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 a hyperbola have equations $y-1=\dfrac{3}{4}(x+3).$ If a focus of the hyperbola has coordinates $(7,1)$, the equation of the hyperbola is

  1. $\dfrac{(x+3)^2}{16}-\dfrac{(y-1)^2}{9} = 1$
  2. $\dfrac{(y-1)^2}{9}-\dfrac{(x+3)^2}{16} = 1$
  3. $\dfrac{(x+3)^2}{64}-\dfrac{(y-1)^2}{36} = 1$
  4. $\dfrac{(y-1)^2}{36}-\dfrac{(x+3)^2}{64} = 1$
  5. $\dfrac{(x+3)^2}{4}-\dfrac{(y-1)^2}{3} = 1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Equation of asymptotes are 

$y-1=\dfrac { 3 }{ 4 } (x+3  )$    ......(i)

$ y-1=-\dfrac { 3 }{ 4 } (x+3)$     .....(ii)

Centre of the hyperbola is point of intersection of asymptotes.

Therefore, by solving (i) and (ii), we get centre as $C(-3,1)$.

Slope of asymptotes $=\dfrac { b }{ a } $

$\Rightarrow \dfrac { b }{ a } =\pm \dfrac { 3 }{ 4 }$      ......(i)

Focus is $(7,1)$.

Focus for hyperbola of form $\dfrac { { (x-h) }^{ 2 } }{ { a }^{ 2 } } -\dfrac { { (y-k) }^{ 2 } }{ { b }^{ 2 } } =1$ is $(h+ae,k)$

$\Rightarrow 7=-3+ae\\ \Rightarrow ae=10\\ \Rightarrow a\dfrac { \sqrt { { a }^{ 2 }+{ b }^{ 2 } }  }{ a } =10\\ \Rightarrow \sqrt { { a }^{ 2 }+{ b }^{ 2 } } =10$

Substituting $b$ from (i), we get

$\Rightarrow \sqrt { { a }^{ 2 }+{ \left( \pm a \dfrac { 3 }{ 4 }  \right)  }^{ 2 } } =10\\ \Rightarrow \dfrac { 5a }{ 4 } =10\\ \Rightarrow a=8\\ \Rightarrow b=\pm \dfrac { 3 }{ 4 } a=\pm 6$

So, the equation of hyperbola is

$\dfrac { { (x+3) }^{ 2 } }{ { 8 }^{ 2 } } -\dfrac { { (y-1) }^{ 2 } }{ { 6 }^{ 2 } } =1$

$\dfrac { { (x+3) }^{ 2 } }{ 64 } -\dfrac { { (y-1) }^{ 2 } }{ 36 } =1$

So, option C is correct.

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

If $PN$ is the perpendicular from a point on a rectangular hyperbola to its asymptotes, the locus, then the midpoint of $PN$ is

  1. circle

  2. parabola

  3. ellipse

  4. hyperbola
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let $xy={ c }^{ 2 }$ be the rectangular hyperbola and let $P\left( { x
} _{ 1 },{ y } _{ 1 } \right) $ be apoint on it. Let $Q(h.k)$ be the
midpoint of $PN$. Then the coordinates of $Q$ are $\left( { { x } _{ 1
},{ y } _{ 1 } }/{ 2 } \right) $
$\therefore \quad { x } _{ 1 }={ h
}\quad \cfrac { { y } _{ 1 } }{ 2 } =k\Rightarrow { x } _{ 1 }={ h }\quad
,\quad { y } _{ 1 }=2k$
But $\left( { x } _{ 1 },{ y } _{ 1 } \right) $ lies on $xy={ c }^{ 2 }$
$\therefore \quad h(2k)={ c }^{ 2 }\Rightarrow hk=\cfrac { { c }^{ 2 } }{ 2 } $
Therefore, the locus of $(h,k)$ is $xy=\cfrac { { c }^{ 2 } }{ 2 } $, which is a hyperbola.
Hence, option 'D' is correct.

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. $x= - 4$
  2. $x= 4$
  3. $y= - 3$
  4. $y= 3$
Reveal answer Fill a bubble to check yourself
B,D 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 physics parametric equations proving properties of curves derivatives - introduction and interpretation introduction to calculus - differentiation

If $y=mx+c$ is the normal at a point $(8,8)$  on the parabola ${ y }^{ 2 }=8x$ Find $m$

  1. $-2 $
  2. $8 $
  3. $10 $
  4. $16 $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given equation $y^2=8x $

Slope of tangent is given as $2y\dfrac{dy}{dx}=8\\dfrac{dy}{dx}=\dfrac{4}{y}\\left.\dfrac{dy}{dx}\right| _{(8,8)}=\dfrac{4}{8}=\dfrac 1{2}$
Slope of normal is $\dfrac{-1}{\dfrac{1}{2}}=-2$

Multiple choice physics parametric equations proving properties of curves derivatives - introduction and interpretation introduction to calculus - differentiation

If line $PQ$, whose equation is $y = 2x + k,$  is a normal to the parabola whose vertex is   $(-2,3)$ and the axis parallel to the $x$-axis with latus rectum equal to $2$, then the possible value of k is

  1. $\dfrac{{58}}{8}$
  2. $\dfrac{{50}}{8}$
  3. $1$
  4. $-1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Normal to parabola $\rightarrow y=2x+k$
parabola $\rightarrow (y-3)^{2}=4a(x+2)$
Latus rectum$=2$
$4a=2$
$a=\dfrac{1}{2}$
parabola $\rightarrow (y-3)^{2}=2(x+2)$
$y=3+\sqrt{2}\sqrt{x+2}$
slope of normal =$\dfrac{1}{-y'}$
$y'=\dfrac{\sqrt{2}}{2\sqrt{x+2}}=\dfrac{1}{\sqrt{2x+4}}$
$m=-\sqrt{-2x+4}$
if $2=-\sqrt{2x+4}$
then $x=0, y=5, 1$
$(0,5$) should also lies on $y=2x+k$
then
$k=5,1$
$C$ is correct
Multiple choice physics parametric equations proving properties of curves derivatives - introduction and interpretation introduction to calculus - differentiation

If the distance between a tangent to the parabola $y^{2} = 4x$ and a parallel normal to the same parabola is $2\sqrt{2}$, then possible values of gradient of either of them are:

  1. $-1$
  2. $+1$
  3. $-\sqrt{\sqrt{5} - 2}$
  4. $+\sqrt{\sqrt{5} - 2}$
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation

$y=\frac { x }{ t } +at\ y=-xs+2as+a{ s }^{ 3 }\ s=-\frac { 1 }{ t } $

For parallel condition.
$ts=-1$

$\left| \dfrac { \left( 2as+a{ s }^{ 3 } \right) -\left( at \right)  }{ \sqrt { 1+{ s }^{ 2 } }  }  \right| =2\sqrt { 2 } \ \ \left| \dfrac { \left( 2as+a{ s }^{ 3 } \right) +\frac { a }{ s }  }{ \sqrt { 1+{ s }^{ 2 } }  }  \right| =\left| \dfrac { a\left( { s }^{ 4 }+2{ s }^{ 2 }+1 \right)  }{ s\sqrt { 1+{ s }^{ 2 } }  }  \right| =\left| \dfrac { a{ \left( { s }^{ 2 }+1 \right)  }^{ 2 } }{ s\sqrt { 1+{ s }^{ 2 } }  }  \right| =\left| \dfrac { a{ \left( { s }^{ 2 }+1 \right)  }^{ \frac { 3 }{ 2 }  } }{ s }  \right| $
As $a=1$
$\left| \dfrac { { \left( { s }^{ 2 }+1 \right)  }^{ \frac { 3 }{ 2 }  } }{ s }  \right| =2\sqrt { 2 } ={ 2 }^{ \frac { 3 }{ 2 }  }$
Putting $s$ as $\pm 1$ the above equation satisfies.
Hence, the answer is $1,-1$.

Multiple choice maths 5-digit numbers expanded form introduction to numbers and number systems numbers in general form

The locus of point of trisections of the focal chords of the parabola, ${y^2} = 4x$ :

  1. ${y^2} = x - 1$
  2. $9{y^2} = 4\left( {3x - 4} \right)$
  3. ${y^2} = 2\left( {1 - x} \right)$

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

The focal chord of y^2 = 4x passes through (1, 0). If the endpoints are (t1^2, 2t1) and (t2^2, 2t2), the condition for a focal chord is t1*t2 = -1. The point of trisection divides the chord in ratio 1:2 or 2:1. Calculating the locus of these points yields 9y^2 = 4(3x - 4).

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

If the equation of a hyperbola is $\frac{{{x^2}}}{9} - \frac{{{y^2}}}{{16}} = 1$, then 

  1. traverse axis is along x-axis of length $6$
  2. traverse axis is along y-axis of length $8$
  3. conjugate axis is along y-axis of length $6$

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

For the hyperbola x^2/9 - y^2/16 = 1, a^2 = 9 and b^2 = 16. The transverse axis is along the x-axis with length 2a = 2 * 3 = 6.

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

For hyperbola  $\dfrac{x^2}{16}-\dfrac{y^2}{25}=1$
vertices are 

  1. $(4,4)$
  2. $(\pm4,0)$
  3. $(\pm4,4)$
  4. $(0,\pm4)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For the hyperbola of the form $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$

Vertices are $(\pm a,0)$
Given equation is $\dfrac {x^2}{16}-\dfrac {y^2}{25}=1$
So, here $a^2=16 \Rightarrow  a=4$
So, the vertices are $(\pm 4,0)$.

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

For hyperbola  $-\dfrac{x^2}{16}+\dfrac{y^2}{25}=1$ equation of directrices are

  1. $y=\pm\dfrac{16}{\sqrt{41}}$
  2. $y=\pm\dfrac{5}{\sqrt{41}}$
  3. $y=\pm\dfrac{2}{\sqrt{41}}$
  4. $y=\pm\dfrac{25}{\sqrt{41}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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


Directrix is at $y=\pm \dfrac{b}{e}$ where $e=\sqrt{\dfrac{b^2}{a^2}+1}$

Here $a^2=25,b^2=16 \implies e=\sqrt{\dfrac{16}{25}+1}=\sqrt{\dfrac{41}{25}}$ 

So equation of directrix is $y=\pm\dfrac{5\sqrt{25}}{\sqrt{41}}=\pm\dfrac{25}{\sqrt{41}}$

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

For hyperbola  $\dfrac{-x^2}{9}+\dfrac{y^2}{16}=1$, centre is 

  1. $(3,3)$
  2. $(5,5)$
  3. $(0,0)$
  4. $(4,5)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The given hyperbola $\dfrac{-x^2}{9} + \dfrac{y^2}{16} = 1$  or $\dfrac{x^2}{9} - \dfrac{y^2}{16} = - 1$ is a standard form of a conjugate hyperbola. 


By comparing it with it's standard form $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = -1$

We can know $a = 3$ and $b = 4$

For a standard form of a conjugate hyperbola, the center lies at origin at $(0,0)$. Hence the correct option is $C$.

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

For hyperbola  $\dfrac{x^2}{16}-\dfrac{y^2}{25}=1$, focus is is on 

  1. x-axis

  2. y-axs

  3. z-axis

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

The given hyperbola $\dfrac{x^2}{16} - \dfrac{y^2}{25} = 1$ is a standard form of hyperbola. 


By comparing it with standard form $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$

We can know $a = 4$ and $b = 5$

For a standard form of hyperbola, the foci lie on the transverse axis. $i.e.$ $x$ - axis. one each side of the hyperbola, at $(ae,0)$ and $(-ae,0)$ respectively.