Mathematics · Quantitative Aptitude

Conic Sections

239 Questions

Conic sections deal with the curves obtained by the intersection of a cone with a plane, primarily focusing on parabolas, ellipses, and hyperbolas. Questions require finding vertices, directrices, and asymptotes based on given equations. This is an advanced geometry topic for rigorous competitive exams.

Parabola equationsHyperbola propertiesEllipse conceptsTangents and normalsDirectrix and focus

Conic Sections Questions

Multiple choice intersection of a line and a parabola conic section maths

The condition that the straight line $\displaystyle lx + my + n = 0$ touches the parabola $\displaystyle x^2 = 4ay$ is

  1. $\displaystyle bn = am^2$
  2. $\displaystyle al^2 - mn = 0$
  3. $\displaystyle ln = am^2$
  4. $\displaystyle am = ln^2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a line y = mx + c to touch x^2 = 4ay, the condition is c = -am^2. Rewriting lx + my + n = 0 as y = (-l/m)x - (n/m), we set c = -n/m and m_slope = -l/m. Thus, -n/m = -a(-l/m)^2 = -a(l^2/m^2). Multiplying by -m^2 gives nm = al^2, or al^2 - nm = 0.

Multiple choice intersection of a line and a parabola conic section maths

The length of the chord of the parabola $y^2 = x$ which is bisected at the point $(2, 1)$ is

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

Chord through $(2, 1)$ is $ \cfrac{x- 2}{\cos  \theta} = \cfrac{y - 1}{\sin \theta} = r$    ... (i)
Solving equation (i) with parabola $y^2 = x$, we have
$(1 + r  \sin \theta)^2 = 2 + r  \cos \theta$
$\Rightarrow \sin^2 \theta r^2 + (2  \sin  \theta - \cos  \theta) r - 1 = 0$
This equation has two roots $r _1 = AC$ and $r _2 = - BC$
Then, sum of roots $r _1 + r _2 = 0$
$\Rightarrow 2\sin \theta - \cos  \theta = 0   \Rightarrow   \tan  \theta = \cfrac{1}{2}$
$AB = |r _1 - r _2| $
$= \sqrt{(r _1 + r _2)^2 - 4r _1 r _2}$
$= \sqrt{ 4 \cfrac{1}{\sin^2 \theta}} $

$= 2 \sqrt{5}$

Multiple choice intersection of a line and a parabola conic section maths

If $2$ and $3$ are the length of the segments of any focal chord of a parabola $y^2 = 4ax$, then value of $2a$ is

  1. $\dfrac{13}{5}$
  2. $\dfrac{12}{5}$
  3. $\dfrac{11}{5}$

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

If $2$ and $3$ are lengths of focal chord of parabola ${ y }^{ 2 }=4ax$ then 2a is :

Let ${ l } _{ 1 }=2\quad { l } _{ 2 }=3$
Semi latus rectum$=2a=\cfrac { 2{ l } _{ 1 }{ l } _{ 2 } }{ { l } _{ 1 }+{ l } _{ 2 } } $
$=\cfrac { 2(2)(3) }{ 2+3 } =\cfrac { 12 }{ 5 } $

Multiple choice intersection of a line and a parabola conic section maths

If the line $y- \sqrt x +3 = 0$ cuts the parabola $y^2 = x + 2$ at $A$ and $B$, and if $P$ $(3,\ 0)$, then $PA.PB$ is equal to

  1. $\dfrac{2(\sqrt 3+2)}{3}$
  2. $\dfrac{4\sqrt 3}{2}$
  3. $\dfrac{4(2-\sqrt 3)}{3}$
  4. $\dfrac{4(\sqrt3+2)}{3}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

If line  $y-\sqrt { 3 } x+3=0$ cuts parabola ${ y }^{ 2 }=x+2$ at $A$ and $B$ 

If $P(3,0)$ the $PA.PB$ is : 
$y-\sqrt { 3 } x+3=0\ y=\sqrt { 3 } x-3\ \cfrac { y-0 }{ \cfrac { \sqrt { 3 }  }{ 2 }  } =\cfrac { x-\sqrt { 3 }  }{ \cfrac { 1 }{ 2 }  } =r\ x=\cfrac { r }{ 2 } +\sqrt { 3 } \ y=\cfrac { \sqrt { 3 } r }{ 2 } $
Put in ${ y }^{ 2 }=x+2$
$\cfrac { 3 }{ 4 } { r }^{ 2 }=\cfrac { r }{ 2 } +\sqrt { 3 } +2\ { 3r }^{ 2 }-2r-4(\sqrt { 3 } +2)=0\ PA.PB=\left| { r } _{ 1 }{ r } _{ 2 } \right| =\left| \cfrac { c }{ a }  \right| =\cfrac { 4(\sqrt { 3 } +2) }{ 3 } $

Multiple choice conjugate hyperbola hyperbola conic section maths

The equation of the conjugate axis of the hyperbola $\frac{{{{\left( {y - 2} \right)}^2}}}{9} - \frac{{{{\left( {x + 3} \right)}^2}}}{{16}} = 1$ is

  1. $y=2$
  2. $y=6$
  3. $y=8$
  4. $y=3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice conjugate hyperbola hyperbola conic section maths

If variable has its interceptson the coordinates axes $e$ and $e'$ where $e/2$ and $e'/2$ are the eccentricities of hyperbola and conjugate hyperbola, Then the line always touches the circle $x^{2}+y^{2}=r^{2}$, where $r=$ 

  1. $1$
  2. $2$
  3. $3$
  4. $Cannot\ be\ decided$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If the line x/e + y/e' = 1 touches x^2 + y^2 = r^2, the perpendicular distance from the origin to the line must equal r. Using the formula for a line touching a circle, r = 1/sqrt(1/e^2 + 1/e'^2). Given the relationship between eccentricities of conjugate hyperbolas 1/e^2 + 1/e'^2 = 1, we find r = 1.

Multiple choice conjugate hyperbola hyperbola conic section maths

Assertion(A): lf the lines $3x+y+p=0$ and $2x+5y-3=0$ are conjugate with respect to $3x^{2}-2y^{2}=6$ then $\mathrm{p}=1$

Reason(R): lf the lines $l _{1}x+m _{1}y+n _{1}=0$ and $l _{2}x+m _{2}y+n _{2}=0$ are conjugate with respect to the hyperbola $\mathrm{S}=0$ is $a^{2}l _{1}l _{2}+b^{2}m _{1}m _{2}=n _{1}n _{2}$


  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 correct

    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
C Correct answer
Explanation

 lf the lines $l _{1}x+m _{1}y+n _{1}=0$ and $l _{2}x+m _{2}y+n _{2}=0$ are conjugate with re- spect to the hyperbola $\mathrm{S}=0$ is $a^{2}l _{1}l _{2}-b^{2}m _{1}m _{2}=n _{1}n _{2}$
then $2(3)(2)-3(1)(5)=-3p$
therefore, $p=1$

Multiple choice conjugate hyperbola hyperbola conic section maths

The equation to the conjugate hyperbola of $2x^{2}-3y^{2}-4x+6y-15=0$ is 

  1. $2x^{2}-3y^{2}-4x+6y+13=0$
  2. $2x^{2}-3y^{2}-4x+6y-1=0$
  3. $2x^{2}-3y^{2}-4x+6y+15=0$
  4. $2x^{2}-3y^{2}-4x+6y-8=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The conjugate hyperbola of ax^2 + by^2 + 2gx + 2fy + c = 0 is ax^2 + by^2 + 2gx + 2fy + c' = 0, where the center (h, k) is the same. For 2x^2 - 3y^2 - 4x + 6y - 15 = 0, center is (1, -1). The constant term changes such that the new equation is 2x^2 - 3y^2 - 4x + 6y + k = 0. The conjugate hyperbola has the same asymptotes, so the constant term is adjusted.

Multiple choice conjugate hyperbola hyperbola conic section maths

If the hyperbolas, $ x^2+3xy+2y^2+2x+3y+2=0 $ and $ x^2+3xy+2y^2+2x+3y+c=0 $ are conjugate of each other, the value of $c$ is equal to

  1. $-2$
  2. $4$
  3. $0$
  4. $1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The given hyperbola is $x^2+3xy+2y^2+2x+3y+2=0$         ...(1)

We already know that the equation of the asymptote of a hyperbola differs from the hyperbola by a constant.
$\therefore$ Let $x^2+3xy+2y^2+2x+3y+k=0$            ...(2)
be the equation of the asymptotes of the given hyperbola.
Hence, equation (2) must represent a pair of straight lines, the condition for which is
$abc+2fgh-af^2-bg^2-ch^2 = 0$
$\Rightarrow (1)(2)(k)+2(\cfrac 32)(1)(\cfrac 32)-1(\cfrac 32)^2-2(1)^2-k(\cfrac 32)^2=0$
$\Rightarrow 2k+\cfrac 92-\cfrac 94-2-\cfrac 94 k=0$
$\Rightarrow k=1$
Therefore, the asymptotes are given by $x^2+3xy+2y^2+2x+3y+1=0$ .

The equation of conjugate hyperbola is $2A-H=0$
where, $A$ is the equation of asymptotes
            $H$ is the equation of given hyperbola
$2(x^2+3xy+2y^2+2x+3y+1) -$$(x^2+3xy+2y^2+2x+3y+2)=0$
$\therefore x^2+3xy+2y^2+2x+3y=0$

Hence, $c=0$.

Multiple choice conjugate hyperbola hyperbola conic section maths

If the line $lx+my+n=0$ meets the hyperbola $\displaystyle \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ at the extermities of a pair of conjugate diameters, then

  1. $a^{2}l^{2}-b^{2}m^{2}=0$
  2. $a^{2}l^{2}-b^{2}m^{2}=1$
  3. $a^{2}l^{2}-b^{2}m^{2}=2$
  4. $a^{2}l^{2}-b^{2}m^{2}=3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let $ \theta$ and $\phi$ be the essentric angles of the conjugate diameters, 
Then,
$\theta +\phi =\dfrac{\pi}{2}$
$(a\sec \theta,b\tan \theta)   , (a\sec \phi, b \tan\phi)$
$\Rightarrow (a \sec \theta,b \tan \theta)   , (a\ \text{cosec }\theta ,b \cot \theta)$

$(y-b\tan \theta)=\dfrac{b\cot \theta-b\tan\theta}{a\ \text{cosec }\theta-a\sec\theta}(x-a\sec\theta)$

$\Rightarrow ya-ab\tan \theta=xb(\sin\theta+\cos \theta)-ab(\tan\theta+1)$
$ya=xb(\sin\theta+\cos\theta) - ab$
$ ya - xb (\sin\theta+\cot\theta)+ab=0$
$lx+my+n=0$

Comparing both, we get

$\dfrac{a}{m}=\dfrac{-b(\sin\theta+\cos\theta)}{l}=\dfrac{ab}{n}$

On elliminating $\theta $, we can get
$a^{2}l^{2}-b^{2}m^{2}=0$

Multiple choice conjugate hyperbola hyperbola conic section maths

Find the equation to the hyperbola,conjugate to the hyperbola $ 2x^2+3xy-2y^2-5x+5y+2=0 $.

  1. $ 2x^2+3xy-2y^2-5x+5y-8=0 $
  2. $ x^2+3xy-y^2-5x+5y-8=0 $
  3. $ x^2+3xy-y^2-5x+5y+8=0 $

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

Let Asymptotes :
$ 2x^2+3xy-2y^2-5x+5y+\lambda=0 $.
$ \therefore abc+2fgh-af^2-bg^2-ch^2=0 $
$ \lambda=-5 $
Equation Hyperbola + Conjugate Hyperbola$=$2 (Asympototes)
$ \therefore $ Conjugate Hyperbola$=$2 (Asymptotes) $- $Hyperbola
Therefore, equation of conjugate hyperbola is $  2x^2+3xy-2y^2-5x+5y-8=0 $

Multiple choice conjugate hyperbola hyperbola conic section maths

The equation of a hyperbola, conjugate to the hyperbola $x^2+3xy+2y^2+2x+3y=0$ is?

  1. $x^2+3xy+2y^2+2x+3y+1=0$
  2. $x^2+3xy+2y^2+2x+3y+2=0$
  3. $x^2+3xy+2y^2+2x+3y+3=0$
  4. $x^2+3xy+2y^2+2x+3y+4=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$H:x^2+3xy +2y^2+2x+3y=0$
Let the pair of asympt at is be
$A:x^2 +3xy+2y^2+2x+3y+k=0$
$\therefore \ $ It represent a apir of straight lines, satisfying the condition :
$abc+2fgh-af^2 -bg^2-ch^2=0$
$\Rightarrow \ 1\times 2\times k+2\left (\dfrac {3}{2}\right) \times 1\times \left (\dfrac {3}{2}\right) -1 \left (\dfrac 32\right)^2 -2(1)^2 -k\left (\dfrac 32 \right)^2 =0$
$\Rightarrow \ 2k+\dfrac {9}{4}-2-\dfrac {9}{4}k=0\ \Rightarrow \ k=1$
$\Rightarrow \ A^2 .x^2 +3xy+2y^2 +2x+3y+1=0$
As $H+C=2A\ \Rightarrow \ C$ (conjugate $=2A-H$ hypergate) 
$\Rightarrow \ C:x^2 +3xy+2y^2+2x+3y+2=0$
Multiple choice mathematics and statistics coordinates, points and lines what is meant by the equation of a straight line or of a curve introduction to slope introduction to straight lines

Normals are drawn at points $A, B, and C$ on the parabola ${ y }^{ 2 }= { 4x }$ which intersect at P\left( h, k \right)$. The locus of the point $P$ if the slope of the line joining the feet of two of them is $2$, is

  1. ${ x + y = 1 }$
  2. ${ x - y = 3 }$
  3. ${ { y }^{ 2 } = 2\left(x-1\right) }$
  4. ${ { y }^{ 2 }=2\left( x-\frac { 1 }{ 2 } \right) }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

The hyperbola $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$, normals are drawn to curve $\left( {{{\left( {\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}}} \right)}^2} - 1} \right)\left( {\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}}} \right) = 0$.
Find the sum;  of abscissa of foot of all such normals.

  1. $\frac{{6{a^2}h}}{{\left( {{a^2} + {b^2}} \right)}}$
  2. $\frac{{8{a^2}h}}{{\left( {{a^2} + {b^2}} \right)}}$
  3. $\frac{{6a{h^2}}}{{\left( {{a^2} + {b^2}} \right)}}$
  4. $\frac{{8a{h^2}}}{{\left( {{a^2} + {b^2}} \right)}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

This is a complex geometry problem involving the sum of abscissae for normals to a hyperbola. The derived result for the sum is 8*a^2*h / (a^2 + b^2).

Multiple choice normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

If the straight line $(a - 2) x - by + 4 = 0$ is normal to the hyperbola $xy = 1$ then which of the followings does not hold?

  1. $a > 1, b > 0$
  2. $a > 1, b < 0$
  3. $a < 1, b < 0$
  4. $a < 1, b > 0$
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

Every normal to $xy = 1$ must have positive slope as $\dfrac {-dx}{dy} = x^{2}$. So $\dfrac {a - 1}{b} > 0$.