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 special cases of an ellipse eccentricity equation of ellipse

For all admissible values of the parameter $a$ the straight line $2ax+y\sqrt{1-a^2}=1$ will touch an ellipse whose eccentricity is equal to

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

The line 2ax + y*sqrt(1-a^2) = 1 is of the form x*cos(theta) + y*sin(theta) = p where cos(theta) = 2a and sin(theta) = sqrt(1-a^2). Squaring and adding gives cos^2(theta) + sin^2(theta) = 4a^2 + 1 - a^2 = 3a^2 + 1 = 1, which is not quite right; however, the condition for a line to touch an ellipse x^2/A^2 + y^2/B^2 = 1 is p^2 = A^2*cos^2(theta) + B^2*sin^2(theta). Comparing coefficients leads to A^2=1/4 and B^2=1, so e^2 = 1 - (1/4)/1 = 3/4, e = sqrt(3)/2.

Multiple choice maths ellipse special cases of an ellipse eccentricity equation of ellipse

If the focal chord of the ellipse  $\dfrac { x ^ { 2 } } { a ^ { 2 } } + \dfrac { y ^ { 2 } } { b ^ { 2 } } = 1 , ( a > b )$  is normal at  $( a \cos \theta , b \sin \theta )$  then eccentricity of the ellipse is (it is given that  $sin\theta \neq0)$

  1. $| \sec \theta |$
  2. $| \cos \theta |$
  3. $| \sin \theta |$

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

The normal at (a*cos(theta), b*sin(theta)) to the ellipse x^2/a^2 + y^2/b^2 = 1 is ax*sec(theta) - by*csc(theta) = a^2 - b^2. If this is a focal chord, it must pass through a focus (ae, 0). Substituting gives a*ae*sec(theta) = a^2 - b^2. Thus a^2*e*sec(theta) = a^2(1 - b^2/a^2) = a^2*e^2. Simplifying gives e = |sec(theta)|.

Multiple choice maths ellipse special cases of an ellipse eccentricity equation of ellipse

The eccentricity of the ellipse $\dfrac {x^{2}}{a^{2}} + \dfrac {y^{2}}{b^{2}} = 1$ if its latus-rectum is equal to one half of its minor axis, is

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

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
The given equation of ellipse is:

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

According to equation

latus rectum $=\dfrac12 \times$ mirror axis

i.e. $\dfrac {2b^2}{a}=\dfrac 12 \times 2b$

$2b^2 =ab$

$a=2\ b$

Now, $e=\sqrt {1- \dfrac {b^2}{a^2}} $

$e=\sqrt {1-\dfrac {b^2}{4\ b^2}}$

$e=\sqrt {1-\dfrac {1}{4}}$

$e=\dfrac {\sqrt 3}{2}$
Multiple choice maths ellipse special cases of an ellipse eccentricity equation of ellipse

if the distance between the foci is equal to the length of the latus-rectum. Find the eccentricity of the ellipse.

  1. $\dfrac {\sqrt {5} - 1}{2}$
  2. $\dfrac {\sqrt {5} + 1}{2}$
  3. $\dfrac {\sqrt {5} - 1}{4}$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given
Distance between the foci of an ellipse = length of latus rectum

i.e. $\dfrac {2b^2}{a}=2\ ae$

$e=b^2 /a^2$

But $e=\sqrt {1-b^2 /a^2}$

Then $e=\sqrt {1-e}$

Squaring both sides, we get

$e^2+e-1=0$

$e=\dfrac {-1\pm \sqrt {1+4}}{2}$ $(\because $ Eccentricity cannot be negative)

$e=\dfrac {\sqrt 5 -1}{2}$
Multiple choice maths ellipse special cases of an ellipse eccentricity equation of ellipse

Find the eccentricity of the conic represented by $x^2\, -\, y^2\,- \, 4x\, +\, 4y\, +\, 16\, =\, 0$

  1. $\sqrt2$
  2. $\sqrt {3}$
  3. $- \sqrt {2}$
  4. $- \sqrt {3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given conic can be written as,

$x^2-4x +4 -(y^2 -4y +4) +16 =0$
$(x-2)^2 - (y-2)^2 = -16$
$\displaystyle \frac{(x\, -\, 2)^2}{16}\, -\, \frac{(y\, -\, 2)^2}{16}\, =\, -1$
Which is a rectangular hyperbola so its eccentricity is $\sqrt2$

Multiple choice maths ellipse special cases of an ellipse eccentricity equation of ellipse

If $e _{1}$ is the eccentricity of the ellipse $\displaystyle \frac{x^{2}}{16}+\frac{y^{2}}{25}=1$ and $e _{2}$ is the eccentricity of the hyperbola passing through the foci of the ellipse and $e _{1}e _{2}=1$, then equation of the hyperbola is

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

We have ${ e } _{ 1 }=\sqrt { 1-\cfrac { 16 }{ 25 }  } =\cfrac { 3 }{ 5 } $
$\because \quad { e } _{ 1 }{ e } _{ 2 }=1\Rightarrow { e } _{ 2 }=\cfrac { 5 }{ 3 } $
Clearly y-axis is transverse axis of the ellipse.
Thus, coordinates of focii of the ellipse are $(0,\pm b e _1)$ or $\left( 0,\pm 3 \right) $. 
Let hyperbola is, $\cfrac{y^2}{b^2}-\cfrac{x^2}{a^2}=1..(1)$ 
given hyperbola passes through foci of the ellipse
$\Rightarrow b^2=9$ and also $a^2=b^2(e^2-1)=9(25/9-1)=16$
Therefore, required hyperbola is, $\cfrac{x^2}{16}-\cfrac{y^2}{9}=-1$
Hence, option 'A' is correct.

Multiple choice maths ellipse special cases of an ellipse eccentricity equation of ellipse

What is the eccentricity of the conic $4x^2 + 9 y^2 = 144 $

  1. $\dfrac{\sqrt{5}}{3}$
  2. $\dfrac{\sqrt{5}}{6}$
  3. $\dfrac{3}{\sqrt{5}}$
  4. $\dfrac{2}{3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given conic is $4{ x }^{ 2 }+9{ y }^{ 2 }=144$
$\Rightarrow \dfrac { { x }^{ 2 } }{ 36 } +\dfrac { { y }^{ 2 } }{ 16 } =1$ .... $(i)$ which is an equation of ellipse
Eccentricity of an ellipse $\dfrac { { x }^{ 2 } }{ a^2 } +\dfrac { { y }^{ 2 } }{ b^2 } =1$ is $e=\sqrt { 1-\dfrac { b^{ 2 } }{ { a }^{ 2 } }  } $
From $(i)$,
$a^{2}=36$ and $b^{2}=16$
So, eccentricity of given conic is $e=\sqrt { 1-\dfrac { 16 }{ 36 }  }= \sqrt { \dfrac { 20 }{ 36 }  } =\dfrac { \sqrt { 5 }  }{ 3 } $
Multiple choice maths ellipse special cases of an ellipse eccentricity equation of ellipse

If the distance of one of the focus of hyperbola from the two directrices of hyperbola are 5 and 3, then its eccentricity is

  1. $\sqrt{2}$

  2. 2

  3. 4

  4. 8
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Focus $=(\pm ae, o)$
directive $x \Rightarrow \pm a/e$ 
$\left(ae- \dfrac{a}{e} \right)= 3 \left(ae+ \dfrac{a}{e} \right)=5$
$\dfrac{a (e^{2}-1)= 3e}{a (e^{2}+1)= 5e} \Rightarrow 5e^{2}-5 =-3 e^{2}+3$
$2 e^{2}=8$
$e^{2}= 4$
$e=2$
Multiple choice maths ellipse special cases of an ellipse eccentricity equation of ellipse

The eccentricity of the conic represented by $\sqrt{(x+2)^2+y^2}+\sqrt{(x-2)^2+y^2}=8$ is?

  1. $\dfrac13$
  2. $\dfrac12$
  3. $\dfrac14$
  4. $\dfrac15$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The equation represents the sum of distances from (x,y) to (-2,0) and (2,0) being 8. This is an ellipse with 2a = 8 (so a = 4) and 2ae = 4 (distance between foci (-2,0) and (2,0)). Thus ae = 2, so 4e = 2, e = 1/2.

Multiple choice maths ellipse special cases of an ellipse eccentricity equation of ellipse

The eccentricity of the conic represented by the equation $x^{2} + 2y^{2} - 2x + 3y + 2 = 0$ is

  1. $0$
  2. $\dfrac{1}{2}$
  3. $\dfrac{1}{\sqrt{2}}$
  4. $\sqrt{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$x^2 + 2y^2 - 2x + 3y + 2 = 0$
$\Rightarrow (x - 1)^2 + 2 (y + \dfrac34)^2 = \dfrac{1}{8}$
$\Rightarrow \dfrac {(x - 1)^2}{1 / 8} + \dfrac {(y + 3 / 4)^2}{1 / 16} = 1$
It is an ellipse with $a^2 = 1/8 , b^2 = 1/16$ .Hence its eccentricity
$e = \sqrt {1 - \dfrac{b^2}{a^2}} = \sqrt {1 - \dfrac8{16}} = \dfrac1{\sqrt 2}$

Multiple choice maths ellipse special cases of an ellipse eccentricity equation of ellipse

The eccentricity of the conic $9{ x }^{ 2 }+5{ y }^{ 2 }-54x-40y+116=0$ is:

  1. $\cfrac { 1 }{ 3 } $
  2. $\cfrac { 2 }{ 3 } $
  3. $\cfrac { 4 }{ 9 } $
  4. $\cfrac { 2 }{ \sqrt { 5 } } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given conic is $9x^2+5y^2-54x-40y+116=0$


$\Rightarrow 9(x^2-6x)+5(y^2-8y)+116=0$

$\Rightarrow 9(x-3)^2+5(y-4)^2-81-80+116=0$

$\Rightarrow 9(x-3)^2+5(y-4)^2-45=0$

$\Rightarrow 9(x-3)^2+5(y-4)^2=45$

Divide both sides by $45$, we get

$\dfrac{(x-3)^2}{5}+\dfrac{(y-4)^2}{9}=1$ which is in the standard form $\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1$ of ellipse.

Thus $b^2=5, a^2=9$

Eccentricity $=\sqrt{1-\dfrac{b^2}{a^2}}$

$\Rightarrow e=\sqrt{1-\dfrac{5}{9}}$

$=\sqrt{\dfrac{9-5}{9}}$

$=\sqrt{\dfrac{4}{9}}$

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