{9}{8}, b^{2}=\frac{27}{8}$

$\displaystyle a=\frac{3}{2\sqrt{2}},b=\frac{3\sqrt{3}}{2\sqrt{2}}$

let $ 2\alpha $ be the angle between asymptotes 

$2\alpha =2\tan^{-1}\dfrac{b}{a}$

$\displaystyle =2\tan^{-1}\frac{\frac{3\sqrt{3}}{2\sqrt{2}}}{\frac{3}{2\sqrt{2}}}$

$=2\tan^{-1}\sqrt{3}$

$= 2\times \dfrac{\pi}{3}$

$\displaystyle =\frac{2\pi }{3}$  or  $120$
Evaluate the following definite integral:
$\displaystyle \int _{0}^1 \dfrac {1-x}{1+x}dx$
  1. $2\log 2+1$

  2. $2\log 2$

  3. $2\log 2-1$

  4. $\log 4$


Correct Option: C
Explanation:

$I=\displaystyle \int _{0}^1 \dfrac {1-x}{1+x}dx$ 

$=\displaystyle \int _{0}^1 \dfrac {2-(1+x)}{1+x} dx$ 

$=\displaystyle \int _{0}^1 \left (\dfrac {2}{1+x} -1 \right)dx $

$=\left [2\log (x+1)-x \right] _0^1$

$\Rightarrow \ I=(2\log 2-1)- (2\log 1-0)$ 

$=2\log 2-1$

If a line intersect a hyperbola at $(-2,-6)$ and $(4,2)$ and one of the asymtote at $(1,-2)$, then the centre of the hyperbola is

  1. $(7,6)$

  2. $(1,-2)$

  3. $(10,10)$

  4. $(-5,-10)$


Correct Option: A

Let product of distances of any point hyperbola (x+y-1) (x-y+3)= 60 to its asymptotes is 'K' then K is divisible by

  1. 2

  2. 3

  3. 4

  4. 5


Correct Option: A

If the cordinate of any point p on the hyperbola $9{x^2} - 16{y^2} = 144$ is produced to cut the asymptotes in the points Q and R. Then the product PQ.PR equals to:

  1. $9$

  2. $\dfrac{12}{5} $

  3. $\dfrac{144}{25}$

  4. $7$


Correct Option: C
Explanation:
$\dfrac{x^{2}}{16}-\dfrac{y^{2}}{9}=1$

Asymplote is $y=\pm\dfrac{3}{4}x$

Let us take $4y=3x, 4y=-3x$

consider a parametric point $(4\sec\theta, 3\tan\theta)$ on the parabola

$Q$ is intersection with $4y=3x$

then $PQ=\left|\dfrac{12\sec\theta-12\tan\theta}{\sqrt{4^{2}+3^{2}}}\right|$

$PQ=\left|\dfrac{12}{5}(\sec \theta-\tan\theta)\right|$

$R$ is intersection with $4y=-3x$

then $PR=\left|\dfrac{12\sec\theta+12\tan\theta}{\sqrt{4^{2}+3^{2}}}\right|$

$PQ.PR=\dfrac{144}{25}(\sec\theta+\tan\theta)(\sec\theta-\tan\theta)$

$=\dfrac{144}{25}(\sec^{2}\theta+\tan^{2}\theta)=\dfrac{144}{25}(1)$

`e`$=\dfrac{144}{25}$

The points of intersection of asymptotes with directrices lies on

  1. Auxillary circle

  2. Director circle

  3. Transverse axis

  4. Conjugate axis


Correct Option: A
Explanation:
  • Asymptotes of a hyperbola are the diagonals of the rectangle formed by the lines drawn through the extremities of each axis parallel to the other axis.
  • A perpendicular drawn from the foci on either asymptote meet it in the same points as the corresponding directrix and the common points of intersection lie on the auxiliary circle.

The area of the triangle formed by the asymptotes and any tangent to the hyperbola ${x}^{2}-{y}^{2}={a}^{2}$ is 

  1. ${4a}^{2}$

  2. ${3a}^{2}$

  3. ${2a}^{2}$

  4. ${a}^{2}$


Correct Option: A

If foci of hyperbola lie on $y=x$ and one of the asymptote is $y=2x$, then equation of the hyperbola, given that is passes through $(3, 4)$ is :

  1. $x^2-y^2-\dfrac {5}{2}xy+5=0$

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

  3. $2x^2+2y^2-5xy+10=0$

  4. None of these


Correct Option: C
Explanation:

Foci of hyperbola lie on $y=x$.
So, the equation of transverse axis is $y-x=0$.
Transverse axis of hyperbola bisects the asymptote
$\Rightarrow$ equation of other asymptote is $y=\dfrac{x}{2}$
or,$x=2y$
$\Rightarrow$ Equation of hyperbola is $(y-2x)(x-2y)+k=0$
Since, it passes through $(3, 4)$
$\Rightarrow k=-10$
Hence, required equation is
$2x^2+2y^2-5xy+10=0$

The combined equation of the asymptotes of the hyperbola $2{x}^{2}+5xy+2{y}^{2}+4x+5y=0$ is

  1. $2{x}^{2}+5xy+2{y}^{2}+4x+5y+2=0$

  2. $2{x}^{2}+5xy+2{y}^{2}+4x+5y-2=0$

  3. $2{x}^{2}+5xy+2{y}^{2}=0$

  4. None of these


Correct Option: A

The ordinate of any point P on the hyperbola, given by  $25x^2-16y^2=400$, is produced to cut its asymptotes in the points Q and R, then $QP.PR=5.$

  1. True

  2. False


Correct Option: B

If the x-y+4=0 and x+y+2=0 are asymptotes of a hyperbola , the its center is 

  1. (-3,1)

  2. (3,1)

  3. (-3,-1)

  4. (3,-1)


Correct Option: A

A chord $AB$ which bisected at $(1,1)$ is drawn to the hyperbola $7x^{2}+8xy-y^{2}-4=0$ with centre $C$. which intersects its asymptotes in $E$ and $F$. If equation of circumcricel of $\triangle CEF$ is $x^{2}+y^{2}-ax-by+c=0$, then value of $\dfrac{23(a-b+c)}{12}$ is equal to 

  1. $1$

  2. $2$

  3. $3$

  4. $4$


Correct Option: A

The product of the lengths of perpendiculars drawn from any point on the hyperbola $x^{2}-2y^{2}-2=0$ to its asymptotes is 

  1. 1/2

  2. 2/3

  3. 3/2

  4. 2


Correct Option: A

The angle between the asymptotes of the hyperbola $\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$, the length of whose latus rectum is $\dfrac{4}{3}$ and hyperbola passes through the point $(4,2)$ is :

  1. $\dfrac{\pi}{6}$

  2. $\dfrac{\pi}{2}$

  3. $\dfrac{\pi}{3}$

  4. $\dfrac{\pi}{4}$


Correct Option: A

The angle between the asymptotes of a hyperbola is $30^{o}$. The eccentricity of the hyperbola may be

  1. $\sqrt{3}\pm 1$

  2. $\sqrt{3}+1$

  3. $\pm\sqrt{2}$

  4. $none\ of\ these$


Correct Option: D

If the equation $3x^{2}+xy-y^{2}-3x+6y+2=0$ represents hyperbola then equation of the asymptotes is given by

  1. $3x^{2}+xy-y^{2}-3x+6y-9=0$

  2. $3x^{2}+xy-y^{2}-3x+6y-7=0$

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

  4. $none of these$


Correct Option: A

If e is the eccentricity of $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ and $'\Theta '$ be the angle between its asymptotes, then $cos(\Theta /2)$ is equal to,

  1. 1/2e

  2. 1/e

  3. $1/e^{2}$

  4. none of these


Correct Option: A

The equation of the line passing through the centre of a rectangle hyperbola is $x-y-1=0$. If one of its asymptotes is $3x-4x-6=0$, the equation of the other asymptote is $

  1. $4x+3y+17=0$

  2. $4x-3y+8=0$

  3. $3x-2y+15=0$

  4. $None of these$


Correct Option: A

if the product of the perpendicular distances from any point on the hyperbola$\frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\quad of\quad eccentrincity\quad e=\sqrt { 3 } $ on its asymptotes is equal to 6 then the length of the transverse axis of the hyperbola is;

  1. 3

  2. 6

  3. 8

  4. 12


Correct Option: A

if the product of the perpendicular distances from any point on the hyperbola $\frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1$ of eccentrincity $e=\sqrt { 3 } $ on the asymptotes is equal to 6 then the length of transverse axis of the hyperbola is

  1. 3

  2. 6

  3. 8

  4. 12


Correct Option: A

If $e$ is the eccentricity of $\dfrac {x^{2}}{a^{2}}-\dfrac {y^{2}}{b^{2}}=1$ and '$\theta $' be the angle between its asymptotes then $\cos (\theta /2)$ is equal to.

  1. $1/ 2e$

  2. $1/ e$

  3. $2/e^{2}$

  4. $none\ of\ these$


Correct Option: B

The asymptotes of the hyperbola $xy-3x+4y+2=0$

  1. $x=-4$

  2. $x=4$

  3. $y=-3$

  4. $y=3$


Correct Option: A,D
Explanation:

Since the equation of a hyperbola and its asymptotes differ in constant terms only. Therefore, the equations of asymptotes of the given hyperbola are given by $xy-3x+4y+k=0$

where $k$ is a constant to be determined  by the condition that $abc+2fgh-{ af }^{ 2 }-{ bg }^{ 2 }-{ ch }^{ 2 }=0$
i.e., $\displaystyle 0+2\times 2\times \left( \frac { -3 }{ 2 }  \right) \times \frac { 1 }{ 2 } -0-0-k\times { \left( \frac { 1 }{ 2 }  \right)  }^{ 2 }=0\Rightarrow k=-12$
$\because $ Asymptotes of the given hyperbola are $xy-3x+4y-12=0$ or $(x+4)(y-3)=0$
i.e., $x=-4$ and $y=3.$

If $x + 2 = 0$ and $y = 1$ are the equation of asymptotes of rectangular hyperbola passing through (1,0).Then which of the following is(are) not the equation(s) of hyperbola :

  1. $xy + 2y -1 = 0$

  2. $xy - 2y + 1 = 0$

  3. $xy - 2y - 1 = 0$

  4. $xy-x+2y+1=0$


Correct Option: A,B,C
Explanation:

Equation of hyperbola is of the form $(x+2)(y-1)=k$
Since, it passes through $(1,0)$
Therefore, $(1+2)(0-1)=k$
$\Rightarrow k=-3$
Therefore, equation of hyperbola is $xy-x+2y+1=0$

If ax + by + c = 0 and $\displaystyle \varphi \chi $ + my + n = 0 are asymptotes of a hyperbola, then: 

  1. $\displaystyle am\neq b\varphi $

  2. $\displaystyle \frac{am+b\varphi }{a\varphi +bm}\neq 0$

  3. $\displaystyle a\varphi \neq bm$

  4. none of these


Correct Option: A
Explanation:

Asymptotes of Hyperbola are Intersecting each other 


So, These line will be anti-parallel or intersecting

Condition for intersecting lines is $\dfrac{a}{b}\neq\dfrac{\varphi}{m}\Rightarrow am \neq b\varphi$

If $\theta$ is the angle between the asymptotes of the hyperbola $\displaystyle \frac{x^2}{a^2}\, -\, \displaystyle \frac{y^2}{b^2}\, =\, 1$ with eccentricity $e$, then $\sec \displaystyle  \frac{\theta}{2}$can be

  1. $e$

  2. $\dfrac{e}2$

  3. $\dfrac{e}3$

  4. $\displaystyle \frac{e}{\sqrt{e^2\, -\, 1}}$


Correct Option: A,D
Explanation:

$\tan\, \displaystyle \frac{\theta}{2}\, =\, \displaystyle \frac{b}{a}\,

\Rightarrow\, e^2\, -\, 1\, =\, \tan^2\, \displaystyle

\frac{\theta}{2}\, \Rightarrow\, \sec \displaystyle \frac{\theta}{2}\,

=\, e$
or $e^2\, -\, 1\, =\, \cot^2\, \displaystyle \frac{\theta}{2}\, \Rightarrow\, co\sec\, \displaystyle \frac{\theta}{2}\, =\, e$
$\Rightarrow\, \sec\, \displaystyle \frac{\theta}{2}\, =\, \displaystyle \frac{e}{\sqrt{e^2\, -\, 1}}$.

The asymptotes of a hyperbola are parallel to lines $2x + 3y = 0$ and $3x + 2y = 0.$ The hyperbola has its centre at $(1, 2)$ and it passes through $(5, 3).$ Find its equation.

  1. $(2x\, +\, 3y\, -\, 8) (3y\, +\, 2y\, -\, 7)\, =\, 154$

  2. $(2x\, +\, 3y\, -\, 7) (3y\, +\, 2y\, -\, 8)\, =\, 154$

  3. $(2x\, +\, 3y\, -\, 7) (3y\, +\, 2y\, -\, 8)\, =\, 127$

  4. $(2x\, +\, 3y\, -\, 8) (3y\, +\, 2y\, -\, 7)\, =\, 127$


Correct Option: A
Explanation:

let the equation of asymtotes be $2x+3y=a$ and $3x+2y=b$
both asymtotes intersect at centre $(1,2)$
Therefore, $a=2+3(2)=8$ and $b=3+2(2)=7$
now, the equation of hyperbola is of the form $(2x+3y-8)(3x+2y-7)=k$
It passes through $(5,3)$
Therefore, $(2(5)+3(3)-8)(3(5)+2(3)-7)=k$
Thus $k=154$

The asymptotes of the hyperbola $xy+3x+2y = 0$ are

  1. $x - 2 = 0$ and $y - 3 = 0$

  2. $x - 3 = 0$ and $y - 2 = 0$

  3. $x + 2 = 0$ and $y + 3 = 0$

  4. $x + 3 = 0$ and $y + 2 = 0$


Correct Option: C
Explanation:

Let equation of asymptotes be $xy\,+ \, 3x\, \, +\, 2y +\, \lambda$ = 0.
Then $abc\, +\, 2fgh\,-\, af^2\,-\, bg^2\,-\, ch^2\, =\, 0$
$\displaystyle \Rightarrow\, \frac{3}{2}\, -\, \frac{\lambda}{4}\, =\, 0\, \Rightarrow\, \lambda\, =\, 6$
$\therefore$ Equation of asymptotes is $xy +3x +2y + 6 = 0$
$\Rightarrow (x+ 2) (y +3) = 0\Rightarrow x+2=0$ and $y+3=0$

Find the asymptotes of the hyperbola $2x^2\, -\, 3xy\,- \, 2y^2\, +\, 3x\,- \, y\, +\, 8\, =\, 0$. Also find the equation to the conjugate hyperbola & the equation of the principal axes of the curve.

  1. $x - 2y + 1 = 0; 2x + y + 1 = 0; 2x^2\,- \, 3xy\, -\, 2y^2\, +\, 3x\,- \, y\,- \, 6\, =\, 0; 3x y + 2 = 0; x - 3y = 0$

  2. $x + 2y - 1 = 0; 2x + y + 1 = 0; 2x^2\,- \, 3xy\, -\, 2y^2\, +\, 3x\,- \, y\,+\, 6\, =\, 0; 3x y + 2 = 0; x + 3y = 0$

  3. $x - 2y + 1 = 0; 2x + y + 1 = 0; 2x^2\,- \, 3xy\, -\, 2y^2\, +\, 3x\,- \, y\,- \, 6\, =\, 0; 3x y + 2 = 0; x + 3y = 0$

  4. $x - 2y + 1 = 0; 2x - y + 1 = 0; 2x^2\,- \, 3xy\, -\, 2y^2\, +\, 3x\,- \, y\,+ \, 6\, =\, 0; 3x y - 2 = 0; x - 3y = 0$


Correct Option: C
Explanation:

Let equation of asymptotes are 
$2x^2\, \, -3xy\, \,- 2y^2\, +\, 3x\, \, -y\, +\, 8\, +\, \lambda\,=\, 0$
As it represents two straight lines
$\displaystyle \therefore\, -4(8\, +\, \lambda)\, +\, \frac{9}{4}\, -\, \frac{1}{2}\, +\, \frac{9}{2}\, -\, (8\, +\, \lambda) \frac{9}{4}\, =\, 0$
$\Rightarrow\, \lambda\, =\, -7$
So asymptotes are $2x^2\, -\, 3xy\, -\, 2y^2\, +\, 3x\, -\, y\, +\, 1\, =\, 0$
$\Rightarrow$ 2y - x - 1 = 0 & 2x + y + 1 = 0
and the equation of conjugate hyperbola will be
$2x^2\, \, -3xy\, \, -2y^2\, +\, 3x\, \, -y\, +\, 8\, \, -14\, =\, 0$.

Any straight line parallel to an asymptote of a hyperbola intersects the hyperbola at 

  1. one point

  2. two points

  3. three points

  4. four points


Correct Option: A
Explanation:

One point. 

Assertion(A): The angle between the asymptotes of $3x^{2}-y^{2}=3$ is $120^{\circ}$
Reason(R): The angle between the asymptotes of $x^{2}-y^{2}=a^{2}$ is $90^{\circ}$

  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.


Correct Option: D
Explanation:

Asymptotes of a hyperbola is given by $y=\pm \frac { x }{ a } $


So for hyperbola $3{ x }^{ 2 }-{ y }^{ 2 }=3$
The asymptotes make an angle of ${ 60 }^{ o }$ and  ${ 120 }^{ o }$ with x-axis which means they make an angle of ${ 60 }^{ o } $ among themself.

Now for hyperbola ${ x }^{ 2 }-{ y }^{ 2 }={a}^{2}$

The asymptotes makes and angle of ${45}^{o}$ and  ${135}^{0}$ which means ${90}^{o}$ among themself.

If $e$ is the eccentricity of $\displaystyle \frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1$ and $\theta$ be the angle between the asymptotes then $\displaystyle \sec { \frac { \theta  }{ 2 }  } $ equals :

  1. ${ e }^{ 2 }$

  2. $\displaystyle \frac { 1 }{ e } $

  3. $2e$

  4. $e$


Correct Option: D
Explanation:

Equation of asymptotes to $\displaystyle \frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1$ are given by 

$\displaystyle y=-\frac { b }{ a } x$
$\displaystyle \therefore { m } _{ 1 }=-\frac { b }{ a } $
Similarly $\displaystyle y=\frac { bx }{ a } $
$\therefore \displaystyle { m } _{ 2 }=\frac { b }{ a } $
Now $\displaystyle \theta =2\tan ^{ -1 }{ \frac { b }{ a }  } $
$\displaystyle \Rightarrow \tan { \frac { \theta  }{ 2 }  } =\frac { b }{ a } \Rightarrow \tan ^{ 2 }{ \frac { \theta  }{ 2 }  } =\frac { { b }^{ 2 } }{ { a }^{ 2 } } ={ e }^{ 2 }-1$
$\displaystyle \Rightarrow \sec ^{ 2 }{ \frac { \theta  }{ 2 }  } ={ e }^{ 2 }\Rightarrow \sec { \frac { \theta  }{ 2 }  } =e$

The equation of hyperbola conjugate to the hyperbola $2x^2 + 3xy - 2y^2 - 5 + 5y + 2 = 0$ is

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

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

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

  4. None of these


Correct Option: A
Explanation:

Let the given hyperbola be $ H =2x^{2}+3xy-2y^{2}-5x+5y+2=0$


Thus the pair of asymptotes be $ A = 2x^{2}+3xy-2y^{2}-5x+5y+\lambda =0 $

Pair of straight line has $ \Delta =0 $
$\Delta = abc+ 2fgh-af^{2}-bg^{2}-ch^{2}=0 $

where,

$a =2 $ 
$b=-2 $
$c=\lambda $
$f=\dfrac{5}{2}$
$g=-\dfrac{5}{2}$
$h=\dfrac{3}{2}$


Thus, $\lambda=-3 $

$A= 2x^{2}+3xy-2y^{2}-5x+5y-3 =0$

Since  $ H+C=2A $  Where $ C $ be the equation of conjugate hyperbola

$C= 2A-H $
So,the equation of conjugate hyperbola be  $ C=2x^{2}+3xy-2y^{2}-5x+5y-8 =0 $

The angle between the asymptotes of the hyperbola ${27x}^{2}-{9y}^{2}=24$ is 

  1. ${30}^{o}$

  2. ${120}^{o}$

  3. ${60}^{o}$

  4. ${90}^{o}$


Correct Option: B
Explanation:
$27 x ^ { 2 } - 9 y ^ { 2 } = 24$
$\Rightarrow \quad \dfrac { x ^ { 2 } } { \left( \dfrac { 24 } { 27 } \right) } - \dfrac { y ^ { 2 } } { \left( \dfrac { 24 } { 9 } \right) } = 1$
$\Rightarrow \dfrac { x ^ { 2 } } { ( \dfrac 89 ) } - \dfrac { y ^ { 2 } } { \dfrac 8 3 } = 1$
$\Rightarrow a ^ { 2 } = \dfrac 8  9 \quad , \quad b ^ { 2 } = \dfrac 8 3$
$\Rightarrow \quad \dfrac { b ^ { 2 } } { a ^ { 2 } } = \dfrac { 8 } { 3 } \times \dfrac { 9 } { 8 } = 3$
$\Rightarrow \quad \dfrac { b } { a } = \sqrt { 3 }$

The angle between the asymptotes 
$\begin{aligned} & = 2 \tan ^ { - 1 } \left( \frac { b } { a } \right) \ = & 2 \tan ^ { - 1 } \left( \sqrt 3 \right) \ = & 2 \cdot 60 \ = & 120 \end{aligned}$

The asymptotes of the hyperbola $xy - 3x + 4y + 2 = 0$ are

  1. $x = - 4,y=3$

  2. $x = 4,y=3$

  3. $x =2, y =- 3$

  4. $x =2, y = 3$


Correct Option: A
Explanation:

Given : Hyperbola,
$xy-3x+4y+2=0$---------------1
For Asymptotes,
Let the Asymptote's Equation be $y=mx+c$
And then finding $\phi _{n}(m)$ by replacing $y\rightarrow m$ and $x\rightarrow 1$
As $n=2$,
$\phi _{2}(m)=m$
putting $\phi _{2}(m)=0$, we get $m=0$
By taking $m=0$, we will get only one asymptote parallel to X-axis, so let's find them with putting the co-efficients of higher terms to zero.
For Asymptote parallel to X-axis, we put co-efficient of highest degree of x to zero that is here 1, so co-efficient of x$=0$
$\Rightarrow (y-3)=0$------------2(from Equation 1)
For Asymptote parallel to Y-axis, we put co-efficient of highest degree of y to zero which is 1 here, co-efficient of y$=0    (from Equation 1)
$\Rightarrow x+4=0$------------3
The Equation 2 & 3 are asymptotes to Equation 1.
$x+4=0$ & $y-3=0$

The curve ${ y }^{ 2 }\left( x-2 \right) ={ x }^{ 2 }\left( 1+x \right) $ has:

  1. An asymtote parallel to $x$-axis

  2. An asymtote parallel to $y$-axis

  3. Asymtotes parallel to both axes

  4. No asymptote


Correct Option: B
Explanation:

To find the asymtote, we need to get one variable in terms of other variable.

By observation, we see that it is very easy to get $y$ in terms of $x$. 
So, that's exactly what we will do:
$y^{ 2 }(x-2)=x^{ 2 }(1+x)$
$\Rightarrow  y^{ 2 }=\dfrac { x^{ 2 }(1+x) }{ x-2 }$ 
By definition, asymtote can be found when for a finite value of one co-ordinate, other tends to $\infty$ or $-\infty$.
So, we see when $x=2$,  $y= \infty$.
So, $x=2$ is an asymptote which is parallel to $y$-axis.

If e is the eccentricity of the hyperbola and $\theta$ is angle between the asymptotes, then $\dfrac{cos\theta}{2}$ = 

  1. $\dfrac{(1-e)}{e}$

  2. $\dfrac{1}{e}-1$

  3. $\dfrac{1}{e}$

  4. None of these


Correct Option: C
Explanation:

Let $\dfrac {x^{2}}{a^{2}}-\dfrac {y^{2}}{b^{2}}=1$  be the hyperbola


It has asymptotes $y=\pm \dfrac {b} {a} x$
Angle between the asymptotes $= 2tan^-1 (\dfrac{b}{a})=\theta$
$\Rightarrow \tan \dfrac {\theta} {2}=\pm \dfrac {b} {a} $

$\Rightarrow sec^{2}\dfrac {\theta} {2}=1+\tan ^{2}\dfrac {\theta} {2}=1+\dfrac {b^{2}}{a^{2}}$

$\Rightarrow \sec ^{2}\dfrac {\theta} {2}=\sqrt {1+\dfrac{b^{2}}{a^{2}}} $

$\Rightarrow \sec ^{2}\dfrac {\theta} {2} =e^{2}$

$\Rightarrow \cos \dfrac {\theta} {2}=\dfrac {1}{e}$

Through any P of the hyperbola $\frac{x^2}{a^2}- \frac{y^2}{b^2} =1 $ a line $PQR$ is drawn with a fixed gradient $m$, meeting the asymptotes in $Q\ &\ R$. Then the product,$ (QP) (PR) =\frac{a^2b^2(1+m^2)}{b^2- a^2m^2}$.

  1. True

  2. False


Correct Option: A

The asymptotes of the hyperbola $6{x^2} + 13xy + 6{y^2} - 7x - 8y - 26 = 0$ are 

  1. $2x + 3y - 1 = 0$,$3x + 2y + 2 = 0$

  2. $2x + 3y = 1,3x + 2y = 2$

  3. $3x + 3y = 0,3x + 2y = 0$

  4. $2x + 3y = 3,3x + 2y = 4$


Correct Option: B

From a point $P (1, 2)$ two tangents are drawn to a hyperbola $H$ in which one tangent is drawn to each arm of the hyperbola. If the equations of asymptotes of hyperbola $H$ are $\sqrt 3x-y+5=0$ and $\sqrt 3x+y-1=0$, then eccentricity of $H$ is :

  1. $2$

  2. $\dfrac {2}{\sqrt 3}$

  3. $\sqrt 2$

  4. $\sqrt 3$


Correct Option: B
Explanation:
Since ${c} _{1}{c} _{2}\left({a} _{1}{a} _{2}+{b} _{1}{b} _{2}\right)<0$

$\therefore$ origin lies in acute angle. 

$P\left(1,2\right)$ lies in obtuse angle

Slope of asymptotes${m} _{1}=\sqrt{3},\,{m} _{2}=-\sqrt{3}$

$\tan{\theta}=\left|\dfrac{{m} _{1}-{m} _{2}}{1+{m} _{1}{m} _{2}}\right|$

$=\left|\dfrac{\sqrt{3}-\left(-\sqrt{3}\right)}{1+\sqrt{3}\times-\sqrt{3}}\right|$

$=\left|\dfrac{2\sqrt{3}}{1-3}\right|$

$=\left|\dfrac{2\sqrt{3}}{-2}\right|$

$\Rightarrow\,\tan{\theta}=\sqrt{3}$

Acute angle between the asymptotes is $\dfrac{\pi}{3}$

Hence eccentricity $e=\sec{\dfrac{\theta}{2}}=\sec{\dfrac{\pi}{6}}=\dfrac{2}{\sqrt{3}}$

The asymptotes of the hyperbola $\dfrac {x^2}{a^2}-\dfrac {y^2}{b^2}=1$ form with any tangent to the hyperbola a triangle whose area is $a^2 \tan\lambda$ in magnitude, then its eccentricity is :

  1. $\sec \lambda$

  2. $\cos ec \lambda$

  3. $\sec^2\lambda$

  4. $\cos ec^2\lambda$


Correct Option: A
Explanation:

Any tangent to hyperbola forms a triangle with the asymptotes which has constant area $ab$.

$\Rightarrow ab=a^2 \tan\lambda$

$\displaystyle \Rightarrow \frac {b}{a}=\tan \lambda$

$\displaystyle e=\sqrt{1+\frac{b^2}{a^2}} $

$\Rightarrow e = \sqrt{1+\tan^2{\lambda}} =\sec{\lambda}$

If $S=0$ be the equation of the hyperbola $x^2+4xy+3y^2-4x+2y+1=0$, then the value of $k$ for which $S+k=0$ represents its asymptotes is :

  1. $20$

  2. $-16$

  3. $-22$

  4. $18$


Correct Option: C
Explanation:

$S+k=x^2+4xy+3y^2-4x+2y+1+k =0$
For equation $S+k=0$ to represent a pair of lines,
$\triangle =0$
$\begin{vmatrix} 1& 2 & -2\ 2 & 3 & 1\ -2 & 1 & 1+k\end{vmatrix}=0$
$\Rightarrow 3(1+k)-1-2(2+2k+2)-2(2+6)=0$
$\Rightarrow k=-22$

One of the asymptotes (with negative slope) of a hyperbola passes through (2, 0) whose transverse axis is given by x - 3y + 2 = 0 then equation of hyperbola if it is given that the line y = 7x - 11 can intersect the hyperbola at only one point (2, 3) is given by

  1. $\displaystyle 7x^{2}+xy-y^{2}+10x-4y-3=0$

  2. $\displaystyle 7x^{2}-xy-y^{2}-10x-5y+2=0$

  3. $\displaystyle 7x^{2}+xy-y^{2}-19x-5y+28=0$

  4. $\displaystyle 7x^{2}+6xy-y^{2}-20x-4y-3=0$


Correct Option: D
Explanation:

As $y=7x-11$ intersects the hyperbola at only one point 


$ \displaystyle \Rightarrow $ it is parallel to one of the asymptotes

$ \displaystyle \Rightarrow $ Equation of one asymoptote can be taken as $7x-y+k=0$ clearly mirror image of $(2,0)$ about transverse axis $x-3y=2 $lies on other asymplote 

$ \displaystyle \Rightarrow \left ( \frac{6}{5},\frac{12}{5} \right )$ lies on $7x-y+k=0$

$ \displaystyle \Rightarrow k=-6$

$ \displaystyle \Rightarrow $other asymptote is $7x-y-6=0$

$ \displaystyle \Rightarrow $ centre is $(1,1)$

$ \displaystyle \Rightarrow $Asymptote through $(2,0)$ is $x+y=2$

Equation of hyperbola is $(7x-y-6)(x+y-2)-(7* 2-3-6)(2+3-2)=0$

$ \displaystyle \Rightarrow 7x^{2}+6xy-y^{2}-20x-4y-3=0 $

The asymptotes of a hyperbola have equations $y-1=\dfrac{3}{4}(x+3).$ If a focus of the hyperbola has coordinates $(7,1)$, the equation of the hyperbola is

  1. $\dfrac{(x+3)^2}{16}-\dfrac{(y-1)^2}{9} = 1$

  2. $\dfrac{(y-1)^2}{9}-\dfrac{(x+3)^2}{16} = 1$

  3. $\dfrac{(x+3)^2}{64}-\dfrac{(y-1)^2}{36} = 1$

  4. $\dfrac{(y-1)^2}{36}-\dfrac{(x+3)^2}{64} = 1$

  5. $\dfrac{(x+3)^2}{4}-\dfrac{(y-1)^2}{3} = 1$


Correct Option: C
Explanation:

Equation of asymptotes are 

$y-1=\dfrac { 3 }{ 4 } (x+3  )$    ......(i)

$ y-1=-\dfrac { 3 }{ 4 } (x+3)$     .....(ii)

Centre of the hyperbola is point of intersection of asymptotes.

Therefore, by solving (i) and (ii), we get centre as $C(-3,1)$.

Slope of asymptotes $=\dfrac { b }{ a } $

$\Rightarrow \dfrac { b }{ a } =\pm \dfrac { 3 }{ 4 }$      ......(i)

Focus is $(7,1)$.

Focus for hyperbola of form $\dfrac { { (x-h) }^{ 2 } }{ { a }^{ 2 } } -\dfrac { { (y-k) }^{ 2 } }{ { b }^{ 2 } } =1$ is $(h+ae,k)$

$\Rightarrow 7=-3+ae\\ \Rightarrow ae=10\\ \Rightarrow a\dfrac { \sqrt { { a }^{ 2 }+{ b }^{ 2 } }  }{ a } =10\\ \Rightarrow \sqrt { { a }^{ 2 }+{ b }^{ 2 } } =10$

Substituting $b$ from (i), we get

$\Rightarrow \sqrt { { a }^{ 2 }+{ \left( \pm a \dfrac { 3 }{ 4 }  \right)  }^{ 2 } } =10\\ \Rightarrow \dfrac { 5a }{ 4 } =10\\ \Rightarrow a=8\\ \Rightarrow b=\pm \dfrac { 3 }{ 4 } a=\pm 6$

So, the equation of hyperbola is

$\dfrac { { (x+3) }^{ 2 } }{ { 8 }^{ 2 } } -\dfrac { { (y-1) }^{ 2 } }{ { 6 }^{ 2 } } =1$

$\dfrac { { (x+3) }^{ 2 } }{ 64 } -\dfrac { { (y-1) }^{ 2 } }{ 36 } =1$

So, option C is correct.

If $PN$ is the perpendicular from a point on a rectangular hyperbola to its asymptotes, the locus, then the midpoint of $PN$ is

  1. circle

  2. parabola

  3. ellipse

  4. hyperbola


Correct Option: D
Explanation:

Let $xy={ c }^{ 2 }$ be the rectangular hyperbola and let $P\left( { x
} _{ 1 },{ y } _{ 1 } \right) $ be apoint on it. Let $Q(h.k)$ be the
midpoint of $PN$. Then the coordinates of $Q$ are $\left( { { x } _{ 1
},{ y } _{ 1 } }/{ 2 } \right) $
$\therefore \quad { x } _{ 1 }={ h
}\quad \cfrac { { y } _{ 1 } }{ 2 } =k\Rightarrow { x } _{ 1 }={ h }\quad
,\quad { y } _{ 1 }=2k$
But $\left( { x } _{ 1 },{ y } _{ 1 } \right) $ lies on $xy={ c }^{ 2 }$
$\therefore \quad h(2k)={ c }^{ 2 }\Rightarrow hk=\cfrac { { c }^{ 2 } }{ 2 } $
Therefore, the locus of $(h,k)$ is $xy=\cfrac { { c }^{ 2 } }{ 2 } $, which is a hyperbola.
Hence, option 'D' is correct.

The asymptotes of the hyperbola $xy - 3x + 4y + 2 = 0$ are

  1. $x= - 4$

  2. $x= 4$

  3. $y= - 3$

  4. $y= 3$


Correct Option: B,D
Explanation:
Given : Hyperbola,
$xy-3x+4y+2=0$---------------1
For Asymptotes,
Let the Asymptote's Equation be $y=mx+c$
And then finding $\phi _{n}(m)$ by replacing $y\rightarrow m$ and $x\rightarrow 1$
As $n=2$,
$\phi _{2}(m)=m$
putting $\phi _{2}(m)=0$, we get $m=0$
By taking $m=0$, we will get only one asymptote parallel to X-axis, so let's find them with putting the co-efficients of higher terms to zero.
For Asymptote parallel to X-axis, we put co-efficient of highest degree of x to zero that is here 1, so co-efficient of x$=0$
$\Rightarrow (y-3)=0$------------2(from Equation 1)
For Asymptote parallel to Y-axis, we put co-efficient of highest degree of y to zero which is 1 here, co-efficient of y$=0    (from Equation 1)
$\Rightarrow x+4=0$------------3
The Equation 2 & 3 are asymptotes to Equation 1.
$x+4=0$ & $y-3=0$




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