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 maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

The maximum distance of the normal to the ellipse $\displaystyle \frac{\mathrm{x}^{2}}{9}+\frac{\mathrm{y}^{2}}{4}=1$ from its centre is:

  1. $\displaystyle \frac{1}{2}$
  2. $2$
  3. $1$
  4. $4$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Ellipse : $\cfrac { { x }^{ 2 } }{ 9 } +\cfrac { { y }^{ 2 } }{ 4 } =1$

Equation of the normal,
$\cfrac { { ax }^{  } }{ \cos  { \theta  }  } -\cfrac { { by }^{  } }{ \sin { \theta  } } ={ a }^{ 2 }-{ b }^{ 2 } \\ \therefore \cfrac { { 3x }^{  } }{ \cos  { \theta  }  } -\cfrac { { 2y }^{  } }{ \sin { \theta  } } =5$

Or, $\ { 3x }^{  }\sin { \theta  }-{ 2y }^{  }\cos  { \theta  } =5\cos  { \theta  } \sin { \theta  }$

Distance from origin d= $\cfrac { \left| 0+0-5\cos  { \theta  } \sin { \theta  } \right|  }{ \sqrt { 9{ \left( \cos  { \theta  }  \right)  }^{ 2 }+{ 4\left( \sin { \theta  } \right)  }^{ 2 } }  } $

Or, d=$\cfrac { 5 }{ \sqrt { 9{ \left( \csc { \theta  }  \right)  }^{ 2 }+{ 4 }{ \left( \sec { \theta  }  \right)  }^{ 2 } }  } $

To maximize d we need to minimize the denominator.
$E=9{ \left( \csc { \theta  }  \right)  }^{ 2 }+{ 4 }{ \left( \sec { \theta  }  \right)  }^{ 2 }\ then,\quad \\\cfrac { dE }{ d\theta  } =-18{ \left( \csc { \theta  }  \right)  }^{ 2 }\cot { \theta  } +8{ \left( \sec { \theta  }  \right)  }^{ 2 }\tan { \theta  } \ For\quad \\Minimizing,\quad \cfrac { dE }{ d\theta  } =0\\ \therefore -18{ \left( \csc { \theta  }  \right)  }^{ 2 }\cot { \theta  } +8{ \left( \sec { \theta  }  \right)  }^{ 2 }\tan { \theta  } =0 \\Or,18{ \left( \csc { \theta  }  \right)  }^{ 2 }\cot { \theta  } =8{ \left( \sec { \theta  }  \right)  }^{ 2 }\tan { \theta  } \\ Or,{ \left( \tan { \theta  }  \right)  }^{ 4 }=\cfrac { 9 }{ 4 } \\ Or,\quad \tan { \theta  } =\sqrt { \cfrac { 3 }{ 2 }  } \\ \therefore \csc { \theta  } =\sqrt { \cfrac { 5 }{ 3 }  } \quad \quad and\quad \quad \sec { \theta  } =\sqrt { \cfrac { 5 }{ 2 }  } $

 On putting the values in d we get,
$d=\cfrac { 5 }{ \sqrt { 15+10 }  } \\ Or,\quad d=1$
Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

If the line $x\cos { \alpha  } +y\sin { \alpha  } =p$ be normal to the ellipse $\dfrac { { x }^{ 2 } }{ { a }^{ 2 } } +\dfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1$, then

  1. ${ p }^{ 2 }\left( { a }^{ 2 }\cos ^{ 2 }{ \alpha } +{ b }^{ 2 }\sin ^{ 2 }{ \alpha } \right) ={ a }^{ 2 }-{ b }^{ 2 }$
  2. ${ p }^{ 2 }\left( { a }^{ 2 }\cos ^{ 2 }{ \alpha } +{ b }^{ 2 }\sin ^{ 2 }{ \alpha } \right) ={ \left( { a }^{ 2 }-{ b }^{ 2 } \right) }^{ 2 }$
  3. ${ p }^{ 2 }\left( { a }^{ 2 }\sec ^{ 2 }{ \alpha } +{ b }^{ 2 }\csc ^{ 2 }{ \alpha } \right) ={ a }^{ 2 }-{ b }^{ 2 }$
  4. ${ p }^{ 2 }\left( { a }^{ 2 }\sec ^{ 2 }{ \alpha } +{ b }^{ 2 }\csc ^{ 2 }{ \alpha } \right) ={ \left( { a }^{ 2 }-{ b }^{ 2 } \right) }^{ 2 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The equation of any normal to $\dfrac { { x }^{ 2 } }{ { a }^{ 2 } } +\dfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1$ is 
       $ax\sec { \phi  } -by\csc { \phi  } ={ a }^{ 2 }-{ b }^{ 2 }$              ......(i)
The straight line $x\cos { \alpha  } +y\sin { \alpha  } =p$ will be a normal to the ellipse $\dfrac { { x }^{ 2 } }{ { a }^{ 2 } } +\dfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1$, if equation (i) and $x\cos { \alpha  } +y\sin { \alpha  } =p$ represent the same line.
$\therefore \dfrac { a\sec { \phi  }  }{ \cos { \alpha  }  } =\dfrac { -b\csc { \phi  }  }{ \sin { \alpha  }  } =\dfrac { { a }^{ 2 }-{ b }^{ 2 } }{ p } $
$\Rightarrow \cos { \phi  } =\dfrac { ap }{ \left( { a }^{ 2 }-{ b }^{ 2 } \right) \cos { \alpha  }  } $
$\sin { \phi  } =\dfrac { -bp }{ \left( { a }^{ 2 }-{ b }^{ 2 } \right) \sin { \alpha  }  } $
$\because \sin ^{ 2 }{ \phi  } +\cos ^{ 2 }{ \phi  } =1$
$\Rightarrow \dfrac { { b }^{ 2 }{ p }^{ 2 } }{ { \left( { a }^{ 2 }-{ b }^{ 2 } \right)  }^{ 2 }\sin ^{ 2 }{ \alpha  }  } +\dfrac { { a }^{ 2 }{ p }^{ 2 } }{ { \left( { a }^{ 2 }-{ b }^{ 2 } \right)  }^{ 2 }\cos ^{ 2 }{ \alpha  }  } =1$
$\Rightarrow { p }^{ 2 }\left( { b }^{ 2 }\csc ^{ 2 }{ \alpha  } +{ a }^{ 2 }\sec ^{ 2 }{ \alpha  }  \right) ={ \left( { a }^{ 2 }-{ b }^{ 2 } \right)  }^{ 2 }$

Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

If the line $x \cos a + y \sin a = p$ be normal to the ellipse $\dfrac{x^2}{a^2}$ $+\dfrac{y^2}{b^2}$ = 1 then

  1. $p^2(a^2\cos^2a+b^2\sin^2a)=a^2-b^2$
  2. $p^2(a^2\cos^2a+b^2\sin^2a)=(a^2-b^2)^2$
  3. $p^2(a^2\sec^2a+b^2\csc^2a)=(a^2-b^2)$
  4. $p^2(a^2\sec^2a+b^2\csc^2a)=(a^2-b^2)^2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A line $y=mx+c$ is normal to ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ if $c^2=m^2\dfrac{(a^2-b^2)^2}{a^2+m^2b^2}$
Given equation $x\cos a+y\sin a=p\Rightarrow y=-(\dfrac{\cos a}{\sin a})x+\dfrac{p}{\sin a}$
Here $m=-\cot a, c=\dfrac{p}{\sin a}$
Substituting in the formulae, we get
$\dfrac{p^2}{\sin ^2a}=\dfrac{\cos ^2a}{sin^2a}\times \dfrac{(a^2-b^2)^2}{a^2+(\cot^2a) b^2}$
After simplification, we get
$p^2\dfrac{(a^2\sin^2a+b^2\cos^2a)}{\sin^2a\cos^2a}=(a^2-b^2)^2$
$p^2(a^2\sec^2a+b^2\csc^2a)=(a^2-b^2)^2$

Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

If the normal at the point $P(\theta)$ to the ellipse $\dfrac {x^{2}}{14} + \dfrac {y^{2}}{5} = 1$ intersects it again at the point $Q(2\theta)$, then $\cos \theta$ is equal to

  1. $2/3$
  2. $-2/3$
  3. $3/4$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Normal at the point P(theta) to the ellipse $\dfrac{x²}{14} + \dfrac{y²}{5 }= 1$ intersects it again at the point $Q(2\  \theta). $

we know, standard equation of ellipse is 

$\dfrac{x²}{a²} + \dfrac{y²}{b²} = 1 $ compare it with given equation

so, $ a² = 14$ then, $a = √14 $

$b² = 5$ then, $b = √5 $

now equation of normal passing through point $P(\theta)$ is given by, 

$\dfrac{ax}{cos \theta} - \dfrac{by}{sin \theta} = a² - b². $

or, $\dfrac{\sqrt14x}{cos \theta} -\dfrac{ \sqrt5y}{sin \theta} = 14 - 5 = 9$ ....(1) 

it again meets the curve at the point $Q(2\theta) $

so, $Q(2\theta) = (√14cos2\theta, √5sin2\theta) $

now, put it in equation (1), 

or, $\dfrac{14cos2\theta}{\cos \theta} - \dfrac{5sin2\theta}{\sin\theta} = 9$ 

or, ${14(2cos² \theta - 1)}{\cos \theta} - \dfrac{10sin\theta cos \theta}{\sin \theta} = 9$

or, $28cos \theta - 14sec \theta - 10cos \theta = 9$

or, $18cos \theta - \dfrac{14}{cos \theta} = 9$

or, $18cos²\theta - 14 - 9cos \theta = 0$

or, $18cos²\theta -21cos \theta + 12cos\theta - 14 = 0$

or, $3cos \theta(6cos \theta - 7) + 2(cos \theta - 7) = 0$

or, $(3cos \theta + 2)(6cos \theta - 7) = 0$

or, $cos \theta = \dfrac{-2}{3} $


Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

Which of the following is/are true?

  1. There are infinite positive integral values of $a$ for which $(13x-1)^2+(13y-2)^2=\left (\dfrac {5x+112y-1}{a}\right )^2$ represents an ellipse
  2. The minimum distance of a point $(1, 2)$ from the ellipse $4x^2+9y^2+8x-36y+4=0$ is $1$
  3. If from a point $P(0, \alpha)$ two normals other than axes are drawn to the ellipse $\dfrac {x^2}{25}+\dfrac {y^2}{16}=1$, then $|\alpha| < \dfrac {9}{4}$
  4. If the length of latus rectum of an ellipse is one-third of its major axis, then its eccentricity is equal to $\dfrac {1}{\sqrt 3}$
Reveal answer Fill a bubble to check yourself
A,B,C Correct answer
Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

Number of distinct normal lines that can be drawn to the ellipse $\displaystyle \frac{x^2}{169} + \frac{y^2}{25} = 1$ from the point $P(0, 6)$ is:

  1. One

  2. Two

  3. Three

  4. Four
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For an ellipse x^2/a^2 + y^2/b^2 = 1, the number of normals from (0, y0) depends on y0. If |y0| < |(a^2-b^2)/a|, there are 3 normals. Here a^2=169, b^2=25, so (a^2-b^2)/a = 144/13 approx 11.07. Since 6 < 11.07, there are 3 normals.

Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

If the normal at any point $P$ on the ellipse $\displaystyle\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ meets the axes in $G$ and $g$ respectively, then $PG:Pg=$

  1. $a:b$
  2. $a^2:b^2$
  3. $b:a$
  4. $b^2:a^2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let $P\equiv (a\cos\theta, b\sin\theta)$
Thus equation of normal to the given ellipse at 'P' is given by,
$ax\sec\theta-by cosec\theta=a^2-b^2$
$\therefore G \equiv ((a-\cfrac{b^2}{a})\cos\theta,0), g\equiv (0,(b-\cfrac{a^2}{b})\sin\theta)$
Thus $PG = \sqrt{\cfrac{b^4}{a^2}\cos^2\theta+b^2\sin^2\theta}=\cfrac{b}{a}\sqrt{b^2\cos^2\theta+a^2\sin^2\theta}$
and $Pg = \sqrt{a^2\cos^2\theta+\cfrac{a^4}{b^2}\sin^2\theta}=\cfrac{a}{b}\sqrt{b^2\cos^2\theta+a^2\sin^2\theta}$
$\therefore PG:Pg = \cfrac{b^2}{a^2} $

Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

The eccentricity of an ellipse whose centre is  at the origin is $\dfrac{1}{2}.$ If one of its directrices is $x =  - 4,$ then the equation of the normal to it at $\left( {1,\dfrac{3}{2}} \right)$ is

  1. $2y-x=2$
  2. $4x-2y=1$
  3. $4x+2y=7$
  4. $x+2y=4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given: Eccentricity of ellipse$=\dfrac{1}{2}$
Now, $\dfrac{a}{e}=-4$
$\Rightarrow a=4\times\dfrac{1}{2}=2$
$\therefore {b}^{2}={a}^{2}\left(1-{e}^{2}\right)$
$\Rightarrow {a}^{2}\left(1-\dfrac{1}{4}\right)=3$
$\Rightarrow \dfrac{3}{4}{a}^{2}=3$
$\therefore {a}^{2}=4$
$\dfrac{{x}^{2}}{4}+\dfrac{{y}^{2}}{3}=1$
Differentiating w.r.t $x$ we get
$\dfrac{2x}{4}+\dfrac{2y}{3}\dfrac{dy}{dx}=0$
$\Rightarrow \dfrac{dy}{dx}=\dfrac{\dfrac{-2x}{4}}{\dfrac{2y}{3}}$
$\Rightarrow \dfrac{dy}{dx}=\dfrac{-3x}{4y}$
$\Rightarrow \left[\dfrac{dy}{dx}\right] _{\left(1,\frac{3}{2}\right)}=\dfrac{-3}{4}\times\dfrac{2}{3}=\dfrac{-1}{2}$
Equation of normal at $\left(1,\dfrac{3}{2}\right)$ is 
$y-\dfrac{3}{2}=2\left(x-1\right)$
$\Rightarrow 2y-3=4x-4$
$\Rightarrow 4x-2y=1$ is the equation of the normal.
Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

The tangent and normal to the ellipse $x^2\, +\, 4y^2\, =\, 4$ at a point $P(\theta)$ on it meet the major axis in $Q$ and $R$ respectively. If $QR = 2$, the eccentric angle $\theta$ of $P$ is given by 

  1. $\cos \theta\, =\, \pm\, \dfrac23$
  2. $\sin \theta\, =\, \pm\, \dfrac23$
  3. $\tan \theta\, =\, \pm\, \dfrac23$
  4. $\cot \theta\, =\, \pm\, \dfrac23$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equation of tangent to ellipse is $bx cos\theta+ay sin\theta=ab$

It meets major axis at $x=\dfrac{a}{cos\theta}$
The equation of normal to ellipse is $ax sec \theta -by cosec \theta=a^{2}-b^{2}$
It meets the major axis at $x=\dfrac{a^{2}-b^{2}}{a sec\theta}$
Here $a=2$ and $b=1$
So, point $Q$ is $\left(\dfrac{2}{cos\theta},0 \right)$ and point $R$ is $\left(\dfrac{3}{2 sec\theta},0\right)$
The distance between them $QR= \dfrac{3cos\theta}{2}-\dfrac{2}{cos\theta}=2$
$\Rightarrow 3\cos ^{ 2 }{ \theta  } -4cos \theta-4=0$
$\Rightarrow cos \theta=-\dfrac{2}{3}$

Multiple choice position of point wrt ellipse ellipse maths

The dist.of a point P on the ellipse $\cfrac{{{x^2}}}{{12}} + \cfrac{{{y^2}}}{4} = 1$ from centre is $\sqrt 6 $ then the eccentric angle of P is 

  1. $\cfrac{\pi }{2}$
  2. $\cfrac{\pi }{6}$
  3. $\cfrac{\pi }{4}$
  4. $\cfrac{\pi }{3}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Let the point be $P = (a\cos\theta , b\sin\theta )$
 
Distance of point $P$ from centre is $\sqrt 6$
$\therefore \sqrt { { a }^{ 2 }\cos^{ 2 }\theta + b^{ 2 }\sin^{ 2 }\theta} = \sqrt { 6 }$
$\Rightarrow { a }^{ 2 }\cos^{ 2 }\theta + b^{ 2 }\sin^{ 2 }\theta = 6$
$\Rightarrow 12\cos^{ 2 }\theta + 4\sin^{ 2 }\theta = 6 \cos^{ 2 }\theta + \sin^{ 2 }\theta = 1$
$\Rightarrow 8\cos^{ 2 }\theta = 2$
$\Rightarrow \cos\theta = \cfrac { 1 }{ 2 }$
Hence, $\theta = \cfrac { \pi }{ 3 }$
Multiple choice position of point wrt ellipse ellipse maths

Point $(1,2)$ lies _____ the ellipse $\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1$.

  1. inside

  2. outside

  3. on

  4. None of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The region (disk) bounded by the ellipse is given by the equation:

$\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}\leq 1$ centered at $(h,k)$ ..... $(1)$
Point $(x,y)$ lies inside the ellipse if it satisfies $(1)$
Take $(x,y)=(1,2)$
Consider, $\dfrac{(x-0)^2}{16}+\dfrac{(y-0)^2}{9}$

                 $=\dfrac{1^{2}}{16}+\dfrac{2^{2}}{9}=\dfrac{73}{144}<1$
Hence, $(1,2)$ lies inside the given ellipse.

Multiple choice position of point wrt ellipse ellipse maths

Eccentric angle of a point on the ellipse $x^{2}+3y^{2}=6$ at a distance $2$ units. from the centre of the ellipse is

  1. $2\pi/3$
  2. $\pi/3$
  3. $4\pi/3$
  4. $none\ of\ these$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice position of point wrt ellipse ellipse maths

Let the equation of the ellipse be $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$. Let $f(x,y) = \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} - 1$. To determine whether the point $(x _1,y _1)$ lies inside the ellipse, the necessary condition is:

  1. $f(x _1,y _1) < 0$
  2. $f(x _1,y _1) > 0$
  3. $f(x _1,y _1) = 0$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$f(x,y) = \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} - 1$ ........ $(1)$ 

The region (disk) bounded by the ellipse is given by the equation:
$\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}\leq 1$ centered at $(h,k)$.

The given equation of ellipse is $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2} = 1$ centered at origin i.e. $(0,0)$.
The region bounded by this ellipse is 
$\dfrac{(x)^2}{a^2}+\dfrac{(y)^2}{b^2}\leq 1$ ...... $(1)$
The point $x _{1},y _{1}$ lies inside the given ellipse if it satisfies $(1)$
i.e. if $\dfrac{(x _{1})^2}{a^2}+\dfrac{(y _{1})^2}{b^2}\leq 1$ ...... $(1)$
if $\dfrac{(x _{1})^2}{a^2}+\dfrac{(y _{1})^2}{b^2} - 1<0$ ...... $(1)$
$\implies$ $f(x _{1}, y _{1})<0$  ....... From $(1)$
Hence, option A is correct.

Multiple choice position of point wrt ellipse ellipse maths

Point $\left(\sqrt5, \dfrac4{\sqrt5}\right)$ lies _____ the ellipse $\dfrac{x^2}{25} + \dfrac{y^2}{4} = 1$.

  1. inside

  2. outside

  3. on

  4. None of the above
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The point $(x,y)$ lies inside the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ if it satisfies the following condition i.e.

$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}\leq 1$
To check $\left(\sqrt{5}, \dfrac{4}{\sqrt{5}}\right)$ lies in the ellipse $\dfrac{x^2}{25}+\dfrac{y^2}{4} = 1$
Here $a=2$ and $b=5$
Take $(x,y)=\left(\sqrt{5}, \dfrac{4}{\sqrt{5}}\right)$
$\therefore$ $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=\dfrac{(\sqrt5)^2}{4}+\dfrac{(\frac{4}{\sqrt5})^2}{25}=\dfrac{5}{4}+\dfrac{16}{5\times 25}=\dfrac{689}{600} < 1$
Hence, $\left(\sqrt{5}, \dfrac{4}{\sqrt{5}}\right)$ lies in the given ellipse.

Multiple choice position of point wrt ellipse ellipse maths

Determine position of a point $(2,3)$ with respect to the ellipse $\dfrac{x^{2}}{16}+\dfrac{y^{2}}{25}=1$.

  1. Outside

  2. Inside

  3. On the ellipse

  4. None of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given equation is $\dfrac { { x }^{ 2 } }{ 16 } +\dfrac { { y }^{ 2 } }{ 25 } =1$

$ \dfrac { { x }^{ 2 } }{ 16 } +\dfrac { { y }^{ 2 } }{ 25 } -1=0$

The curve is defined by, 

$f(x,y)=\dfrac { { x }^{ 2 } }{ 16 } +\dfrac { { y }^{ 2 } }{ 25 } -1\\ f(2,3)=\dfrac { { (2) }^{ 2 } }{ 16 } +\dfrac { { (3) }^{ 2 } }{ 25 } -1\\ f(2,3)=\dfrac { 1 }{ 4 } +\dfrac { 9 }{ 25 } -1=\dfrac { 25+36-100 }{ 100 } =-\dfrac { 39 }{ 100 } \\ f(2,3)<0$

So, the point lies inside the ellipse.

Option B is correct.