Mathematics · Quantitative Aptitude

Conic Sections

239 Questions

Conic sections deal with the curves obtained by the intersection of a cone with a plane, primarily focusing on parabolas, ellipses, and hyperbolas. Questions require finding vertices, directrices, and asymptotes based on given equations. This is an advanced geometry topic for rigorous competitive exams.

Parabola equationsHyperbola propertiesEllipse conceptsTangents and normalsDirectrix and focus

Conic Sections Questions

Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $(3,5)$



Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $(h,k)=(2,3)$
Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

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. $\lambda\, =\, 3$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

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.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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

Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

$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. $None\quad of\quad these$
Reveal answer Fill a bubble to check yourself
D Correct answer
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$(\cfrac{-38}{5}, \cfrac{+52}{10})$.
Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

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
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $(\dfrac{19}{5},\dfrac{-13}{5})$



Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

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
Reveal answer Fill a bubble to check yourself
D Correct answer
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$

Multiple choice mathematics and statistics hyperbola auxiliary circle auxiliary circle, director circle director circle and auxiliary circle of a hyperbola

For the hyperbola $\dfrac{x^2}{64}-\dfrac{y^2}{36}=1$, the equation of director circle is 

  1. $x^2+y^2=100$
  2. $2x^2+2y^2=100$
  3. $x^2+y^2=28$
  4. $x^2-y^2=100$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The Director circle of a hyperbola is defined as the locus of the point of intersection of two perpendicular tangents to the hyperbola. For any standard hyperbola $\dfrac{x^2}{a^2} -\dfrac {y^2}{b^2} = 1$,


The equation of Director circle is given by $x^2 + y^2 = a^2 - b^2$


Here the given hyperbola is $\dfrac{x^2}{64} -\dfrac {y^2}{36} = 1$,

Here $a =8$ and $b =6$

So equation of the director circle will be $x^2 + y^2 = (8)^2 - (6)^2$

$\Rightarrow x^2 + y^2 = 28$

Correct option is $C$.

Multiple choice mathematics and statistics hyperbola auxiliary circle auxiliary circle, director circle director circle and auxiliary circle of a hyperbola

The equation of auxillary circle of hyperbola is $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$

  1. $x^2+y^2=a^2$
  2. $x^2+y^2=2a^2$
  3. $x^2+y^2=a^2+b^2$
  4. $x^2+y^2=a^2-b^2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For any Hyperbola of the form $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$,


The circle drawn taken major axis as a diameter also called the Auxiliary circle of the Hyperbola, will have a diameter of $2a$, equal to the length of major axis and center same as center of Hyperbola.

Hence equation of Auxiliary circle of any standard Hyperbola will be $x^2+y^2=a^2$

So the correct option is $A$

Multiple choice mathematics and statistics hyperbola auxiliary circle auxiliary circle, director circle director circle and auxiliary circle of a hyperbola

The circle passing through the vertices of hyperbola is called 

  1. director circle

  2. auxillary circle

  3. nine point circle

  4. none
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For any Hyperbola of the form $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$,


The circle drawn taken major axis as a diameter also called auxiliary circle of the Hyperbola, will have a diameter of $2a$, equal to the length of major axis and center same as center of Hyperbola.

Hence equation of Auxiliary circle of any standard Hyperbola will be $x^2+y^2=a^2$

As center of the Auxiliary circle is same as the center of the  hyperbola i.e. origin and the diameter is $2a$, hence the circle touches the two vertices of hyperbola $(a,0)$ and $(-a,0)$

Hence we can say that the circle passing through the two vertices of the hyperbola is Auxiliary circle. 
So the correct option is $B$.

Multiple choice mathematics and statistics hyperbola auxiliary circle auxiliary circle, director circle director circle and auxiliary circle of a hyperbola

Find the range of $p$ such that no perpendicular tangents can be drawn to the hyperbola $\dfrac{x^2}{(-p^2 + 6p + 5)} - \dfrac{y^2}{(-p - 3)} = 1$, i.e. the director circle of the given hyperbola is imaginary.

  1. $R - [-1 , 8]$
  2. $(5 , 6)$
  3. $(3 , 4)$
  4. $(-7 , 4)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The director circle is the locus of the point of intersection of a pair of perpendicular tangents to a hyperbola.

Equation of the director circle of the hyperbola $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ is $x^2 + y^2 = a^2  b^2$ i.e. a circle whose center is origin and radius is $(a^2  b^2)$.

Hence, for the director circle to be imaginary, $a^2 < b^2$ .

$-p^2 + 6p + 5 < -p - 3$ ---> $p^2 -7p -8 > 0$ - $(p-  8)(p + 1) > 0$ $\Rightarrow$ $p$ lies in $R - [-1 , 8]$. Hence, option (A).

Multiple choice mathematics and statistics hyperbola auxiliary circle auxiliary circle, director circle director circle and auxiliary circle of a hyperbola

For the hyperbola $\dfrac{x^2}{49}-\dfrac{y^2}{25}=1$, the equation of auxillary circle is

  1. $x^2+y^2=49$
  2. $x^2+y^2=25$
  3. $x^2+y^2=10$
  4. $x^2+y^2=10074$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For any Hyperbola of the form $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$,


The circle drawn taken major axis as a diameter also called the Auxiliary circle of the Hyperbola, will have a diameter of $2a$, equal to the length of major axis and center same as center of Hyperbola.

Hence equation of Auxiliary circle of any standard Hyperbola will be $x^2+y^2=a^2$

Here the given hyperbola is $\dfrac{x^2}{49}-\dfrac{y^2}{25}=1$, which is similar to standard form of the hyperbola.

Here $a = 7$ and $b = 5$

The equation of Auxiliary circle for the given hyperbola will also be $x^2 + y^2 = (7)^2$

$\Rightarrow x^2 + y^2 = 49$

So the correct option is $A$

Multiple choice mathematics and statistics hyperbola auxiliary circle auxiliary circle, director circle director circle and auxiliary circle of a hyperbola

For the hyperbola $\dfrac{x^2}{15}-\dfrac{y^2}{10}=1$, the equation of auxillary circle is

  1. $x^2+y^2=15$
  2. $x^2+y^2=10$
  3. $x^2+y^2=35$
  4. $x^2+y^2=5$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For any Hyperbola of the form $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$,


The circle drawn taken major axis as a diameter also called the Auxiliary circle of the Hyperbola, will have a diameter of $2a$, equal to the length of major axis and center same as center of Hyperbola.

Hence equation of Auxiliary circle of any standard Hyperbola will be $x^2+y^2=a^2$

Here the given hyperbola is $\dfrac{x^2}{15}-\dfrac{y^2}{10}=1$, which is similar to standard form of the hyperbola.

Here $a = \sqrt{15}$ and $b = \sqrt{10}$

The equation of Auxiliary circle for the given hyperbola will also be $x^2 + y^2 = (\sqrt{15})^2$

$\Rightarrow x^2 + y^2 = 15$

So the correct option is $A$