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 position of point wrt ellipse ellipse maths

Which of the following points is an exterior point of the ellipse $\displaystyle 16 x^{2} + 9y^{2} - 16x - 32 = 0$.

  1. $\displaystyle \left ( \frac{1}{2}, \: 2 \right )$
  2. $\displaystyle \left ( \frac{1}{4}, \: 2 \right )$
  3. $\displaystyle \left ( 3, \: 2 \right )$

  4. none of these
Reveal answer Fill a bubble to check yourself
B,C Correct answer
Explanation

Let   $S = \displaystyle 16 x^{2} + 9y^{2} - 16x - 32 $
Now $S(\dfrac12,2)=4+36-8-32 = 0 \Rightarrow $ point on the ellipse.
$S(\dfrac14,2) = 1+36-4-32> 0 \Rightarrow $ point is exterior to the ellipse.
$S(3,2) = 144+36-48-32>0 \Rightarrow $ point is exterior to the ellipse.

Multiple choice position of point wrt ellipse ellipse maths

An ellipse with foci $(0,\pm 2)$ has length of minor axis as $4$ units. Then the ellipse will pass through the point

  1. $\left( 2,\sqrt { 2 } \right) $
  2. $\left( \sqrt { 2 } ,2 \right) $
  3. $\left( 2,2\sqrt { 2 } \right) $
  4. $\left( 2\sqrt { 2 } ,2 \right) $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let $\cfrac { { x }^{ 2 } }{ { a }^{ 2 } } +\cfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1(a<b)\quad $ is the equation of ellipse, foci $(0,\pm 2)$
(be $=2$)
Given: $2a=4\Rightarrow a=2$
${ e }^{ 2 }=1-\cfrac { { a }^{ 2 } }{ { b }^{ 2 } } \Rightarrow { b }^{ 2 }{ e }^{ 2 }={ b }^{ 2 }-{ a }^{ 2 }\quad $
$\quad 4={ b }^{ 2 }-4\Rightarrow { b }^{ 2 }=8$
$\therefore$ equation of ellipse is $\cfrac { { x }^{ 2 } }{ 4 } +\cfrac { { y }^{ 2 } }{ 8 } =1\quad $
It passes through $\left( \sqrt { 2 } ,2 \right) $

Multiple choice position of point wrt ellipse ellipse maths

Let $E$ be the ellipse $\displaystyle \frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 4 } =1$ and $C$ be the circle ${ x }^{ 2 }+{ y }^{ 2 }=9$. Let $P$ and $Q$ be the points $(1,2)$ and $(2,1)$ respectively. Then

  1. $Q$ lies inside $C$ but outside $E$
  2. $Q$ lies outside both $C$ and $E$
  3. $P$ lies inside both $C$ and $E$
  4. $P$ lies inside $C$ but outside $E$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Since $\displaystyle \frac { { 1 }^{ 2 } }{ 9 } +\frac { { 2 }^{ 2 } }{ 4 } -=\frac{1}{9}>0$

$\therefore P(1,2)$ lies outside $E$
Since $\displaystyle \frac { { 2 }^{ 2 } }{ 9 } +\frac { { 1 }^{ 2 } }{ 4 } -1<0$
$\therefore Q(2,1)$ lies inside $E$
Since ${ 1 }^{ 2 }+{ 2 }^{ 2 }-9<0$
$\therefore P(1,2)$ lies inside $C$
Since ${ 2 }^{ 2 }+{ 1 }^{ 2 }-9<0$
$\therefore Q(2,1)$ also lies inside $C$
$\therefore P$ lies inside $C$ but outside $E$.

Multiple choice position of point wrt ellipse ellipse maths

Find the equation of the ellipse whose eccentricity is $\dfrac{4}{5}$ and axes are along the coordinate axes and foci at $(0, \pm 4)$.

  1. $\dfrac{x^2}{9}+\dfrac{y^2}{25}=1$
  2. $\dfrac{x^2}{4}+\dfrac{y^2}{16}=1$
  3. $\dfrac{x^2}{9}+\dfrac{y^2}{16}=1$
  4. $\dfrac{x^2}{9}+\dfrac{y^2}{36}=1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the required equation of the ellipse be $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$.


According to the problem, the coordinates of the foci are $(0, \pm 4)$.

We know, coordinates of foci are $(0, \pm be)$.

Therefore, $be =4$

$b\left (\dfrac{4}{5}\right )=4$

$b=5$

$b^2=25$

Now, $a^2=b^2(1-e^2)$

$a^2=5^2\left (1-\dfrac{16}{25}\right )$

$a^2=9$

Thus, the required equation of ellipse is $\dfrac{x^2}{9}+\dfrac{y^2}{25}=1$.

Multiple choice position of point wrt ellipse ellipse maths

Consider the ellipse with the equation $x^{2}+3y^{2}-2x-6y-2=0.$ The eccentric angle of a point on the ellipse at a distance 2 units from the contra of the ellipse is

  1. $\dfrac{\pi }{4}$
  2. $\dfrac{\pi }{2}$
  3. $\dfrac{\pi }{6}$
  4. $\dfrac{\pi }{3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We have,  $x^{2}+3y^{2}-2x-6y-2=0.$

$\Rightarrow$ $x^{2}-2x+1-1+3y^{2}-6y+3-3-2=0.$

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

$\Rightarrow$ $(x-1)^{2}+3(y-1)^{2}=6$ which becomes $x^{2}+3y^{2}=6$ on shifting the origin to $(1, 1)$. Any point with eccentric angle $\theta $ is $(\sqrt{6}cos\theta ,\sqrt{2}sin\theta )$ 

$\Rightarrow$ $4=6cos^{2}\theta +2sin^{2}\theta \Rightarrow 4cos^{2}\theta =2\Rightarrow cos\theta =\pm \dfrac{1}{\sqrt{2}}$

$\Rightarrow$ Hence $\theta =\dfrac{\pi }{4}$ 

Multiple choice position of point wrt ellipse ellipse maths

Find the set of value(s) of $\alpha$ for which the point $\left ( 7\,-\, \displaystyle \frac{5}{4}\alpha,\,\alpha \right )$ lies inside the ellipse $\displaystyle \frac{x^2}{25}\,+\,\frac{y^2}{16}\,=\, 1.$

  1. $ \displaystyle\left( \frac{17}{5} \dfrac{12}{5}\right) $
  2. $ \left(\dfrac{12}{5},\dfrac{16}{5}\right) $
  3. $ \dfrac{-16}{5} $

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

$\displaystyle \frac{x^2}{25}\,+\,\frac{y^2}{16}\,=\, 1$


Since point $(7\,-\, \displaystyle \frac{5}{4}\alpha,\,\alpha)$ lies inside the ellipse

$\therefore S _1\,<\,0$

$\Rightarrow 16 (7\,-\, \displaystyle \frac{5}{4}\alpha)^2\,+\, 25.\alpha^2\,<\,400$

$\Rightarrow\,(28\, -\,5\alpha)^2\,+\, 25\alpha^2\,<\, 400$

$\Rightarrow\, 50\alpha^2\, -\, 280\,\alpha\,+\, 384\, < \,0$

$\Rightarrow\, 25\alpha^2\, -\, 140\,\alpha\,+\, 192\, < \,0$

$\Rightarrow (5\alpha-12)(5\alpha-16)<0$

$\Rightarrow \, \alpha \, \in \, \left( \dfrac { 12 }{ 5 } ,\, \dfrac { 16 }{ 5 }  \right) $

Multiple choice position of point wrt ellipse ellipse maths

$\mathrm{A}$ssertion ($\mathrm{A}$): The point $(5,-2)$ lies outside the ellipse $24x^{2}+7y^{2}=12$.
Reason (R): lf the point $(x _{1},y _{1})$ lie outside the ellipse $\mathrm{S}=0$ then $S _{11}>0$ 

  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 coorect 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

$24x^{2}+7y^{2}-12=5$

$S(5,-2)=24\times 25+7\times 4-12$

$S(5,-2)>0$

$\therefore $ lies outside the ellipse.
$S=0$
$S(x _{1}y _{1})>0$ the point is outside the ellipse.

Multiple choice position of point wrt ellipse ellipse maths

The point $(2\cos \theta , 3\sin \theta)$ lies ____________ the ellipse $\dfrac{x^2}{4}+\dfrac{y^2}{9}=1$.

  1. outside

  2. inside

  3. on the periphery of

  4. on the auxillary of
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given point is $\left ( 2\cos\theta, 3\sin\theta  \right )$


Substituting given point in the ellipse equation:
$\dfrac{x^{2}}{4}+\dfrac{y^{2}}{9}-1$
$= \dfrac{4\cos^{2}\theta}{4}+\dfrac{9\sin^{2}\theta }{9}-1$
$= \cos^{2}\theta + \sin^{2}\theta -1$
$= 1-1=0$

$\therefore$ Given point satisfies the ellipse 
$\Rightarrow$ point lie on periphery of ellipse.

Multiple choice position of point wrt ellipse ellipse maths

The distance of point '$\theta$' on the ellipse $\dfrac {x^2}{a^2} + \dfrac {y^2}{b^2}=1$ from a focus is:

  1. $a(e + \cos \theta)$
  2. $a(e - \cos \theta)$
  3. $a(1 + e \cos \theta)$
  4. $a(1 + 2e \cos \theta)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given equation of ellipse is
$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$
Any point on the ellipse will be $(a cos\theta, b sin \theta)$
$P=(acos\theta, b sin \theta)$
Centre of the ellipse is $(0,0)$
Ellipse is parallel to horizontal axis
Foci of the ellipse is
$F=(h-ae,k)$ if $(h,k) $ is the centre
$F=(0-ae,0)=(-ae,o)$
Distance $FP=\sqrt{(-ae-cos\theta)^2+(bsin\theta-0)^2}$
$=a \sqrt{(e^2+cos^2 \theta+2ecos\theta+(1-e^2)sin \theta-0)^2}$
$a\sqrt{(e^2+cos^2 \theta+2ecos\theta+sin^2 \theta-e^2sin^2\theta)}$
$a\sqrt{(1+2ecos\theta+e^2(1-sin^2\theta)}$
$a\sqrt{(1+2ecos \theta+e^2cos^2\theta)}$
$a\sqrt{(1+ecos\theta)^2}$
$FP=a(1+ecos\theta)$



Multiple choice position of point wrt ellipse ellipse maths

$(2,3)$ lies _______  the ellipse $16 x^{2} + 9y^{2} - 16x - 32 = 0$

  1. Inside

  2. Outside

  3. On

  4. None of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given ellipse $16x^2+9y^2-16x-32=0$
Let $S=16x^2+9y^2-16x-32=0$
Put $(2,3) $ in $S$ we get
$S(2,3)=16(2)^2+9(3)^2-16(2)-32=81 $
$S(2,3)>0$
So it $(2,3) $ lies outside the ellipse.
Multiple choice position of point wrt ellipse ellipse maths

The point $(4\cos \theta , 4\sin \theta)$ lies ____________ the ellipse $\dfrac{x^2}{16}+\dfrac{y^2}{9}=1$

  1. outside

  2. inside

  3. on the periphery

  4. on the auxiliary circle
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Equation of ellipse is $\dfrac { { x }^{ 2 } }{ 16 } +\dfrac { { y }^{ 2 } }{ 9 } =1$

Equation of auxiliary circle of ellipse is
${ x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }\ { x }^{ 2 }+{ y }^{ 2 }=16$
Substiuting $(4\cos\theta,4\sin\theta$) in the equation, we get
${ (4\cos { \theta  } ) }^{ 2 }+{ (4\sin { \theta  } ) }^{ 2 }=16\ 16(\cos ^{ 2 }{ \theta  } +\sin ^{ 2 }{ \theta  } )=16\ 16=16$
The point satisfies the equation of auxiliary circle.
So, option D is the correct.

Multiple choice position of point wrt ellipse ellipse maths

$(3,2)$ lies _______  the ellipse $16 x^{2} + 9y^{2} - 16x - 32 = 0$

  1. Inside

  2. Outside

  3. On

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

Ellipse : $16{ x }^{ 2 }+9{ y }^{ 2 }-16x-32=0$

Let $S=16{ x }^{ 2 }+9{ y }^{ 2 }-16x-32$
Putting point (3,2) in S we get
         S(3,2) $=16\times { (3) }^{ 2 }+9\times { (2) }^{ 2 }-16\times 3-32$
                   $=144+36-48-32$
                   $=180-80$
                   $=100$
          $S(3,2) > 0$
Hence, point lies outside ellipse.

Multiple choice position of point wrt ellipse ellipse maths

The point $(1,1)$ lies ____________ the ellipse $\dfrac{x^2}{4}+\dfrac{y^2}{9}=1$

  1. outside

  2. inside

  3. on the periphery

  4. on the auxillary circle
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We will substitute the given point in ellipse equation

$\Rightarrow$ $\dfrac{x^{2}}{4}+\dfrac{y^{2}}{9}-1$
$=\dfrac{1}{4}+\dfrac{1}{9}-1$
$\Rightarrow  \dfrac{13}{36}-1$
$\Rightarrow \dfrac{-23}{36}<0$

By Substituting the point, we are getting less than $0$ 

$\therefore $ point lies inside the ellipse .

Multiple choice position of point wrt ellipse ellipse maths

The distance of a point $(\sqrt 6 \cos \theta, \sqrt 2 \sin \theta)$ on the ellipse $\dfrac {x^2}{6} + \dfrac {y^2}{2}=1$ from the centre is $2$, if:

  1. $\theta =\dfrac {\pi}{2}$
  2. $\theta =\dfrac {3\pi}{2}$
  3. $\theta =\dfrac {5\pi}{2}$
  4. $\theta = \dfrac{\pi}{4}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$P(\sqrt { 6 } \cos { \theta  } ,\sqrt { 2 } \sin { \theta  } )$

$Centre(0,0)$
$PC=\sqrt { 6\cos ^{ 2 }{ \theta  } +2\sin ^{ 2 }{ \theta  }  } $
$2=\sqrt { 4\cos ^{ 2 }{ \theta  } +2 } $
$2=4\cos ^{ 2 }{ \theta  } $
$\cos { \theta  } =\pm \dfrac { 1 }{ \sqrt { 2 }  } $
$\theta =\dfrac { \pi  }{ 4 } $

Multiple choice position of point wrt ellipse ellipse maths

Let $\dfrac {(x-3) ^2}9+\dfrac {(y-4) ^2}{16}=1$ then   $(0,0)$ is

  1. On the ellipse.

  2. Outside the ellipse.

  3. Inside the ellipse.

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

Equation of ellipse : $\dfrac { { \left( x-3 \right)  }^{ 2 } }{ 9 } +\dfrac { { \left( y-4 \right)  }^{ 2 } }{ 16 } -1=0$

putting point (0,0) in above ellipse,
          $=\dfrac { { \left( -3 \right)  }^{ 2 } }{ 9 } +\dfrac { { \left( -4 \right)  }^{ 2 } }{ 16 } -1$
          $=1+1-1$
          $=1>0$
Hence, point lies outside ellipse.