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 parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

The equation of the hyperbola whose foci are $(8,3)$ and $(0,3)$ and eccentricity$=\cfrac { 4 }{ 3 } $ is

  1. $ 7{\left(x-4 \right ) }^{2} -9{\left(y-3 \right) }^{2}=63$
  2. ${ 7x }^{ 2 }-{ 9y }^{ 2 }=63$
  3. $ 9{ \left( x-4 \right) }^{ 2 }-9{ \left( y-3 \right) }^{ 2 }=63$
  4. $7{ \left( x+4 \right) }^{ 2 }-9{ \left( y+3 \right) }^{ 2 }=63$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The centre of the hyperbola is the mid-point of the line joining the two
foci. So, the coordinates of the centre are $\left( \cfrac { 8+0 }{ 2 },\cfrac { 3+3 }{ 2 }  \right) \quad $ i.e $(4,3)$
Let $2a,2b$ be the length of the transverse and conjugate axes and let $e$ be the
eccentricity. Then, the equation of the hyperbola is
$\cfrac { {\left( x-4 \right)  }^{ 2 } }{ { a }^{ 2 } } -\cfrac { { \left( y-3 \right)  }^{ 2 } }{ { b }^{ 2 } } =1\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (i)$
Distance between the two foci$ = 2ae$
$\Rightarrow \sqrt { { \left( 8-0 \right)  }^{ 2 }+{ \left( 3-3 \right)  }^{ 2 } } =2ae$
$\Rightarrow ae=4\Rightarrow a=3$
$\therefore\quad { b }^{ 2 }={ a }^{ 2 }\left( { e }^{ 2 }-1 \right) \Rightarrow {b }^{ 2 }=9\left( \cfrac { 16 }{ 9 } -1 \right) =7$
Substituting the value of $a$ and $b$ in $(i)$, we find that the equation of the hyperbola is
$\cfrac{ { \left( x-4 \right)  }^{ 2 } }{ 9 } +\cfrac { { \left( y-3 \right)  }^{ 2 } }{ 7 } =1\quad or\quad 7{ \left( x-4 \right)  }^{ 2 }-9{ \left( y-3 \right)  }^{ 2 }=63$
Hence, option 'A' is correct.

Multiple choice mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

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

  1. $\cfrac { { x }^{ 2 } }{ 9 } -\cfrac { { y }^{ 2 } }{ 16 } =1$
  2. $\cfrac { { x }^{ 2 } }{ 16 } -\cfrac { { y }^{ 2 } }{ 9 } =-1$
  3. $\cfrac { { x }^{ 2 } }{ 9 } -\cfrac { { y }^{ 2 } }{ 25 } =1$

  4. none of these
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 foci of the ellipse are $(0,\pm b e _1)$ or $\left( 0,\pm 3 \right) $.
Let the equation of hyperbola is, $\dfrac{y^2}{b^2}-\cfrac{x^2}{a^2}=1$ ..... $(1)$
Since, 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, the required hyperbola is, $\cfrac{x^2}{16}-\cfrac{y^2}{9}=-1$
Hence, option 'B' is correct.

Multiple choice mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

The equation of the conic with focus at $(1, -1)$, directrix along $x - y + 1= 0$ and with eccentricity $\sqrt{2}$ is

  1. $x^2 - y^2 = 1$
  2. $xy = 1$
  3. $2xy - 4x + 4y + 1 = 0$
  4. $2xy + 4x - 4y - 1 = 0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using Definition of hyperbola
$PS^2=e^2\cdot PM^2$
$(x-1)^2+(y+1)^2=2\left(\cfrac{x-y+1}{\sqrt{2}}\right)^2$
$(x^2+y^2-2x+2y+2)=(x^2+y^2+1-2xy+2x-2y)$
$\Rightarrow 2 xy - 4 x + 4y + 1 = 0$
Hence, option 'C' is correct.

Multiple choice mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

Eccentricity of the hyperbola satisfying the differential equation $2xy\dfrac{dy}{dx}=x^2+y^2$ and passing through $(2,1)$ is

  1. $\sqrt2$
  2. $2\sqrt2$
  3. $3\sqrt2$
  4. $5\sqrt2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Solving the differential equation 2xy dy/dx = x^2 + y^2 leads to the hyperbola x^2 - y^2 = c. Passing through (2, 1) gives 4 - 1 = c, so c = 3. The hyperbola is x^2 - y^2 = 3. For a rectangular hyperbola, e = sqrt(2).

Multiple choice mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

If a hyperbola passes through the focii of the ellipse$\dfrac { { x }^{ 2 } }{ 25 } +\dfrac { { y }^{ 2 } }{ 16 } =1.$ Its transverse and conjugate axes coincide respectively with the major and minor axes of the ellipse and if the product of eccentricities hyperbola and ellipse is 1, then

  1. the equation of hyperbola is $\dfrac { x^{ 2 } }{ 9 } -\dfrac { { y }^{ 2 } }{ 16 } =1\\ \quad \quad $
  2. the equation of hyperbola is $\dfrac { x^{ 2 } }{ 9 } -\dfrac { { y }^{ 2 } }{ 25 } =1\\ \quad \quad $

  3. focus of hyperbola is (5,0)
  4. focus of hyperbola is $\left( 5\sqrt { 3, } 0 \right) $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Formula,

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

$=1-\dfrac{16}{25}$

$\therefore e=\dfrac{3}{5}$

$e _2 \times e =1$

$\Rightarrow e _2=\dfrac{5}{3}$

Equation,

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

Given,

$\Rightarrow (3,0)$

$\dfrac{3^2}{a^2}=1$

$\Rightarrow a^2=9$

we have,

$e _2^2=1+\dfrac{b^2}{a^2}$

$\dfrac{25}{9}=1+\dfrac{b^2}{9}$

$\Rightarrow b^2=16$

$\dfrac{x^2}{9}-\dfrac{y^2}{16}=1$

Hence the required equation.
Multiple choice mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

The hyperbola $\displaystyle \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}}=1$ passes through the point $\displaystyle \left ( 2, : 3 \right )$ and has the eccentricity $2$. Then the transverse axis of the hyperbola has the length

  1. $1$
  2. $3$
  3. $2$
  4. $4$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given hyperbola is,  $\displaystyle \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}}=1$
It passes through $(2,3)$
$\cfrac{4}{a^{2}} - \cfrac{9}{b^{2}}=1 ..(1)$
Also eccentricity is $2$,
$\Rightarrow e^2=1+\cfrac{b^2}{a^2}=4\Rightarrow \cfrac{b^2}{a^2}=3   ..(2)$
Solving (1) and (2) we get $a=1, b=\sqrt{3}$
Hence, length of transverse axis is $=2a=2$

Multiple choice mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

If in a hyperbola the eccentricity is $\displaystyle \sqrt{3}$, and the distance between the foci is $9$ then the equation of the hyperbola in the standard form is

  1. $\displaystyle \dfrac{x^{2}}{\left ( \dfrac{\sqrt{3}}{2} \right )^{2}} - \dfrac{y^{2}}{\left ( \sqrt{\dfrac{3}{2}} \right )^{2}} = 1$
  2. $\displaystyle \dfrac{x^{2}}{\left ( \dfrac{3 \sqrt{3}}{2} \right )^{2}} - \dfrac{y^{2}}{\left ( \dfrac{3\sqrt{3}}{\sqrt{2}} \right )^{2}} = 1$
  3. $\displaystyle \dfrac{x^{2}}{\left ( \dfrac{3\sqrt{3}}{\sqrt{2}} \right )^{2}} - \dfrac{y^{2}}{\left ( \dfrac{3\sqrt{2}}{2} \right )^{2}} = 1$

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

Given eccentricity of the hyperbola $e=\sqrt{3}$
and distance between focii is 9. $\Rightarrow 2ae=9\Rightarrow a=\cfrac{3\sqrt{3}}{2}$
also $b^2=a^2(e^2-1)=\cfrac{27}{4}(3-1)=\cfrac{27}{2}\Rightarrow b=\cfrac{3\sqrt{3}}{\sqrt{2}}$
Hence equation of required hyperbola is,
$\cfrac{x^2}{\left(\cfrac{3\sqrt{3}}{2}\right )^2}-\cfrac{y^2}{\left (\cfrac{3\sqrt{3}}{\sqrt{2}}\right )^2}=1$
Hence, option 'B' is correct.

Multiple choice mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

If the eccentricity of the hyperbola $\displaystyle \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ is $e$ then the eccentricity of the hyperbola $\displaystyle \frac{y^{2}}{b^{2}} - \frac{x^{2}}{a^{2}} = 1$ is :

  1. $e$
  2. $\displaystyle \frac{e}{\sqrt{e^{2} - 1}}$
  3. $\displaystyle e \sqrt{e^{2} - 1}$
  4. $\displaystyle e^{2} - e$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For hyperbola $\displaystyle \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$
eccentricity $\Rightarrow e=\sqrt{1+\cfrac{b^2}{a^2}}\Rightarrow \cfrac{b^2}{a^2}=e^2-1$
For hyperbola $\displaystyle \frac{x^{2}}{b^{2}} - \frac{y^{2}}{a^{2}} = 1$
Required eccentricity $ e'=\sqrt{1+\cfrac{a^2}{b^2}}=\sqrt{1+\cfrac{1}{e^2-1}}=\displaystyle \cfrac{e}{\sqrt{e^{2} - 1}}$

Multiple choice mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

A hyperbola, having the transverse axis of length $\displaystyle 2\sin \theta$, is confocal with the ellipse $\displaystyle 3x^{2}+4y^{2}=12$, then its equation is

  1. $\displaystyle x^{2}\text{cosec} ^{2}\theta -y^{2}\sec ^{2}\theta=1$
  2. $\displaystyle x^{2} \sec ^{2}\theta -y^{2}\text{cosec}^{2}\theta=1$
  3. $\displaystyle x^{2} \sin ^{2}\theta -y^{2}\cos ^{2}\theta=1$
  4. $\displaystyle x^{2} \cos ^{2}\theta -y^{2}\sin ^{2}\theta=1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given ellipse may be written as $\cfrac{x^2}{4}+\cfrac{y^2}{3}=1$
$\Rightarrow a^2=4, b^2=3$

$\Rightarrow e= \sqrt{1-\dfrac{3}{4}}=\cfrac{1}{2}$
$\therefore $ Focus of the ellipse $=(\pm ae,0)=(\pm 1, 0)$
Given required hyperbola is confocal to the ellipse
Let $a',b',e'$ are transverse axis, conjugate axis an eccentricity of the hyperbola
$a'e'=1\Rightarrow \sin\theta. e'=1\Rightarrow e'=\cfrac{1}{\sin\theta}$
Using $b'^2=a'^2(e^2-1)\Rightarrow b'^2=1-\sin^2\theta=\cos^2\theta$
Therefore required hyperbola is $\cfrac{x^2}{a'^2}-\cfrac{y^2}{b'^2}=1$
$\Rightarrow \cfrac{x^2}{\sin^2\theta}-\cfrac{y^2}{\cos^2\theta}=1$
$\Rightarrow x^2 \text{cosec}^2\theta-y^2\sec^2\theta=1$

Hence, option 'A' is correct.

Multiple choice mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

Find Directrix, foci and eccentricity of the conics:

$\displaystyle x^{2}+2x-y^{2}+5= 0$

  1. Directrices $\displaystyle y= \pm \sqrt{2}$
  2. foci $\displaystyle \left ( -1,\pm 2\sqrt{2} \right )$
  3. $e= \sqrt{2}$
  4. $e=2$
Reveal answer Fill a bubble to check yourself
A,B,C Correct answer
Explanation

$\displaystyle x^{2}+2x-y^{2}+5= 0$
$\displaystyle x^{2}+2x+1-y^{2}= -4$
$\displaystyle (x+1)^2-y^{2}= -4$
$\displaystyle \frac{y^2}{4}-\frac{(x+1)^2}{4}= 1$
Clearly this equation of rectangular hyperbola with $y-$axis as major axis
eccentricity $e = \sqrt{2}$ directrices $:y = \cfrac{a}{e}=\pm \sqrt{2}$ foci $(-1,\pm ae)=(-1,\pm 2\sqrt{2})$

Multiple choice mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

Let the eccentricity of the hyperbola $  \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1  $ be reciprocal to that of the ellipse $  x^{2}+4 y^{2}=4 .  $ If thehyperbola passes through a focus of the ellipse, then __________________.

  1. (A) the equation of the hyperbola is $ \frac{x^{2}}{3}-\frac{y^{2}}{2}=1 $
  2. (B) a focus of the hyperbola is $ (2,0) $
  3. (C) the eccentricity of the hyperbola is $ \sqrt{\frac{5}{3}} $
  4. (D) the equation of the hyperbola is $ x^{2}-3 y^{2}=3 $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Ellipse x^2/4 + y^2/1 = 1 has a^2=4, b^2=1. e_e = sqrt(1 - 1/4) = sqrt(3)/2. Hyperbola eccentricity e_h = 2/sqrt(3). Hyperbola x^2/a^2 - y^2/b^2 = 1 passes through focus (ae_e, 0) = (sqrt(3), 0). So 3/a^2 = 1 => a^2 = 3. Then b^2 = a^2(e_h^2 - 1) = 3(4/3 - 1) = 1. Equation: x^2/3 - y^2/1 = 1, or x^2 - 3y^2 = 3.

Multiple choice mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

The eccentricity of the hyperbola $\displaystyle \dfrac { \sqrt { 1999 }  }{ 3 } \left( { x }^{ 2 }-{ y }^{ 2 } \right) =1$ is:

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

Equation of hyperbola is $\displaystyle \frac { { x }^{ 2 } }{ 3/\sqrt { 1999 }  } -\frac { { y }^{ 2 } }{ 3/\sqrt { 1999 }  } =1$

Here $\displaystyle { a }^{ 2 }={ b }^{ 2 }=\frac { 3 }{ \sqrt { 1999 }  } $
$\therefore$ Eccentricity $\displaystyle e=\sqrt { 1+\frac { { b }^{ 2 } }{ { a }^{ 2 } }  } =\sqrt { 1+1 } =\sqrt { 2 } $

Multiple choice mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

The equation of the hyperbola whose foci are $(6,5), (-4, 5)$ and eccentricity $\dfrac54$ is:

  1. $\displaystyle \frac{(x\, -\, 1)^2}{16}\, -\, \frac{(y\, -\, 5)^2}{9}\, =\, 1$
  2. $\displaystyle \frac{x^2}{16}\, -\, \frac{y^2}{9}\, =\, 1$
  3. $\displaystyle \frac{(x\, -\, 1)^2}{16}\, -\, \frac{(y\, -\, 5)^2}{9}\, =\, -1$
  4. $\displaystyle \frac{(x\, -\, 1)^2}{4}\, -\, \frac{(y\, -\, 5)^2}{9}\, =\, 1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Centre of the ellipse $=$ mid point of foci $=(1,5)$

Distance between foci $=\sqrt{(6-(-4))^2+(5-5^2)}$ $= 10$

$2ae=6-(-4)=10\Rightarrow a=5/e=4$

$\Rightarrow b^2 = a^2(e^2-1) = 9$

Hence required hyperbola is $\cfrac{(x-1)^2}{16}-\cfrac{(y-5)^2}{9}=1$

Multiple choice mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

The eccentricity of the hyperbola $4x^2\, -\, 9y^2\, -\, 8x\, =\, 32$ is

  1. $\displaystyle \frac{\sqrt{5}}{3}$
  2. $\displaystyle \frac{\sqrt{13}}{3}$
  3. $\displaystyle \frac{4}{3}$
  4. $\displaystyle \frac{3}{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$4x^2\, -\, 9y^2\, -\, 8x\, =\, 32$
$\Rightarrow 4(x^2-2x)-9y^2=32$
$\Rightarrow 4(x^2-2x+1)-9y^2=32+4=36$
$\Rightarrow \cfrac{(x-1)^2}{9}-\cfrac{y^2}{4}=1$
$\Rightarrow a^2=9, b^2=4$
$\therefore e=\sqrt{1+\cfrac{b^2}{a^2}}=\cfrac{\sqrt{13}}{3} $
Hence, option 'B' is correct.