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Introduction to hyperbola - class-XI

Description: introduction to hyperbola
Number of Questions: 41
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Tags: conic section conic sections mathematics and statistics hyperbola circles and conics section two dimensional analytical geometry-ii maths
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Consider
the set of hyperbola $xy = {\text{ }}K,{\text{ K}} \in {\text{R,}}$  let ${e _1}$  be eccentricity
when $K = \sqrt {2017} $  and ${e _2}$ be the
eccentricity when $K = \sqrt {2018} $ , then ${e _1} = {e _2}$   is equal to 

  1. -1

  2. 0

  3. 2

  4. 1


Correct Option: A

The exhaustive interval of $\lambda$ for which the equation $\dfrac{x^2}{(\lambda^2-2\lambda-3)}+\dfrac{y^2}{\lambda^2+2\lambda-8}=1$ represents a hyperbola is 

  1. $ \lambda \varepsilon (- \infty, -4) \cup (3, \infty)$

  2. $ \lambda \varepsilon (-4, -1) \cup (2, 3)$

  3. $ \lambda \varepsilon (- \infty, -1) \cup (2, \infty)$

  4. $ \lambda \varepsilon (-4,-1)$


Correct Option: A

Length of the latus rectum of the hyperbola $xy=c^{2}$, is

  1. $2c$

  2. $\sqrt{2}c$

  3. $2\sqrt{2}c$

  4. $4c$


Correct Option: C

If area of quadrilateral formed by tangents drawn at ends of latus rectum of hyperbola $\dfrac { { x }^{ 2 } }{ { a }^{ 2 } } -\dfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1$ is equal to square of distance between centre and one focus of hyperbola,then ${ e }^{ 3 }$ is (e is eccentricity of hyperbola)

  1. $2\sqrt { 2 } $

  2. 2

  3. 3

  4. 8


Correct Option: A

Eccentricity of a hyperbola is always less than 1.

  1. True

  2. False


Correct Option: B
Explanation:

Standard equation of the hyperbola is $\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}$

The eccentricity of the hyperbola is given by
$e=\sqrt { 1+\dfrac { { b }^{ 2 } }{ { a }^{ 2 } }  }$ which is always greater than $1$.
Thus, the given statement is false.
Hence, option B is correct.

Which of the following equations does not represent a hyperbola?

  1. $xy = 4$

  2. $
    \dfrac{1}
    {x^2} + \dfrac{1}
    {y^2} = \dfrac{1}
    {4}
    $

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

  4. $
    x^2 - 4xy + 3y^2 = 1
    $


Correct Option: A

Equation of the latus rectum of the hyperbola $(10x - 5)^{2} + (10y - 2)^{2} = 9(3x + 4y - 7)^{2}$ is

  1. $y - 1/5 =-3/4(x - 1/2)$

  2. $x - 1/5 =-3/4(y - 1/2)$

  3. $y + 1/5 =-3/4(x + 1/2)$

  4. $x + 1/5 =-3/4(y + 1/2)$


Correct Option: A

The equation $\frac{x^2}{1-k}-\frac{y^2}{1+k}=1$, $k<1$ represents 

  1. $circle$

  2. $ellipse$

  3. $hyperbola$

  4. $none$


Correct Option: C
Explanation:
Equating the above equation with the second-degree equation
$A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0$ with $\dfrac{{x}^{2}}{1-k}-\dfrac{{y}^{2}}{1+k}=1$
we get $A=\dfrac{1}{1-k}, B=0, C=\dfrac{1}{1+k},D=0,E=0$ and $F=-1$
$(i)$For the second degree equation to represent a circle , the coefficients must satisfy the discriminant condition ${B}^{2}-4AC=0$ and also $A=C$
$\Rightarrow -4\times \dfrac{1}{1-k}\times \dfrac{1}{1+k}=0$
$\Rightarrow \dfrac{1}{1-{k}^{2}}=0$
This case does not exist
$(ii)$For the second degree equation to represent a ellipse , the coefficients must satisfy the discriminant condition ${B}^{2}-4AC<0$ and also $A\neq C$
$\Rightarrow -4\times \dfrac{1}{1-k}\times \dfrac{1}{1+k}<0$
$\Rightarrow \dfrac{1}{1-{k}^{2}}>0$
$\Rightarrow 1-{k}^{2}<0$
$\Rightarrow -{k}^{2}<-1$
$\Rightarrow {k}^{2}>1$ does not exist since it is given that $k<1$
$(iii)$For the second degree equation to represent a hyperbola, the coefficients must satisfy the discriminant condition ${B}^{2}-4AC>0$ and also $A\neq C$
$\Rightarrow -4\times \dfrac{1}{1-k}\times \dfrac{1}{1+k}>0$
$\Rightarrow \dfrac{1}{1-{k}^{2}}<0$
$\Rightarrow 1-{k}^{2}>0$
$\Rightarrow -{k}^{2}>-1$
$\Rightarrow {k}^{2}<1$ 
$\therefore k<1$
Hence the above equation represents a hyperbola.

The equation $\displaystyle\frac{x^2}{10-\lambda}+\frac{y^2}{6-\lambda}=1$ represents

  1. a hyperbola if $\lambda < 6$

  2. an ellipse if $\lambda>6$

  3. a hyperbola if $6 < \lambda < 10$

  4. an ellipse if $0 < \lambda < 6$


Correct Option: C,D
Explanation:

The general equation of an ellipse is $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ and that of a hyperbola is $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$
Using this, we get that the above equation is an ellipse if $10 - \lambda > 0$ and $6 - \lambda > 0$. The combined solution gives $\lambda < 6.$
For a hyperbola, the coefficient of $x^2$ and $y^2$ must be of opposite sign. Hence, 
 $10 - \lambda > 0$ and $6 - \lambda < 0$ which gives $6 < \lambda < 10$

The point to which the axes are to be translated to eliminate $x$ and $y$ terms in the equation $3x^{2}-4xy-2y^{2}-3x-2y-1=0$ is 

  1. $\left(\dfrac{5}{2},3\right)$

  2. $(-4,\dfrac{3}{2})$

  3. $ (-2,3)$

  4. $ (2,3)$


Correct Option: A
Explanation:
Given equation is $3x^{2}-4xy-2y^{2}-3x-2y-1=0$

Let $\left({x} _{1},{y} _{1}\right)$ be a point to which the origin is shifted by translation

Let $\left(X,Y\right)$ be the new coordinates of the point $\left(x,y\right)$

$\therefore\,$ the equations of the transformation are $x=X+{x} _{1},\,y=Y+{y} _{1}$

Now the transformed equation is 
$3{\left(X+{x} _{1}\right)}^{2}-4\left(X+{x} _{1}\right)\left(Y+{y} _{1}\right)-2{\left(Y+{y} _{1}\right)}^{2}-3\left(X+{x} _{1}\right)-2\left(Y+{y} _{1}\right)-1=0$

$\Rightarrow\,3\left({X}^{2}+2X{x} _{1}+{{x} _{1}}^{2}\right)-4\left(XY+X{y} _{1}+Y{x} _{1}+{x} _{1}{y} _{1}\right)-2\left({Y}^{2}+{{y} _{1}}^{2}+2Y{Y} _{1}\right)-3X-{x} _{1}-2Y-2{y} _{1}-1=0$

$\Rightarrow\,3{X}^{2}+3{{x} _{1}}^{2}+6X{x} _{1}-4X{y} _{1}-4{x} _{1}Y-4{x} _{1}{y} _{1}-2{Y}^{2}-2{{y} _{1}}^{2}+4Y{y} _{1}-3X-3{x} _{1}-2Y-2{y} _{1}-1=0$

$\Rightarrow\,\left(3{X}^{2}-4XY-2{Y}^{2}\right)+\left(3{{x} _{1}}^{2}-2{{y} _{1}}^{2}-4{x} _{1}{y} _{1}-3{x} _{1}-2{y} _{1}-1\right)+2X\left(3{x} _{1}-2{y} _{1}-\dfrac{3}{2}\right)+2Y\left(-2{x} _{1}+2{y} _{1}-1\right)=0$

Solving the first degree terms,we have
$3{x} _{1}-2{y} _{1}=\dfrac{3}{2}$

$-2{x} _{1}+2{y} _{1}=1$

Adding the above equations, we get
$3{x} _{1}-2{y} _{1}-2{x} _{1}+2{y} _{1}=\dfrac{3}{2}+1$

$\Rightarrow\,{x} _{1}=\dfrac{5}{2}$

From equation ,$-2{x} _{1}+2{y} _{1}=1$

$\Rightarrow\,2{y} _{1}=1+2{x} _{1}=1+2\times\dfrac{5}{2}=1+5=6$

$\Rightarrow\,{y} _{1}=\dfrac{6}{2}=3$

$\therefore\,\left({x} _{1},{y} _{1}\right)=\left(\dfrac{5}{2},3\right)$


Hence the point is $\left(\dfrac{5}{2},3\right)$

General solution of the equation $ y=x\dfrac{dy}{dx}+\dfrac {dx}{dy}$ represents _____________.

  1. a straight line or hyperbola

  2. a straight line or parabola

  3. a parabola or hyperbola

  4. circles


Correct Option: C

Eccentricity of hyperbola$ \dfrac { { x }^{ 2 } }{ k } -\dfrac { { y }^{ 2 } }{ k } =1$

  1. $\ \sqrt { 1+k } $

  2. $\ \sqrt { 1-k } $

  3. $\ \sqrt {2 } $

  4. $\2 \sqrt {2 } $


Correct Option: C
Explanation:

$Standard\, hyperbola\, : $


$\dfrac { { { x^{ 2 } } } }{ k } -\dfrac { { { y^{ 2 } } } }{ k } =1$

$Now, \ \dfrac { { { { \left( { x-h } \right)  }^{ 2 } } } }{ { { a^{ 2 } } } } -\dfrac { { { { \left( { y-k } \right)  }^{ 2 } } } }{ { { b^{ 2 } } } } =1$

$Therefore\, Hyperbola\, properties\, are \ (h,k)=\left( { 0,0 } \right) ,\, \, a=\sqrt { k } ,\, b=\sqrt { k }  $

$=\dfrac { { \sqrt { { { \left( \sqrt k \right)  }^{ 2 } }+{ { \left( { \sqrt { k }  } \right)  }^{ 2 } } }  } }{ { \sqrt { k }  } } $

$=\sqrt { 2 }  \ Hence,\, the\, option\, C\, is\, the\, correct\, answer$

A hyperbola passes through the focus of the ellipse $\dfrac{x^2}{25}+\dfrac{y^2}{16}=1,$ and its transverses and conjugate axes coincide with the major and minor axes of the ellipse. If the product of the eccentricites of the two curve is $1$, then the focus of the hyperbola is

  1. $(5\sqrt3,0)$

  2. $(5,0)$

  3. $\left(\dfrac{5}{3},0\right)$

  4. none of these


Correct Option: A

The foci of the hyperbola $xy=4$ are 

  1. $(2\surd{2},2\surd{2})$

  2. $(-2\surd{2},-2\surd{2})$

  3. $(-2\surd{2},2\surd{2})$

  4. $None of these$


Correct Option: A

If eccentricity of the hyperbola $\dfrac {x^{2}}{\cos^{2}\theta}-\dfrac {y^{2}}{\sin^{2}\theta}=1$ is more then $2$ when $\theta\ \in \ \left(0,\dfrac {\pi}{2}\right)$. Find the possible values of length of latus rectum 

  1. $(3,\infty)$

  2. $(1,3/2)$

  3. $(2,3)$

  4. $(-3,-2)$


Correct Option: A

The latus rectum of the hyperbola $16{x^2} - 9{y^2} = 144$ is-

  1. $\dfrac{13}{6}$

  2. $\dfrac{32}{3}$

  3. $\dfrac{8}{3}$

  4. $\dfrac{4}{3}$


Correct Option: B
Explanation:

We have,


$16{x^2} - 9{y^2} = 144$

$\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$

Here, $a=3,  b=4$

We know that the latus rectum 

$=\dfrac{2b^2}{a}$

Therefore,

$=\dfrac{2\times 16}{3}$

$=\dfrac{32}{3}$

Hence, this is the answer.

The Vertex of the parabola $y^{2} - 10y + x + 22=0$ is.

  1. (3,4)

  2. (3,5)

  3. (5,3)

  4. none of these


Correct Option: B
Explanation:
Given,

$y^2-10y+x+22=0$

$\Rightarrow x=-y^2+10y-22$

$x=-\left(y-5\right)^2+3$

$x-3=-\left(y-5\right)^2$

$-\left(x-3\right)=\left(y-5\right)^2$

$4\left(-\frac{1}{4}\right)\left(x-3\right)=\left(y-5\right)^2$

$\left(h,\:k\right)=\left(3,\:5\right),\:p=-\frac{1}{4}$

Vertex of parabola $(3,5)$



The centre of the hyperbola 9x$^2$ - 36 x - 16y$^2$ + 96y - 252 = 0 is

  1. $(2,3)$

  2. $(-2,-3)$

  3. $(-2, 3)$

  4. none of these


Correct Option: A
Explanation:
Given,

$9x^2-36x-16y^2+96y-252=0$

$9x^2-36x-16y^2+96y=252$

$9\left(x^2-4x\right)-16\left(y^2-6y\right)=252$

$\left(x^2-4x\right)-\dfrac{16}{9}\left(y^2-6y\right)=28$

$\dfrac{1}{16}\left(x^2-4x\right)-\dfrac{1}{9}\left(y^2-6y\right)=\dfrac{7}{4}$

$\dfrac{1}{16}\left(x-2\right)^2-\dfrac{1}{9}\left(y-3\right)^2=\dfrac{7}{4}+\dfrac{1}{16}\left(4\right)-\dfrac{1}{9}\left(9\right)$

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

$\dfrac{\left(x-2\right)^2}{4^2}-\dfrac{\left(y-3\right)^2}{3^2}=1$

Center $(h,k)=(2,3)$

Find the locus of a point which moves so that the difference of its distances from the points, $(5, 0)$ and $(-5, 0)$ is $2$ is:

  1. $\dfrac{x^2}{1}+\dfrac{y^2}{24}=1$

  2. $\dfrac{x^2}{24}+\dfrac{y^2}{1}=1$

  3. $\dfrac{x^2}{24}-\dfrac{y^2}{2}=1$

  4. $\dfrac{x^2}{1}-\dfrac{y^2}{24}=1$


Correct Option: D
Explanation:

The locus is nothing but hyperbola.
Difference of distance of a point from foci $=2a$ 

Given distance is $2 \Rightarrow a=1$
Distance between foci $=2ae=2\sqrt{a^2+b^2}=\sqrt{(5+5)^2}$
                                                 $\Rightarrow a^2+b^2 =25$
                                                  $\Rightarrow b^2=24$
Therefore, locus is $\dfrac{x^2}{1}-\dfrac{y^2}{24}=1$ 

If $e$ and $e'$ be the eccentricities of two conics $S$ and $S'$ such that $\displaystyle e^{2}+(e')^{2}= 3,$  then both $S$ and $S'$ are

  1. Ellipses

  2. Parabolas

  3. Hyperbolas

  4. None of these


Correct Option: C
Explanation:

For a parabola the eccentricity is $1$

$\therefore e^2 + e'^2 = 1 + 1 = 2$
For an ellipse the eccerntricity is less than $1$
$\therefore$ for a hyperbola the eccentricity is greater than $1$
So, the conics can be hyperbolas
Hence, hyperbola correct.

The eccentricity the hyperbola $x=\left( t+\dfrac { 1 }{ t }  \right) ,y=\dfrac { a }{ 2 } \left( t-\dfrac { 1 }{ t }  \right) $ is ____________.

  1. $\sqrt { 2 } $

  2. $\sqrt { 3 } $

  3. $2\sqrt { 3 } $

  4. $3\sqrt { 2 } $


Correct Option: A

The equation $ \displaystyle 3x^{2}-2xy+y^{2}=0 $ represents:

  1. a circle

  2. hyperbola

  3. a pair of lines

  4. none of these


Correct Option: D
Explanation:

Given expression,$\displaystyle 3{ x }^{ 2 }-2xy+{ y }^{ 2 }=0$ 
As Coefficient of $\displaystyle xy$ is not zero,It will not be a circle and hyperbola.
Let $\displaystyle \frac { y }{ x } =m$

We get $\displaystyle { m }^{ 2 }-2m+3=0$ will not have any real solutions as discriminant is less than zero.
$\displaystyle \therefore $ They will not be pair of lines too.

The graph between $\log {(\theta-{\theta} _{0})}$ and time $(t)$ is a straight line in the experiment based on Newton's law cooling. What is the shape of graph between $\theta$ and $t$?

  1. A straight line

  2. A parabola

  3. A hyperbola

  4. A circle


Correct Option: C
Explanation:

It is given that
$log(\theta-\theta _{0})=kt $ where k is a constant of proportionality.
Hence
$e^{kt}=\theta-\theta _{0}$
Or
$\triangle \theta=e^{kt}$
The graph of $\triangle\theta$ vs $t$ will therefore be a hyperbola.

Equation $(2\, +\, \lambda)x^2\, -\, 2 \lambda xy\, +\, (\lambda\, -\, 1)y^2\, -\, 4x\, -\, 2\, =\, 0$ represents a hyperbola if

  1. $\lambda\, =\, 4$

  2. $\lambda\, =\, 1$

  3. $\lambda\, =\, \dfrac43$

  4. $\lambda\, =\, 3$


Correct Option: B
Explanation:

Given equation will represent hyperbola if
$h^2> ab$
$\Rightarrow \lambda^2\, >\, (\lambda\, +\, 2)\, (\lambda\, -\, 1)$
$\Rightarrow\, \lambda\, <\, 2$
Also $\Delta\, \neq\, 0$
$\Rightarrow\, -2(\lambda^2\, +\, \lambda\, -\, 2)\, -\, 4(\lambda\, -\, 1)\, +\, 2 \lambda^2\, \neq\, 0$
$\Rightarrow\, \lambda\, \neq\, \displaystyle \frac{4}{3}$.
Hence option 'B' is correct.

The difference between the length $2a$ of the transverse axis of a hyperbola of eccentricity $e$ and the length of its latus rectum is :
  1. $2a\left| 3-{ e }^{ 2 } \right| $

  2. $2a\left| 2-{ e }^{ 2 } \right| $

  3. $2a\left( { e }^{ 2 }-1 \right) $

  4. $a\left( 2{ e }^{ 2 }-1 \right) $


Correct Option: B
Explanation:

Let the equation of hyperbola be $\displaystyle \frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1$

Length of transverse axis is $2a$ and 
Length of latus rectum is $\displaystyle \frac { 2{ b }^{ 2 } }{ a } $
Now, difference $\displaystyle =\left| 2a-\frac { 2{ b }^{ 2 } }{ a }  \right| =\frac { 2 }{ a } \left| 2{ a }^{ 2 }-{ a }^{ 2 }{ e }^{ 2 } \right| $
$\therefore$ Difference $=2a\left| 2-{ e }^{ 2 } \right| $

Assertion(A): The difference of the focal distances of any point on the hyperbola $\displaystyle \frac{x^{2}}{36}-\frac{y^{2}}{9}=1$ is 12.
Reason(R): The difference of the focal distances of any point on the hyperbola is equal to the length of it transverse axis

  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 the correct

    explanation of A.

  3. A is true but R is false.

  4. A is false but R is true.


Correct Option: A
Explanation:

Clearly, $|SP-S'P|=2a=12$
Thus statement 1 is correct.
Also statement 2 is correct and followed by statement 1.

The asymptotes of a hyperbola $4x^2 - 9y^2=36$ are

  1. $2x \pm 3y = 1$

  2. $2x \pm 3y = 0$

  3. $3x \pm 2y = 1$

  4. None


Correct Option: B
Explanation:

The equation of hyperbola is $\displaystyle \frac{x^{2}}{9}+\frac{y^{2}}{4}=1$

So the equation of asymptotes is $\displaystyle \frac{x^{2}}{9}-\frac{y^{2}}{4}=0$
$\Rightarrow 4x^{2}-9y^{2}=0$
$\Rightarrow 2x \pm 3y=0$
Therefore option $B$ is correct

The equation of hyperbola whose coordinates of the foci are $(\pm8,0)$ and the lenght of latus rectum is $24$ units, is

  1. $3{ x }^{ 2 }-{ y }^{ 2 }=48$

  2. $4{ x }^{ 2 }-{ y }^{ 2 }=48$

  3. ${ x }^{ 2 }-3{ y }^{ 2 }=48$

  4. ${ x }^{ 2 }-4{ y }^{ 2 }=48$


Correct Option: A
Explanation:

The Foci of hyperbola are $(\pm8,0)$, hence Foci lie on $x$ - axis. 


We know that foci of hyperbola lie at $(\pm ae,0)$, So $ae = 8$ ...$(1)$

squaring both sides of equation $(1)$, we get,

$\Rightarrow a^2e^2 = 64$

Eccentricity of hyperbola $e^2 =1 + \dfrac {b^2}{a^2}$  

$\Rightarrow a^2(1+\dfrac{b^2}{a^2}) = 64$

$\Rightarrow a^2 + b^2 = 64$ ...$(2)$

Now the length of latus rectum is given as 24 units.

length of latusrectum of hyperbola $ = \dfrac{2b^2}{a} = 24$

$\Rightarrow b^2 = 12a$ ...$(3)$

putting value of $b^2$ in eq. $(2)$, we get,

$\Rightarrow a^2 +12a -64 = 0$

Hence $a  = 4, -16$

As $a$ is always taken as positive value so $a =4$ 

from eq. $(3)$,  $b = \sqrt{48}$

Hence equation of hyperbola is $\dfrac{x^2}{16} -\dfrac {y^2}{48} = 1$

Or $3x^2 -y^2 = 48$, So correct option is $A$.

If $ e$ and $e'$ be the eccentricities of a hyperbola and its conjugate, them $ \dfrac {1}{e^2} + \dfrac {1}{e'^{2}} $ is equal to :

  1. 0

  2. 1

  3. 2

  4. None of these


Correct Option: B
Explanation:

$ e^2 = \dfrac {a^2 + b^2}{a^2} $ and $ e'^{2} = \dfrac {a^2 + b^2}{b^2} $

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

The equation of the hyperbola whose foci are $(6, 5), (-4, 5)$ and eccentricity $5/4$ is?

  1. $\displaystyle\frac{(x-1)^2}{16}-\frac{(y-5)^2}{9}=1$

  2. $\displaystyle\frac{x^2}{16}-\frac{y^2}{9}=1$

  3. $\displaystyle\frac{(x-1)^2}{16}-\frac{(y-5)^2}{9}=-1$

  4. $\displaystyle\frac{(x-1)^2}{4}-\frac{(y-5)^2}{9}=1$


Correct Option: A
Explanation:

Let the centre of hyperbola be $(\alpha , \beta)$

As $y=5$ line has the foci, it also has the major axis.
$\therefore \dfrac{(x-\alpha)^2}{a^2}-\dfrac{(y-\beta)^2}{b^2}=1$
Midpoint of foci = centre of hyperbola
$\therefore \alpha =1, \beta =5$
Given, $e=\dfrac{5}{4}$.
We know that foci is given by $(\alpha \pm ae, \beta)$
$\therefore \alpha +ae=6$
$\Rightarrow 1+\dfrac{5}{4}a=6\Rightarrow a=4$
Using $b^2=a^2(e^2-1)$
$\Rightarrow b^2=16\left(\dfrac{25}{16}-1\right)=9$
$\boxed{\therefore\ Equation\ of\ hyperbola \Rightarrow\ \dfrac{(x-1)^2}{16}-\dfrac{(y-5)^2}{9}=1}$

Find the locus of the point of intersection of the lines $\sqrt{3}x-y-4\sqrt{3} \lambda=0$ and $\sqrt{3}\lambda x+\lambda y-4\sqrt{3}=0$ for different values of $\lambda$.

  1. $4x^2-y^2=48$

  2. $x^2-4y^2=48$

  3. $3x^2-y^2=48$

  4. $y^2-3x^2=48$


Correct Option: C
Explanation:

Let $(h,k)$ be the point of intersection of the given lines. Then,


$\sqrt{3}h-k-4\sqrt{3} \lambda=0$ and $\sqrt{3} \lambda h+\lambda k-4\sqrt{3}=0$


$\sqrt{3}h-k=4\sqrt{3}\lambda$ and $\lambda(\sqrt{3}h+k)=4\sqrt{3}$

$(\sqrt{3}h-k)\lambda(\sqrt{3}h+k)=(4\sqrt{3}\lambda)(4\sqrt{3})$

$3h^2-k^2=48$

Hence, the locus of (h,k) is $3x^2-y^2=48$.

The AFC Curve passes through the Origin statement is -

  1. True

  2. False

  3. Partially True

  4. Nothing can be said


Correct Option: B
Explanation:

AFC curve is downward sloping because fixed costs are distributed over large volume when quantity produced increases.

So, it doesn't pass through origin.

$Center\quad of\quad the\quad hyperbola\quad { x }^{ 2 }+4{ y }^{ 2 }+6xy+8x-2y+7=0\quad is\quad $

  1. $(1,1)$

  2. $(0,2)$

  3. $(2,0)$

  4. $None\quad of\quad these$


Correct Option: D
Explanation:
Given Hyperbola: $x^{2}+4y^{2}+6xy+8x-2y+7=0$
Centre: Point of intersection of asymptotes of hyperbola.
Now finding Asymptotes of given equation, taking equation of asymptote $y=mx+c$ by replacing $x\rightarrow 1, y\rightarrow m$ in $\phi _n(m)$
When $n=2$, $\phi _{2}(m)=1+4m^{2}+6m$
$\phi _{1}(m)=8-2m$
$\phi _{0}(m)=7$
$\phi _{2}^{1}(m)=8m+6$
$\phi _{1}(m)=8-2m$

Putting $\phi _{2}(m)=0$
$\Rightarrow 1+6m+4m^{2}=0$
$\Rightarrow  m=\cfrac{-6\pm \sqrt {36-16}}{2(4)}$
$\Rightarrow m=\cfrac{-6\pm\sqrt 20}{2(4)}$
$\Rightarrow m=\cfrac{-3\pm \sqrt 5}{4}$
So, m$=\cfrac{-3+\sqrt 5}{4}, \cfrac{-3-\sqrt 5}{4}$
Value of $c$, when $m$ is different
$c= \cfrac{-\phi _{1}(m)}{\phi _{2}^{'}(m)}=\cfrac{8-2m}{8m+6}$

For $m=\cfrac{-3+\sqrt 5}{4}, c=\cfrac{8-2(\cfrac{-3+\sqrt 5}{4})}{8\cfrac{-3+\sqrt 5}{4})+6}=\cfrac{3-\sqrt 5+16}{-6+2\sqrt 5+6}=\cfrac{19\sqrt 5-5}{10}$

For $m=\cfrac{-3-\sqrt 5}{4}, c=\cfrac{8-2(\cfrac{-3-\sqrt 5}{4})}{8\cfrac{-3-\sqrt 5}{4})+6}=\cfrac{19+15}{-2 \sqrt 5}=\cfrac{-(19\sqrt 5+5)}{10}$
Equation of Asymptotes : $y=\cfrac{-3+\sqrt 5}{4}x +\cfrac{19\sqrt 5-5}{10}$ & $y=\cfrac{-3-\sqrt 5}{4}x-(\cfrac{5+19\sqrt 5}{10})$
On solving them for x & y, putting LHS-RHS
$\Rightarrow 0=\cfrac{\sqrt 5}{2}x+\cfrac{19\sqrt 5}{5} \Rightarrow x=\cfrac{-38}{5}$
Now putting values of x in Asymptotes equation, we get
$y=\cfrac{-3-\sqrt 5(-19)}{10}+\cfrac{+5+19\sqrt 5}{10}=\cfrac{+52}{10}$
Centre$(\cfrac{-38}{5}, \cfrac{+52}{10})$.

Circles are drawn on chords of the rectangular hyperbola $xy=4$ parallel to the line $y=x$ as diameters.All such circles pass through two fixed points whose coordinates are 

  1. $\left(2,2\right)$

  2. $\left(2,-2\right)$

  3. $\left(-2,2\right)$

  4. $\left(-2,-2\right)$

Correct Option: A,D
Explanation:
Given:Rectangular hyperbola $xy=4={c}^{2}$
$\Rightarrow\,{c}^{2}=4$
$\Rightarrow\,c=2$
Let $P$ and $Q$ be the end points on the Rectangular hyperbola where $P\left(2{t} _{1},\dfrac{2}{{t} _{1}}\right)$ and $Q\left(2{t} _{2},\dfrac{2}{{t} _{2}}\right)$
Using the end points of diameter the equation of the circle

$C:\left(x-2{t} _{1}\right)\left(x-2{t} _{2}\right)+\left(y-\dfrac{2}{{t} _{1}}\right)\left(y-\dfrac{2}{{t} _{2}}\right)=1$         ..........$(1)$

Now,Slope of the $PQ=\dfrac{\dfrac{2}{{t} _{2}}-\dfrac{2}{{t} _{1}}}{2{t} _{2}-2{t} _{1}}$

$=\dfrac{2\left(\dfrac{1}{{t} _{2}}-\dfrac{1}{{t} _{1}}\right)}{2\left({t} _{2}-{t} _{1}\right)}$

$=\dfrac{\dfrac{{t} _{1}-{t} _{2}}{{t} _{1}{t} _{2}}}{\left({t} _{2}-{t} _{1}\right)}$

$=\dfrac{\dfrac{{t} _{1}-{t} _{2}}{{t} _{1}{t} _{2}}}{\left({t} _{2}-{t} _{1}\right)}$

$=\dfrac{-1}{{t} _{1}{t} _{2}}$

Hence $PQ$ is the diameter for circle and it is parallel to the line $y=x$

Slope of $PQ=$Slope of the line $y=x$

$\Rightarrow\,\dfrac{-1}{{t} _{1}{t} _{2}}=1$

$\Rightarrow\,{t} _{1}{t} _{2}=-1$

$(1)\Rightarrow\,\left(x-2{t} _{1}\right)\left(x-2{t} _{2}\right)+\left(y-\dfrac{2}{{t} _{1}}\right)\left(y-\dfrac{2}{{t} _{2}}\right)=1$ 

$\Rightarrow\,x\left(x-2{t} _{2}\right)-2{t} _{1}\left(x-2{t} _{2}\right)+y\left(y-\dfrac{2}{{t} _{2}}\right)-\dfrac{2}{{t} _{1}}\left(y-\dfrac{2}{{t} _{2}}\right)=1$
 
$\Rightarrow\,{x}^{2}-2x{t} _{2}-2x{t} _{1}+4{t} _{1}{t} _{2}+{y}^{2}-\dfrac{2y}{{t} _{2}}-\dfrac{2y}{{t} _{1}}+\dfrac{4}{{t} _{1}{t} _{2}}=1$

$\Rightarrow\,{x}^{2}-2x{t} _{2}-2x{t} _{1}+4\times -1+{y}^{2}-\dfrac{2y}{{t} _{2}}-\dfrac{2y}{{t} _{1}}+\dfrac{4}{\times -1}=1$ using ${t} _{1}{t} _{2}=-1$

$\Rightarrow\,{x}^{2}+{y}^{2}-2x\left({t} _{2}+{t} _{1}\right)-4-2y\left(\dfrac{1}{{t} _{2}}+\dfrac{1}{{t} _{1}}\right)-4=1$

$\Rightarrow\,{x}^{2}+{y}^{2}-8-2x\left({t} _{2}+{t} _{1}\right)-2y\left(\dfrac{{t} _{1}+{t} _{2}}{{t} _{1}{t} _{2}}\right)=1$

$\Rightarrow\,{x}^{2}+{y}^{2}-8-2x\left({t} _{2}+{t} _{1}\right)-2y\left(\dfrac{{t} _{1}+{t} _{2}}{-1}\right)=1$ using ${t} _{1}{t} _{2}=-1$

$\Rightarrow\,{x}^{2}+{y}^{2}-8-2x\left({t} _{2}+{t} _{1}\right)+2y\left({t} _{1}+{t} _{2}\right)=1$ 

$\Rightarrow\,{x}^{2}+{y}^{2}-8+\left(2y-2x\right)\left({t} _{2}+{t} _{1}\right)=1$ is of the form $C+\lambda\,L$ 

where $C={x}^{2}+{y}^{2}-8=0$ is the equation of a circle.
and $L=2y-2x=0$ is the equation of a line.
$\Rightarrow\,y-x=0$ or $x=y$

Substituting $x=y$ in the equation ${x}^{2}+{y}^{2}-8=0$ we get
$\Rightarrow\,2{x}^{2}-8=0$

$\Rightarrow\,2\left({x}^{2}-4\right)=0$

$\Rightarrow\,\left(x-2\right)\left(x+2\right)=0$

$\therefore\,x=2,-2$

$\Rightarrow\,y=2,-2$ since $x=y$

Hence the coordinates of the fixed points are $\left(2,2\right)$ and $\left(-2,-2\right)$ 

Centre of the hyperbola ${x^2} + 4{y^2} + 6xy + 8x - 2y + 7 = 0$ is 

  1. $(1,1)$

  2. $(0,2)$

  3. $(2,0)$

  4. None of these


Correct Option: D
Explanation:

Consider equation of a hyperbola as $F=ax^2+2by+cy^2+2dx+2ey+f=0$

The centre of this hyperbola can be found by applying the concepts of partial differentiation
We first find $\dfrac{\delta F}{\delta x} $ and $\dfrac{\delta F}{\delta y}$
We then solve $\dfrac{\delta F}{\delta x} =0$ and $\dfrac{\delta F}{\delta y}=0 $ to find $x,y$ which is the centre of the hyperbola .

Given that,
$F=x^2+4y^2+6xy+8x-2y+7$

$\Rightarrow \dfrac{\delta F}{\delta x}=2x+0+6y+8+0+0$

$\Rightarrow \dfrac{\delta F}{\delta x}=2x+6y+8$         ...$(1)$

$\Rightarrow \dfrac{\delta F}{\delta y}=0+8y+6x+0-2+0$

$\Rightarrow \dfrac{\delta F}{\delta y}=6x+8y-2$      ....$(2)$
 
$(1) \rightarrow 2x+6y+8=0$                    

$(2) \rightarrow 6x+8y-2=0$

Solving $(1), (2)$ we get,

$\Rightarrow x=\dfrac{19}{5}, y=\dfrac{-13}{5}$

Therefore the centre of hyperbola is $(\dfrac{19}{5},\dfrac{-13}{5})$



The eccentricity of the hyperbola whose latus-return is $8$ and length of the conjugate axis is equal to half the distance between the foci, is

  1. $\dfrac43$

  2. $\dfrac4{\surd 3}$

  3. $\dfrac2{\surd 3}$

  4. $None\ of\ these$


Correct Option: C
Explanation:

Given that the length of the latus rectum is $8$ and length of the conjugate axis is equal to half the distance between the foci.


$\Rightarrow \dfrac{2b^2}{a}=8$ and $2b=\dfrac{1}{2}(2ae)$

$\therefore \dfrac{2}{a}\left(\dfrac{ae}{2}\right)^2=8$

$\Rightarrow ae^2=16$ ...(1)

We have $\dfrac{2b^2}{a}=8$

$\Rightarrow b^2=4a$

$\Rightarrow a^2(e^2-1)=4a$

$\Rightarrow ae^2-a=4$

$\Rightarrow 16-a=4$ (by (1))

$\Rightarrow a=12$

Substitute $a=12$ in (1)

$\Rightarrow 12e^2=16$

$\Rightarrow e^2=\dfrac{4}{3}$

$\therefore e=\dfrac{2}{\sqrt{3}}$

From any point on the hyperbola $\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ tangents are drawn to the hyperbola $\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = 2$. The area cut-off by the chord of contact on the asymptotes is equal to

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

  2. ab

  3. 2 ab

  4. 4 ab


Correct Option: D
Explanation:

Let $P\left( { x } _{ 1 },{ y } _{ 1 } \right) $ be a point on the hyperbola $\cfrac { { x }^{ 2 } }{ { a }^{ 2 } } -\cfrac { { y }^{ 2 }
}{ { b }^{ 2 } } =1$. Then,
$\cfrac { { { x } _{ 1 } }^{ 2 } }{ { a }^{ 2 } } -\cfrac { { { y } _{ 1 } }^{ 2 } }{ { b }^{ 2 } } =1$
The chord of contact of tangents from $P$ to the hyperbola $\cfrac { { x

}^{ 2 } }{ { a }^{ 2 } } -\cfrac { { y }^{ 2 } }{ { b }^{ 2 } } =2$ is
$\cfrac

{ { x }{ x } _{ 1 } }{ { a }^{ 2 } } -\cfrac { { y }{ y } _{ 1 } }{ { b

}^{ 2 } } =2\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (i)$

The equations of the asymptotes are
$\cfrac { x }{ a } -\cfrac { y }{ b } =0$ and $\cfrac { x }{ a } +\cfrac { y }{ b } =0$

The points of intersection of $(i)$ with the two asymptotes are given by
${

x } _{ 1 }=\cfrac { 2a }{ \cfrac { { x } _{ 1 } }{ a } -\cfrac { { \quad y

} _{ 1 } }{ b }  } \quad ,{ \quad y } _{ 1 }=\cfrac { 2b }{ \cfrac { { x

} _{ 1 } }{ a } -\cfrac { { \quad y } _{ 1 } }{ b }  } $
${ x } _{ 2

}=\cfrac { 2a }{ \cfrac { { x } _{ 1 } }{ a } -\cfrac { { \quad y } _{ 1 }

}{ b }  } \quad ,{ \quad y } _{ 2 }=\cfrac {- 2b }{ \cfrac { { x } _{ 1 }

}{ a } -\cfrac { { \quad y } _{ 1 } }{ b }  } $
$\quad \therefore $ Area of the triangle
$\cfrac

{ 1 }{ 2 } \left( { x } _{ 1 }{ y } _{ 2 }-{ x } _{ 2 }{ y } _{ 1 } \right)

=\cfrac { 1 }{ 2 } \left( \cfrac { 4ab\times 2 }{ \cfrac { { { x } _{ 1 }

}^{ 2 } }{ { a }^{ 2 } } +\cfrac { { { y } _{ 1 } }^{ 2 } }{ { b }^{ 2 }

}  }  \right) =4ab$

Let $a, b$ be non-zero real numbers. The equation $\displaystyle \left ( ax^{2}+by^{2}+c \right )\left ( x^{2}-5xy+6y^{2} \right )$ represents

  1. four straight lines, when $\displaystyle c=0$ and $a, b$ are of the same sign

  2. two straight lines and a circle, when $\displaystyle a=b$ and $c$ is of sign opposite to that of $a$

  3. two straight lines and a hyperbola, when $a$ and $b$ are of the same sign and $c$ is of sign opposite to that of $a$

  4. a circle and an ellipse, when $a$ and $b$ are of the same sign


Correct Option: B
Explanation:

From given expression, $\displaystyle { x }^{ 2 }-5xy+6{ y }^{ 2 }=0$ are pair of straight lines $\displaystyle y=\frac { x }{ 2 } $ and $ y=\dfrac { x }{ 3 } $
Now, $\displaystyle a{ x }^{ 2 }+b{ y }^{ 2 }+c=0$ will be cirlce of radius $\displaystyle \sqrt { -\frac { c }{ a }  } $.

If $\displaystyle a=b$ and $\displaystyle c$ is of opposite sign of $\displaystyle a  $ and $  b$, then $\displaystyle { x }^{ 2 }+{ y }^{ 2 }=-\frac { c }{ a } $

If a hyperbola passes through the foci of the ellipse $\displaystyle \frac {x^2}{25} + \frac {y^2}{16} = 1$ and its traverse and conjugate axis coincide with major and minor axes of the ellipse, and product of the eccentricities is 1, then:

  1. Equations of the hyperbola is $\displaystyle \frac {x^2}{9} - \frac {y^2}{16} = 1$

  2. Equations of the hyperbola is $\displaystyle \frac {x^2}{9} - \frac {y^2}{25} = 1$

  3. Focus of the hyperbola is $\displaystyle (5, 0)$

  4. Focus of the hyperbola is $\displaystyle (5 \sqrt 3, 0)$


Correct Option: A,C
Explanation:

Given ellipse is, $\displaystyle \frac {x^2}{25} + \frac {y^2}{16} = 1$
Eccentricity of the ellipse is, $\displaystyle e _e  =\sqrt{1-\frac{b^2}{a^2}}=\frac{3}{5}$
So the foci of the ellipse is, $(\pm ae _e,0)=(\pm 3 , 0)$
Let eccentricity of the required  hyperbola is $e _h$ and semi major and minor axes are $a$ and $b$, so the equation of hyperbola is, $\displaystyle \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
Given hyperbola passes trough $(\pm 3,0)\Rightarrow \displaystyle \frac{9}{a^2}-\frac{0}{b^2}=1\Rightarrow a^2 = 9$
Also given that $\displaystyle  e _e\times e _h = 1$ $\Rightarrow e _h=\cfrac{5}{3}$ $\Rightarrow$ $b^2 =a^2(e _h^2-1)=16 $
Hence required hyperbola is, $\displaystyle \frac{x^2}{9}-\frac{y^2}{16}=1$
And foci of the hyperbola is, $(\pm 5,0)$

The equation ${x}^{2}+9=2{y}^{2}$ is an example of which of the following curves?

  1. hyperbola

  2. circle

  3. ellipse

  4. parabola

  5. line


Correct Option: A
Explanation:

Given, ${x}^{2}+9=2{y}^{2}$ 

$\Rightarrow 2{y}^{2}-{x}^{2}=9$
$\Rightarrow \dfrac { { y }^{ 2 } }{ 9/2 } -\dfrac { { x }^{ 2 } }{ 9 } =1$
It is in the form of $\dfrac { { y }^{ 2 } }{ { a }^{ 2 } } -\dfrac { { x }^{ 2 } }{ { b }^{ 2 } } =1$ which is the equation of hyperbola.
Therefore, the given equation is a equation of hyperbola.

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