Mathematics

Conic Sections

278 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 ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

Find the  Lactus Rectum of  $\displaystyle 9y^{2}-4x^{2}=36$ 

  1. $ 9$
  2. $6$
  3. $11$
  4. $15$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\displaystyle \frac{y^{2}}{4}-\frac{x^{2}}{9}= 1.$ 
Here the coefficient of $\displaystyle y^{2}$ is + ive and that of $\displaystyle x^{2}$ is -ive and hence it represents a hyperbola whose transerse axis is vertical, i.e.
$\displaystyle a^{2}=4, b^{2}=9.$
$\displaystyle b^{2}= a^{2}\left ( e^{2}-1 \right )$
or $\displaystyle \frac{9}{4}+1=e^{2}\therefore e= \frac{\sqrt{13}}{2}$ 
Foci lie on y-axis $\displaystyle \left ( 0, \pm ae \right )$ i.e $\displaystyle \left ( 0, \pm ae \sqrt{13} \right )$ 
$\displaystyle L.R.= \frac{2b^{2}}{a}= 2.\frac{9}{2}= 9$

Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

The difference between the lengths of the major axis and the latus-rectum of an ellipse is

  1. $ae$
  2. $2ae$
  3. $ae^{2}$
  4. $2ae^{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We know that the length of major axis is $2a$ and latus rectum is $\dfrac {2b^2}{a}$

for the ellipse

$\dfrac {x^2}{a^2}+\dfrac {y^2}{b^2}=1$

Let $d$ be the difference

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

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

We know thta $b^2 =a^2 (1-e^2)$

$d=\dfrac {2a^2 e^2}{a}$

$d=2ae^2$
Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

The latus-rectum of the conic $3x^{2} + 4y^{2} - 6x + 8y - 5 = 0$ is

  1. $3$
  2. $\dfrac {\sqrt {3}}{2}$
  3. $\dfrac {2}{\sqrt {3}}$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given equation of conic is:

$3x^2+4y^2-6x+8y-5=0$

$3(x^2-2x)+4(y^2+2y)=5$

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

$3(x-1)^2 +4(y+1)^2 =12$

$\dfrac {3(x-1)^2}{12}+\dfrac {4(y+1)^2}{12}=1 \Rightarrow \dfrac {(x-1)^2}{4}+\dfrac {(y+1)^2}{3}=1$

so, $a=2, b=\sqrt 3$

$\therefore \ $ Latus rectum $=\dfrac {2b^2}{a}$

$=\dfrac {2[\sqrt 3]^2}{2}$

$=3$
Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

The equation $\dfrac {x^{2}}{2 - \lambda} + \dfrac {y^{2}}{\lambda - 5} - 1 = 0$ represents an ellipse, if

  1. $\lambda < 5$
  2. $\lambda < 2$
  3. $2 < \lambda < 5$
  4. $\lambda < 2$ or $\lambda < 5$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
General equation of ellipse is $\dfrac {x^2}{a^2}+\dfrac {y^2}{b^2}=1$

So both denominator should be positive as they are squares
 
In the given equation

$\dfrac {x^2}{2-\lambda} +\dfrac {y^2}{\lambda -5}-1=0$

So,
 
$2-\lambda > 0, +(\lambda -5) >o$

$\Rightarrow \ \lambda < 2, \lambda > 5$

$\Rightarrow \ 2 < \lambda < 5$
Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

An ellipse has its centre at $(1, -1)$ and semi-major axis $= 8$ and it passes through the point $(1, 3)$. The equation of the ellipse is

  1. $\dfrac {(x + 1)^{2}}{64} + \dfrac {(y + 1)^{2}}{16} = 1$
  2. $\dfrac {(x - 1)^{2}}{64} + \dfrac {(y + 1)^{2}}{16} = 1$
  3. $\dfrac {(x - 1)^{2}}{16} + \dfrac {(y + 1)^{2}}{64} = 1$
  4. $\dfrac {(x + 1)^{2}}{64} + \dfrac {(y - 1)^{2}}{16} = 1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given that

centre is at $(1, -1)$

semi major axis $(a)=8$

so, equation of ellipse can be written as

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

It passes through point $(1,3)$

i.e, $x=1, y=3$

Putting these value in equation $(1)$ we get

$\dfrac {(1-1)^2}{a^2} +\dfrac {(3+1)^2}{b^2}=1$

$\dfrac {16}{b^2}=1$

$b^2=16\ \Rightarrow b=4$

Substituting the values of $a$ and $b$ in equation $(1)$ we get

$\dfrac {(x-1)^2}{64}+\dfrac {(y+1)^2}{16}=1$

This is the required equation of ellipse
Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

The equation $\displaystyle \frac {x^2}{8-t}\, +\, \displaystyle \frac {y^2}{t-4}\, =\, 1$ will represent an ellipse if

  1. $t\, \in\, (1,\, 5)$
  2. $t\, \in\, (2,\, 8)$
  3. $t\, \in\, (4,\, 8)\, -\, \{6\}$
  4. $t\, \in\, (4,\, 10)\, -\, \{6\}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Consider Equation, $\displaystyle\frac{x^2}{a^2}+\displaystyle\frac{y^2}{b^2}=1$ to represent an ellipse equation.
$a>0,b>0,a\neq b$
Given,equation $\displaystyle\frac{x^2}{(8-t)}+\displaystyle\frac{y^2}{(t-4)}=1$
$\Rightarrow (8-t)>0\;$ and $\;(t-4)>0,(8-t)\neq(t-4)$
$\Rightarrow t\in(-\infty,8) \cap (4,\infty) \cap$ {$t\neq6$}
$\Rightarrow t\in(4,8)-${$6$}

Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

$\mathrm{S}$ and $\mathrm{S}^{'}$ are the foci of the ellipse $25x^{2}+16y^{2}=1600$, then the sum of the distances from $\mathrm{S}$ and $\mathrm{S}'$ to the point $(4\sqrt{3},5)$ is:

  1. $20$
  2. $15$
  3. $40$
  4. $30$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given ellipse is $25{ x }^{ 2 }+16{ y }^{ 2 }=1600$

This can be written as $\displaystyle \frac { { x }^{ 2 } }{ 64 } +\frac { { y }^{ 2 } }{ 100 } =1$
Compare it to standard form of ellipse to get $a=8,b=10$
here $a<b$
So eccentricity e$=\sqrt { 1-\displaystyle \frac { { a }^{ 2 } }{ { b }^{ 2 } }  } =\sqrt { 1-\displaystyle\frac { 64 }{ 100 }  } =\displaystyle \frac { 3 }{ 5 } $
Now the foci are given by $\left( 0,\pm \sqrt { { b }^{ 2 }-{ a }^{ 2 } }  \right) $
So $foci:\left( 0,\pm \sqrt { 36 }  \right) $
$S:\left( 0,6 \right) ,{ S } _{ 1 }:\left( 0,-6 \right) $
Now find distance of $S$ and ${ S } _{ 1 }$ from the given point $\left( 4\sqrt { 3 } ,5 \right) $
So sum$=\sqrt { { \left( 4\sqrt { 3 }  \right)  }^{ 2 }+1 } +\sqrt { { \left( 4\sqrt { 3 }  \right)  }^{ 2 }+{ \left( 11 \right)  }^{ 2 } } =7+13=20$

Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

The equation of the ellipse having vertices at $\displaystyle \left( \pm 5,0 \right) $ and foci $\displaystyle \left( \pm 4,0 \right) $ is

  1. $\displaystyle \frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 16 } =1$
  2. $\displaystyle 9{ x }^{ 2 }+25{ y }^{ 2 }=225$
  3. $\displaystyle \frac { { x }^{ 2 } }{ 9 } +\frac { { y }^{ 2 } }{ 25 } =1$
  4. $\displaystyle 4{ x }^{ 2 }+5{ y }^{ 2 }=20$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The vertices and foci of an ellipse are $\displaystyle \left( \pm 5,0 \right) $ and $\displaystyle \left( \pm 4,0 \right) $ respectively.
$\displaystyle \therefore \quad a=5$ and $\displaystyle ae=4$
$\displaystyle \Rightarrow \quad e=\frac { 4 }{ 5 } $
We know that,
$\displaystyle e=\sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } }  } $
$\displaystyle \Rightarrow \quad \frac { 16 }{ 25 } =1-\frac { { b }^{ 2 } }{ 25 } \Rightarrow { b }^{ 2 }=9$
Hence, equation of an ellipse is
$\displaystyle \frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 9 } =1\Rightarrow 9{ x }^{ 2 }+25{ y }^{ 2 }=225$

Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

The foci of an ellipse are located at the points $(2, 4)$ and $(2, -2)$. The points $(4, 2)$ lies on the ellipse. If $a$ and $b$ represent the lengths of the semi-major and semi-minor axes respectively, then the value of $(ab)^{2}$ is equal to

  1. $68 + 22\sqrt {10}$
  2. $6 + 22\sqrt {10}$
  3. $26 + 10\sqrt {10}$
  4. $6 + 10\sqrt {10}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The distance between the foci is $6$, so $c = 3$.
The sum of the distance from $(4, 2)$ to each of the foci is the major axis length,
so
$2a = \sqrt {(4 - 2)^{2} + (2 - 4)^{2}} + \sqrt {(4 - 2)^{2} + (2 + 2)^{2}}$
$= \sqrt {4 + 4} + \sqrt {4 + 16} = \sqrt {8} + \sqrt {20}$
$= 2\sqrt {2} + 2\sqrt {5} \Rightarrow a = \sqrt {2} + \sqrt {5}$
Also, for an ellipse,
$b^{2} = a^{2} - c^{2} = (\sqrt {2} + \sqrt {5})^{2} - 3^{2}$
$= 7 + 2\sqrt {10} = -2 + 2\sqrt {10}$.
Thus, we have
$(ab)^{2} = (7 + 2\sqrt {10})(-2 + 2\sqrt {10})$
$= -14 + 14\sqrt {10} - 4\sqrt {10} + 40$
$= 26 + 10\sqrt {10}$.

Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

Which of the following is/are not false?

  1. The mid point of the line segment joining the foci is called the centre of the ellipse.

  2. The line segment through the foci of the ellipse is called the major axis.

  3. The end points of the major axis are called the vertices of the ellipse.

  4. Ellipse is symmetric with respect to Y-axis only.
Reveal answer Fill a bubble to check yourself
A,B,C Correct answer
Explanation

(A) Midpoint of the line segment joining the foci is called the centre of ellipse: TRUE


(B) Line segment through the foci is called major axis: TRUE

(C) End point of major axis are called vertices of ellipse: TRUE

(D) Ellipse is symmetric with respect to Y-axis only.: FALSE
(Ellipse is symmetric to both x-axis and y-axis)

Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

The equation of ellipse whose major axis is along the direction of x-axis, eccentricity is $e=2/3$

  1. $36x^2+20y^2=405$
  2. $20x^2+36y^2=405$
  3. $30x^2+22y^2=411$
  4. $22x^2+32y^2=409$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$\begin{array}{l} e=\frac { 2 }{ 3 } =\sqrt { \frac { { { a^{ 2 } }-{ b^{ 2 } } } }{ c }  }  \\ \Rightarrow { \left( { \frac { 2 }{ 3 }  } \right) ^{ 2 } }=\frac { { { a^{ 2 } }-{ b^{ 2 } } } }{ { { a^{ 2 } } } }  \\ \Rightarrow \frac { { 4{ a^{ 2 } } } }{ a } ={ a^{ 2 } }-{ b^{ 2 } } \\ \Rightarrow { b^{ 2 } }={ a^{ 2 } }-4{ a^{ 2 } }=\frac { { 5{ a^{ 2 } } } }{ 9 } \to \left( i \right)  \end{array}$
Equation of Ellipse are
$\begin{array}{l} \Rightarrow \frac { { { x^{ 2 } } } }{ { { a^{ 2 } } } } +\frac { { { y^{ 2 } } } }{ { { b^{ 2 } } } } =1 \\ \Rightarrow \frac { { { x^{ 2 } } } }{ { { a^{ 2 } } } } +\frac { { 9{ y^{ 2 } } } }{ { 5{ a^{ 2 } } } } =1\to \left( { ii } \right)  \\ Put\, \, { a^{ 2 } }=\frac { { 405 } }{ { 20 } } \, \, \left( { From\, \, option\, \, in\, \, equation\left( i \right)  } \right)  \\ Then,\, \, { b^{ 2 } }=\frac { { 405 } }{ { 360 } }  \end{array}$
Hence, equation of ellipse is
$ \Rightarrow \frac{{{x^2}}}{{\left( {\frac{{405}}{{20}}} \right)}} + \frac{{{y^2}}}{{\left( {\frac{{405}}{{36}}} \right)}} =  - 1,20{x^2} + 36{y^2} = 405$
Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

Eccentricity of ellipse $\frac{{{x^2}}}{{{a^2} + 1}} + \frac{{{y^2}}}{{{a^2} + 2}} = 1\,is\,\frac{1}{{\sqrt 3 }}$ then length of Latus rectum is 

  1. $\frac{2}{{\sqrt 3 }}$
  2. $\frac{4}{{\sqrt 3 }}$
  3. $2\sqrt 3 $
  4. $\frac{{\sqrt 3 }}{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Let ${ A }^{ 2 }={ a }^{ 2 }+1$,  ${ B }^{ 2 }={ a }^{ 2 }+2$    {Here, ${ B }^{ 2 }>{ A }^{ 2 }$}
So, $e=\sqrt { 1-\dfrac { { A }^{ 2 } }{ { B }^{ 2 } }  } =\sqrt { 1-\dfrac { { a }^{ 2 }+1 }{ { a }^{ 2 }+2 }  } =\dfrac { 1 }{ \sqrt { { a }^{ 2 }+2 }  } =\dfrac { 1 }{ \sqrt { 3 }  } $
So,  $\sqrt { { a }^{ 2 }+2 } =\sqrt { 3 } \Rightarrow a=\pm 1$
So, length of lotus return $=2\dfrac { { A }^{ 2 } }{ B } $
Length $=\dfrac { 2\left( { a }^{ 2 }+1 \right)  }{ \sqrt { { a }^{ 2 }+2 }  } =\dfrac { 2\left( 2 \right)  }{ \sqrt { 3 }  } =\dfrac { 4 }{ \sqrt { 3 }  } $
Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

If the latus rectum of an ellipse $x ^ { 2 } \tan ^ { 2 } \varphi + y ^ { 2 } \sec ^ { 2 } \varphi =$ $1$ is $1 / 2 ,$ then $\varphi$ is

  1. $\pi / 2$
  2. $\pi / 6$
  3. $\pi / 3$
  4. $5$ $\pi/ 12$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given $x^2 tan^2 \phi + y^2 \, sec^2 \phi = 1$

$\rightarrow \dfrac{x^2}{(1/tan^2 \phi)} + \dfrac{y^2}{(1/sec^2 \phi)} = 1$
$a = \pm \dfrac{1}{tan \phi} , b = \pm \dfrac{1}{sec \phi}$
and $\rightarrow e^2 = 1 - \dfrac{b^2}{a^2}$
$\rightarrow e^2 = 1 - \dfrac{1/sec^2 \phi}{1/tan^2 \phi} = 1 - \dfrac{tan^2 \phi}{sec^2 \phi}$
$\rightarrow e^2 = 1 - sin^2 \phi = cos^2 \phi$
length of latus rectum 
$(LL') = \dfrac{2 b^2}{a} = 2a (1 - e^2)$
$\rightarrow 2a (1 - cos^2 \phi) = 2a. sin^2 \phi = \dfrac{1}{2} $ (Given)
$\therefore 2. \dfrac{cos \phi}{sin \phi} sin^2 \phi = \dfrac{1}{2}$
$\rightarrow 2 cos \phi \, sin \phi = \dfrac{1}{2}$
$\rightarrow sin^2 \phi = \dfrac{1}{2} $
$\rightarrow 2 \phi = \dfrac{\pi}{6} , \dfrac{5 \pi}{6}$
$\therefore \phi = \dfrac{\pi}{12}$    or 
$\phi = \dfrac{ 5 \pi}{12}$

Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

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

  1. $4x+2y=7$
  2. $x+2y=4$
  3. $2y-x=2$
  4. $4x-2y=1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

With eccentricity e = 1/2 and directrix x = -4, the focus and center can be found to determine the ellipse equation. Using the point (1, 3/2) on the ellipse, the normal equation can be derived, matching option D.