Mathematics

Conic Sections

285 Questions

Conic sections deal with the geometry of curves like ellipses, hyperbolas, and parabolas formed by the intersection of a plane with a cone. This is an important topic in advanced mathematics sections of competitive exams. Practice these questions to understand eccentricity, focal distances, and equations of tangents and normals.

Ellipse equationsHyperbola eccentricityFocal distance calculationsTangent and normal equationsConic standard forms

Conic Sections Questions

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

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. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
C Correct answer
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)

$\Rightarrow 12e^2=16$

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

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

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

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 $\displaystyle (5 \sqrt 3, 0)$
Reveal answer Fill a bubble to check yourself
A,C Correct answer
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, $(\pm 5,0)$

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

If $\theta$ is eliminated from the equations $a\sec\theta - x\tan\theta = y \mbox{ and } b\sec\theta + y\tan\theta = x$ ($a$ and $b$ are constant), then the eliminant denotes the equation of 

  1. the director circle of the hyperbola $\displaystyle\frac{x^2}{a^2} - \displaystyle\frac{y^2}{b^2} = 1$
  2. auxiliary circle of the ellipse $\displaystyle\frac{x^2}{a^2} + \displaystyle\frac{y^2}{b^2} = 1$
  3. director circle of the ellipse $\displaystyle\frac{x^2}{a^2} + \displaystyle\frac{y^2}{b^2} = 1$
  4. director circle of the circle $x^2 + y^2 = \displaystyle\frac{a^2 + b^2}{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Solving given equation we get,
$\displaystyle \tan\theta=\frac{ax-by}{ay+bx}$
and $\displaystyle \sec\theta = \frac{x^2+y^2}{ay+bx}$
Eliminating $\theta$ we get, $x^2+y^2=a^2+b^2$ which is director  circle of the ellipse $\displaystyle\frac{x^2}{a^2} + \displaystyle\frac{y^2}{b^2} = 1$

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

The radius of the director circle of the ellipse $9{x^2} + 25{y^2} - 18x - 100y - 116 = 0$ is 

  1. $\sqrt {34} \,$
  2. $\sqrt {29} \,\,\,$

  3. 5

  4. 8
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equation is 9(x^2 - 2x) + 25(y^2 - 4y) = 116. Completing the square: 9(x-1)^2 + 25(y-2)^2 = 116 + 9 + 100 = 225. Dividing by 225 gives (x-1)^2/25 + (y-2)^2/9 = 1. The director circle radius is sqrt(a^2 + b^2) = sqrt(25 + 9) = sqrt(34).

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

The diametre of director circle of hyperbola $\dfrac{x^2}{25}-\dfrac{y^2}{16}=1$ 

  1. $3$
  2. $9$
  3. $6$
  4. $2$
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 $\dfrac{x^2}{25} -\dfrac {y^2}{16} = 1$,

Here $a =5$ and $b =4$

So equation of the director circle will be $x^2 + y^2 = (5)^2 - (4)^2$

$\rightarrow x^2 + y^2 = 9$

Hence the radius of the director circle is $\sqrt9 = 3$. So the diameter will be $6$.

So correct option is $C$.

Multiple choice mathematics and statistics parabola tracing of the parabola definitions related to parabola introduction to parabola

The equation of the conic with focus $\displaystyle S \left( \frac{3}{2}, 0 \right) $ and the directrix 2x + 3 = 0 having eccentricity 1, is

  1. $y^2 = 4x$
  2. $y^2 = 5x$
  3. $y^2 = 6x$
  4. $y^2 = 8x$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using the definition of a parabola (distance from focus equals distance to directrix), for focus (3/2, 0) and directrix x = -3/2, the equation is sqrt((x-3/2)^2 + y^2) = |x + 3/2|. Squaring both sides gives (x-3/2)^2 + y^2 = (x+3/2)^2, which simplifies to y^2 = 6x.

Multiple choice mathematics and statistics parabola tracing of the parabola definitions related to parabola introduction to parabola
If the focus is $ \displaystyle (\alpha, \beta) $ & the directrix is $ \displaystyle ax+by+c=0 $ then the equation of conic whose eccentricity $=e $ is given by $ \displaystyle \left ( x-\alpha \right )^{2}+\left ( y-\beta \right )^{2}=e^{2}\frac{\left ( ax+by+c \right )^{2}}{a^{2}+b^{2}}$. If $e=1$ then conic is called parabola, for $ e < 1 $ (conic is an ellipse) and for $e > 1,$ conic is a hyperbola. 
Now consider the conic
$\displaystyle 169 \left \{\left (x-1  \right )^{2}+\left (y-3  \right )^{2}  \right \}=\left (5x-12y+17  \right )^{2} $ ......$(*)$
On the basis of above information answer the following question:
 The focus of the conic $(*)$ is
  1. $(-1, -3)$
  2. $(1,3)$
  3. $(5, -12)$
  4. $(-1, 3)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

A conic is the locus of a point '$P$' which moves in such a way that its distances from a fixed point'$ S $' always bears a constant ratio to its distances from a fixed straight line. 
The fixed point '$S$' is called focus. 

The fixed straight line is called the directrix & the constant ratio is known as eccentricity denoted by $e$.
$\displaystyle \therefore e=\cfrac {PS}{PM}$
Now $\displaystyle 169\left { \left ( x-1 \right )^{2}+\left ( y-3 \right )^{2} \right }=\left ( 5x-12y+17 \right )^{2}$
$\displaystyle \Rightarrow \left { \left ( x-1 \right )^{2}+\left ( y-3 \right )^{2} \right }=\frac{\left ( 5x-12y+17 \right )^{2}}{5^{2}+(-12)^{2}}=\left ( \frac{5x-12y+17}{13} \right )^{2}$
$\displaystyle \therefore \left ( x-1 \right )^{2}+\left ( y-3 \right )^{2}=e^{2}\left ( \frac{5x-12+17}{13} \right )$ where $e=1$

Focus of conic $(* )$ is $(1, 3)$

Multiple choice mathematics and statistics parabola tracing of the parabola definitions related to parabola introduction to parabola

 If the focus is $ \displaystyle (\alpha, \beta) $ & the directrix is $ \displaystyle ax+by+c=0 $ then the equation of conic whose eccentricity $=e $ is given by $ \displaystyle \left ( x-\alpha \right )^{2}+\left ( y-\beta \right )^{2}=e^{2}\frac{\left ( ax+by+c \right )^{2}}{a^{2}+b^{2}}$. If $e=1$ then conic is called parabola, for $ e < 1 $ (conic is an ellipse) and for $e > 1,$ conic is a hyperbola. 

Now consider the conic
$\displaystyle 169 \left {\left (x-1  \right )^{2}+\left (y-3  \right )^{2}  \right }=\left (5x-12y+17  \right )^{2} $ ......$()$
On the basis of above information answer the following question:
The equation of axis of the conic $()$  is

  1. $\displaystyle 12x+5y+20=0 $
  2. $\displaystyle 12x- 5y+17 =0 $
  3. $\displaystyle 12x+5y+17 =0 $
  4. $\displaystyle 12x+5y-27=0 $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

A conic is the locus of a point'$P$' which moves in such a way that
its distances from a fixed point'$ S $' always bears a constant ratio
to its distances from a fIxed straight line.
The fixed point '$S$'
is called focus. The fixed straight line is called directrix' & the
constant ratio' is known as eccentricity denoted by $e$.
$\displaystyle \therefore e=PS/PM$
Now $\displaystyle 169\left { \left ( x-1 \right )^{2}+\left ( y-3 \right )^{2} \right }=\left ( 5x-12y+17 \right )^{2}$
$\displaystyle\Rightarrow \left { \left ( x-1 \right )^{2}+\left ( y-3 \right )^{2}
\right }=\frac{\left ( 5x-12y+17 \right )^{2}}{5^{2}+(-12)^{2}}=\left (
\frac{5x-12y+17}{13} \right )^{2}$
$\displaystyle \therefore \left
( x-1 \right )^{2}+\left ( y-3 \right )^{2}=e^{2}\left (
\frac{5x-12y+17}{13} \right )^{2}$ where $e=1$
Any line passing through focus $(1, 3)$ and perpendicular to $
\displaystyle 5x-12y+17=0$ is the axis of conic (*). Now any line
$\displaystyle \perp$er to $ \displaystyle 5x-12y+17=0$ is given by
$\displaystyle 12x+5y+\lambda =0$ but it passes through focus
$\displaystyle \therefore \lambda =-12\left ( 1 \right )-5\left ( 3 \right )=-27$
$\displaystyle \therefore $ equation of axis is $\displaystyle 12x+5y-27=0$

Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

The point of intersection of the two ellipse $x^2+2y^2-6x-12y+23=0$ and $4x^2+2y^2-20x-12y+35=0$

  1. lie on a circle centered at $\displaystyle \left( \frac { 8 }{ 3 } ,3 \right) $ and of radius $\displaystyle \frac { 1 }{ 3 } \sqrt { \frac { 47 }{ 2 } } $
  2. lie on a circle centered at $\displaystyle \left( -\frac { 8 }{ 3 } ,-3 \right) $ and of radius $\displaystyle \frac { 1 }{ 3 } \sqrt { \frac { 47 }{ 2 } } $
  3. lie on a circle centered at $\displaystyle \left( 8 ,9 \right) $ and of radius $\displaystyle \frac { 1 }{ 3 } \sqrt { \frac { 47 }{ 3 } } $

  4. are not concyclic
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If ${S} _{1}=0$ and ${S} _{2}=0$ are the equations, then, $\lambda { S } _{ 1 }+{ S } _{ 2 }=0$ is a second degree curve passing through the points of intersection of ${S} _{1}=0$ and ${S} _{2}=0.$

$\Rightarrow \left( \lambda +4 \right) { x }^{ 2 }+2\left( \lambda +1 \right) { y }^{ 2 }-2\left( 3\lambda +10 \right) x-12\left( \lambda +1 \right) y+\left( 23\lambda +35 \right) =0$   ...(1)
For it to be a circle, choose $\lambda$ such that the coefficients of ${x}^{2}$ and ${y}^{2}$ are equal: $\Rightarrow \lambda+4=2\lambda+2$
$\therefore \lambda=2$
This gives the equation of the circle as $6\left( { x }^{ 2 }+{ y }^{ 2 } \right) -32x-36y+81=0$  {(using (1))}
$\displaystyle \Rightarrow { x }^{ 2 }+{ y }^{ 2 }-\frac { 16 }{ 3 } x-6y+\frac { 27 }{ 2 } =0$ 
Its centre is $\displaystyle C\left( \frac { 8 }{ 3 } ,3 \right) $ and radius is $\displaystyle r=\sqrt { \frac { 64 }{ 9 } +9-\frac { 27 }{ 2 }  } =\frac { 1 }{ 3 } \sqrt { \frac { 47 }{ 2 }  } .$

Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

The points of intersection of the two ellipses $x^{2}+2y^{2}-6x-12y+23=0$ and $4x^{2}+2y^{2}-20x-12y+35=0$.

  1. lie on a circle centred at $\left(\dfrac83, 3\right)$ and of radius $\displaystyle \frac{1}{3}\sqrt{\frac{47}{2}}$
  2. lie on a circle centred at $\left(-\dfrac83, 3\right)$ and of radius $\displaystyle \frac{1}{3}\sqrt{\frac{47}{2}}$
  3. lie on a circle centred at $(8, 9)$ and of radius $\displaystyle \frac{1}{3}\sqrt{\frac{47}{2}}$

  4. are not cyclic.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Equation of any curve passing through the intersection of the given ellipse is
   $4x^{2}+2y^{2}-20x-12y+35+\lambda \left ( x^{2}+2y^{2}-6x-12y+23 \right )=0$
which represents a circle is
   $4+\lambda =2+2\lambda \Rightarrow \lambda =2$
and the equation of the circle is thus,
   $6x^{2}+6y^{2}-32x-36y+81=0$
$\Rightarrow $   $\displaystyle x^{2}+y^{2}-\left ( \frac{16}{3} \right )x-6y+\frac{81}{6}=0$
centre of the circle is $\left(\dfrac83, 3\right)$
and the radius is $\displaystyle \sqrt{\left ( \frac{8}{3} \right )^{2}+\left ( 3 \right )^{2}-\frac{81}{6}}$
   $\displaystyle =\sqrt{\frac{128+162-243}{18}}=\frac{1}{3}\sqrt{\frac{47}{2}}$

Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

The points of intersection of the two ellipses ${ x }^{ 2 }+2{ y }^{ 2 }-6x-12y+23=0$ and $4{ x }^{ 2 }+2{ y }^{ 2 }-20x-12y+35=0$

  1. lies on a circle centered at $\displaystyle \left( \frac { 8 }{ 3 } ,3 \right) $ and of radius $\displaystyle \frac { 1 }{ 3 } \sqrt { \frac { 47 }{ 2 } } $
  2. lies on a circle centered at $\displaystyle \left( -\frac { 8 }{ 3 } ,-3 \right) $ and of radius $\displaystyle \frac { 1 }{ 3 } \sqrt { \frac { 47 }{ 2 } } $
  3. lies on a circle centered at $\displaystyle \left( 8,9 \right) $ and of radius $\displaystyle \frac { 1 }{ 3 } \sqrt { \frac { 47 }{ 3 } } $

  4. are not cyclic

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

If ${ S } _{ 1 }=0$ and ${ S } _{ 2 }=0$ are the equations, then $\lambda { S } _{ 1 }+{ S } _{ 2 }=0$ is a second degree curve passing through the points of intersection of ${ S } _{ 1 }=0$ and ${ S } _{ 2 }=0$

$\Rightarrow \left( \lambda +4 \right) { x }^{ 2 }+2\left( \lambda +1 \right) { y }^{ 2 }-2\left( 3\lambda +10 \right) x-12\left( \lambda +1 \right) y+\left( 23\lambda +35 \right) =0$
For it to be a circle, choose $\lambda$ such that the coefficients of ${ x }^{ 2 }$ and ${ y }^{ 2 }$ are equal:
$\Rightarrow \lambda +4=2\lambda +2\Rightarrow \lambda =2$
This gives the equation of the circle as
$\displaystyle 6\left( { x }^{ 2 }+{ y }^{ 2 } \right) -32x-36y+81=0$    (Using (1))
$\displaystyle \Rightarrow { x }^{ 2 }+{ y }^{ 2 }-\frac { 16 }{ 3 } x+6y+\frac { 27 }{ 2 } =0$
Its center is $\displaystyle C\left( \frac { 8 }{ 3 } ,3 \right) $ and radius is $\displaystyle r=\sqrt { \frac { 64 }{ 9 } +9-\frac { 27 }{ 2 }  } =\frac { 1 }{ 3 } \sqrt { \frac { 47 }{ 2 }  } $

Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

If the ellipse $\displaystyle \frac{x^{2}}{4}+\frac{y^{2}}{b^{2}}=1$ meets the ellipse $\displaystyle \frac{x^{2}}{1}+\frac{y^{2}}{a^{2}}=1$ in four distinct points and $\displaystyle a^{2} = b^{2} -4b + 8$, then $b$ lies in

  1. $(- \infty ,0)$
  2. $(- \infty ,2)$
  3. $(2,\infty)$
  4. $[2, \infty)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given,
$\displaystyle \dfrac{x^{2}}{1}+\dfrac{y^{2}}{a^{2}}=1$   -- (i)

$\displaystyle \dfrac{x^{2}}{4}+\dfrac{y^{2}}{b^{2}}=1$   -- (ii) 
are the two equations of the ellipses

Eliminating $y^2$ from both the equations we get, 
$ x^2 \left( \dfrac{b^2-4a^2}{4a^2b^2} \right) =  \dfrac{b^2-a^2}{a^2b^2} $

$\dfrac{x^2}{4}  = \dfrac{b^2-a^2}{b^2-4a^2} $

Substituting the value of $a^2$ we get, 
$ \dfrac{x^2}{4} = \dfrac{4b-8}{-3b^2+16b-32} $

The denominator is always negative as the discriminant of the expression is negative and the coefficient of $b^2$ is also negative. 

Hence, $4b-8 < 0$
$\Rightarrow b <2 $

Eliminating $x^2$ from the two equations we get, 

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

$ \dfrac{y^2}{3} = \dfrac {a^2b^2} { 4a^2 - b^2} $

Hence, $4a^2 -b^2 >0 $
$\Rightarrow 3b^2 -16b +32 > 0 $. 

The discriminant of the expression is less than 0 and the coefficient of $b^2$ is greater than 0. Hence, the inequality holds true for all values of $b$. 

Hence the common set of the values of $b$ is $ (-\infty,  2) $. 
Hence, option B is correct