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 mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

The asymptotes of the hyperbola $xy-3x+4y+2=0$

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

Since the equation of a hyperbola and its asymptotes differ in constant terms only. Therefore, the equations of asymptotes of the given hyperbola are given by $xy-3x+4y+k=0$

where $k$ is a constant to be determined  by the condition that $abc+2fgh-{ af }^{ 2 }-{ bg }^{ 2 }-{ ch }^{ 2 }=0$
i.e., $\displaystyle 0+2\times 2\times \left( \frac { -3 }{ 2 }  \right) \times \frac { 1 }{ 2 } -0-0-k\times { \left( \frac { 1 }{ 2 }  \right)  }^{ 2 }=0\Rightarrow k=-12$
$\because $ Asymptotes of the given hyperbola are $xy-3x+4y-12=0$ or $(x+4)(y-3)=0$
i.e., $x=-4$ and $y=3.$

Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

If $x + 2 = 0$ and $y = 1$ are the equation of asymptotes of rectangular hyperbola passing through (1,0).Then which of the following is(are) not the equation(s) of hyperbola :

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

Equation of hyperbola is of the form $(x+2)(y-1)=k$
Since, it passes through $(1,0)$
Therefore, $(1+2)(0-1)=k$
$\Rightarrow k=-3$
Therefore, equation of hyperbola is $xy-x+2y+1=0$

Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

If ax + by + c = 0 and $\displaystyle \varphi \chi $ + my + n = 0 are asymptotes of a hyperbola, then: 

  1. $\displaystyle am\neq b\varphi $
  2. $\displaystyle \frac{am+b\varphi }{a\varphi +bm}\neq 0$
  3. $\displaystyle a\varphi \neq bm$

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

Asymptotes of Hyperbola are Intersecting each other 


So, These line will be anti-parallel or intersecting

Condition for intersecting lines is $\dfrac{a}{b}\neq\dfrac{\varphi}{m}\Rightarrow am \neq b\varphi$

Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

The asymptotes of a hyperbola are parallel to lines $2x + 3y = 0$ and $3x + 2y = 0.$ The hyperbola has its centre at $(1, 2)$ and it passes through $(5, 3).$ Find its equation.

  1. $(2x\, +\, 3y\, -\, 8) (3y\, +\, 2y\, -\, 7)\, =\, 154$
  2. $(2x\, +\, 3y\, -\, 7) (3y\, +\, 2y\, -\, 8)\, =\, 154$
  3. $(2x\, +\, 3y\, -\, 7) (3y\, +\, 2y\, -\, 8)\, =\, 127$
  4. $(2x\, +\, 3y\, -\, 8) (3y\, +\, 2y\, -\, 7)\, =\, 127$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

let the equation of asymtotes be $2x+3y=a$ and $3x+2y=b$
both asymtotes intersect at centre $(1,2)$
Therefore, $a=2+3(2)=8$ and $b=3+2(2)=7$
now, the equation of hyperbola is of the form $(2x+3y-8)(3x+2y-7)=k$
It passes through $(5,3)$
Therefore, $(2(5)+3(3)-8)(3(5)+2(3)-7)=k$
Thus $k=154$

Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

The asymptotes of the hyperbola $xy+3x+2y = 0$ are

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

Let equation of asymptotes be $xy\,+ \, 3x\, \, +\, 2y +\, \lambda$ = 0.
Then $abc\, +\, 2fgh\,-\, af^2\,-\, bg^2\,-\, ch^2\, =\, 0$
$\displaystyle \Rightarrow\, \frac{3}{2}\, -\, \frac{\lambda}{4}\, =\, 0\, \Rightarrow\, \lambda\, =\, 6$
$\therefore$ Equation of asymptotes is $xy +3x +2y + 6 = 0$
$\Rightarrow (x+ 2) (y +3) = 0\Rightarrow x+2=0$ and $y+3=0$

Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

Find the asymptotes of the hyperbola $2x^2\, -\, 3xy\,- \, 2y^2\, +\, 3x\,- \, y\, +\, 8\, =\, 0$. Also find the equation to the conjugate hyperbola & the equation of the principal axes of the curve.

  1. $x - 2y + 1 = 0; 2x + y + 1 = 0; 2x^2\,- \, 3xy\, -\, 2y^2\, +\, 3x\,- \, y\,- \, 6\, =\, 0; 3x y + 2 = 0; x - 3y = 0$
  2. $x + 2y - 1 = 0; 2x + y + 1 = 0; 2x^2\,- \, 3xy\, -\, 2y^2\, +\, 3x\,- \, y\,+\, 6\, =\, 0; 3x y + 2 = 0; x + 3y = 0$
  3. $x - 2y + 1 = 0; 2x + y + 1 = 0; 2x^2\,- \, 3xy\, -\, 2y^2\, +\, 3x\,- \, y\,- \, 6\, =\, 0; 3x y + 2 = 0; x + 3y = 0$
  4. $x - 2y + 1 = 0; 2x - y + 1 = 0; 2x^2\,- \, 3xy\, -\, 2y^2\, +\, 3x\,- \, y\,+ \, 6\, =\, 0; 3x y - 2 = 0; x - 3y = 0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let equation of asymptotes are 
$2x^2\, \, -3xy\, \,- 2y^2\, +\, 3x\, \, -y\, +\, 8\, +\, \lambda\,=\, 0$
As it represents two straight lines
$\displaystyle \therefore\, -4(8\, +\, \lambda)\, +\, \frac{9}{4}\, -\, \frac{1}{2}\, +\, \frac{9}{2}\, -\, (8\, +\, \lambda) \frac{9}{4}\, =\, 0$
$\Rightarrow\, \lambda\, =\, -7$
So asymptotes are $2x^2\, -\, 3xy\, -\, 2y^2\, +\, 3x\, -\, y\, +\, 1\, =\, 0$
$\Rightarrow$ 2y - x - 1 = 0 & 2x + y + 1 = 0
and the equation of conjugate hyperbola will be
$2x^2\, \, -3xy\, \, -2y^2\, +\, 3x\, \, -y\, +\, 8\, \, -14\, =\, 0$.

Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

The equation of hyperbola conjugate to the hyperbola $2x^2 + 3xy - 2y^2 - 5 + 5y + 2 = 0$ is

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

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

Let the given hyperbola be $ H =2x^{2}+3xy-2y^{2}-5x+5y+2=0$


Thus the pair of asymptotes be $ A = 2x^{2}+3xy-2y^{2}-5x+5y+\lambda =0 $

Pair of straight line has $ \Delta =0 $
$\Delta = abc+ 2fgh-af^{2}-bg^{2}-ch^{2}=0 $

where,

$a =2 $ 
$b=-2 $
$c=\lambda $
$f=\dfrac{5}{2}$
$g=-\dfrac{5}{2}$
$h=\dfrac{3}{2}$


Thus, $\lambda=-3 $

$A= 2x^{2}+3xy-2y^{2}-5x+5y-3 =0$

Since  $ H+C=2A $  Where $ C $ be the equation of conjugate hyperbola

$C= 2A-H $
So,the equation of conjugate hyperbola be  $ C=2x^{2}+3xy-2y^{2}-5x+5y-8 =0 $

Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

The angle between the asymptotes of the hyperbola ${27x}^{2}-{9y}^{2}=24$ is 

  1. ${30}^{o}$
  2. ${120}^{o}$
  3. ${60}^{o}$
  4. ${90}^{o}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$27 x ^ { 2 } - 9 y ^ { 2 } = 24$
$\Rightarrow \quad \dfrac { x ^ { 2 } } { \left( \dfrac { 24 } { 27 } \right) } - \dfrac { y ^ { 2 } } { \left( \dfrac { 24 } { 9 } \right) } = 1$
$\Rightarrow \dfrac { x ^ { 2 } } { ( \dfrac 89 ) } - \dfrac { y ^ { 2 } } { \dfrac 8 3 } = 1$
$\Rightarrow a ^ { 2 } = \dfrac 8  9 \quad , \quad b ^ { 2 } = \dfrac 8 3$
$\Rightarrow \quad \dfrac { b ^ { 2 } } { a ^ { 2 } } = \dfrac { 8 } { 3 } \times \dfrac { 9 } { 8 } = 3$
$\Rightarrow \quad \dfrac { b } { a } = \sqrt { 3 }$

The angle between the asymptotes 
$\begin{aligned} & = 2 \tan ^ { - 1 } \left( \frac { b } { a } \right) \ = & 2 \tan ^ { - 1 } \left( \sqrt 3 \right) \ = & 2 \cdot 60 \ = & 120 \end{aligned}$

Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

The asymptotes of the hyperbola $xy - 3x + 4y + 2 = 0$ are

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

Given : Hyperbola,
$xy-3x+4y+2=0$---------------1
For Asymptotes,
Let the Asymptote's Equation be $y=mx+c$
And then finding $\phi _{n}(m)$ by replacing $y\rightarrow m$ and $x\rightarrow 1$
As $n=2$,
$\phi _{2}(m)=m$
putting $\phi _{2}(m)=0$, we get $m=0$
By taking $m=0$, we will get only one asymptote parallel to X-axis, so let's find them with putting the co-efficients of higher terms to zero.
For Asymptote parallel to X-axis, we put co-efficient of highest degree of x to zero that is here 1, so co-efficient of x$=0$
$\Rightarrow (y-3)=0$------------2(from Equation 1)
For Asymptote parallel to Y-axis, we put co-efficient of highest degree of y to zero which is 1 here, co-efficient of y$=0    (from Equation 1)
$\Rightarrow x+4=0$------------3
The Equation 2 & 3 are asymptotes to Equation 1.
$x+4=0$ & $y-3=0$

Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

Through any P of the hyperbola $\frac{x^2}{a^2}- \frac{y^2}{b^2} =1 $ a line $PQR$ is drawn with a fixed gradient $m$, meeting the asymptotes in $Q\ &\ R$. Then the product,$ (QP) (PR) =\frac{a^2b^2(1+m^2)}{b^2- a^2m^2}$.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The product of the segments cut by the asymptotes on a line with gradient m is given by the formula (a^2b^2(1+m^2)) / (b^2 - a^2m^2). This is a standard result for hyperbolas.

Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

The asymptotes of the hyperbola $6{x^2} + 13xy + 6{y^2} - 7x - 8y - 26 = 0$ are 

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

The asymptotes of a hyperbola S=0 are given by S - k = 0, where k is a constant such that the equation represents a pair of lines. For 6x^2 + 13xy + 6y^2 - 7x - 8y - 26 = 0, the homogeneous part factors as (2x+3y)(3x+2y). The asymptotes are of the form (2x+3y+c1)(3x+2y+c2)=0. Expanding and matching coefficients with the original equation (with a constant adjustment) yields the lines 2x+3y=1 and 3x+2y=2.

Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

If $S=0$ be the equation of the hyperbola $x^2+4xy+3y^2-4x+2y+1=0$, then the value of $k$ for which $S+k=0$ represents its asymptotes is :

  1. $20$
  2. $-16$
  3. $-22$
  4. $18$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$S+k=x^2+4xy+3y^2-4x+2y+1+k =0$
For equation $S+k=0$ to represent a pair of lines,
$\triangle =0$
$\begin{vmatrix} 1& 2 & -2\ 2 & 3 & 1\ -2 & 1 & 1+k\end{vmatrix}=0$
$\Rightarrow 3(1+k)-1-2(2+2k+2)-2(2+6)=0$
$\Rightarrow k=-22$

Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

One of the asymptotes (with negative slope) of a hyperbola passes through (2, 0) whose transverse axis is given by x - 3y + 2 = 0 then equation of hyperbola if it is given that the line y = 7x - 11 can intersect the hyperbola at only one point (2, 3) is given by

  1. $\displaystyle 7x^{2}+xy-y^{2}+10x-4y-3=0$
  2. $\displaystyle 7x^{2}-xy-y^{2}-10x-5y+2=0$
  3. $\displaystyle 7x^{2}+xy-y^{2}-19x-5y+28=0$
  4. $\displaystyle 7x^{2}+6xy-y^{2}-20x-4y-3=0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

As $y=7x-11$ intersects the hyperbola at only one point 


$ \displaystyle \Rightarrow $ it is parallel to one of the asymptotes

$ \displaystyle \Rightarrow $ Equation of one asymoptote can be taken as $7x-y+k=0$ clearly mirror image of $(2,0)$ about transverse axis $x-3y=2 $lies on other asymplote 

$ \displaystyle \Rightarrow \left ( \frac{6}{5},\frac{12}{5} \right )$ lies on $7x-y+k=0$

$ \displaystyle \Rightarrow k=-6$

$ \displaystyle \Rightarrow $other asymptote is $7x-y-6=0$

$ \displaystyle \Rightarrow $ centre is $(1,1)$

$ \displaystyle \Rightarrow $Asymptote through $(2,0)$ is $x+y=2$

Equation of hyperbola is $(7x-y-6)(x+y-2)-(7* 2-3-6)(2+3-2)=0$

$ \displaystyle \Rightarrow 7x^{2}+6xy-y^{2}-20x-4y-3=0 $

Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

The second-degree curve and pair of asymptotes differ by a constant. Let the second-degree curve $S = 0$ represent the hyperbola then respective pair of asymptote is given by.$\displaystyle S+\lambda =0\left ( \lambda \in R \right )$ which represent a pair of straight lines so $\lambda$  can be determined. The equation of asymptotes is $\displaystyle A=s+\lambda =0$ if equation of conjugate hyperbola of the curve $S =0$ be represents by $S _{1}$, then $A$ is arithmetic mean of the curves $S _{1}$, & $ S $.

A hyperbola passing through origin has $\displaystyle 2x-y+3=0$ and $\displaystyle x-2y+2=0$ as its asymptotes, then equation of its transverse and conjugate axes are:

  1. $\displaystyle x-y+2=0$ and $\displaystyle 3x-3y+5=0$
  2. $\displaystyle x+y+2=0$ and $\displaystyle 3x-3y+5=0$
  3. $\displaystyle x-y+1=0$ and $\displaystyle 3x-3y+5=0$
  4. $\displaystyle2 x-2y+1=0$ and $\displaystyle 3x-3y+5=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The transverse axis of hyperbola is the bisector of the angle between the asymptotes containing the origin and the conjugate axis is the other bisector.


 And equation of bisector of angle of the asymptotes are given by

$\displaystyle \frac{2x-y+3}{\sqrt{5}}=\pm \frac{x-2y+2}{\sqrt{5}}$

$\displaystyle \Rightarrow  2x-y+3 =\pm \left ( x-2y+2 \right )$

$\displaystyle \Rightarrow  2x-y+3 =x-2y+2$

and $\displaystyle  2x-y+3 =x-2y+2= -\left ( x-2y+2 \right )$

$\displaystyle  \Rightarrow x+y+1=0 \ and \ 3x-3y+5=0$

Hence, option 'C' is correct.