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

The centre of the hyperbola $\dfrac {x^{2} + 4x + 4}{25} - \dfrac {y^{2} - 6x + 9}{16} = 1$ is: 

  1. $(-4, -9)$
  2. $(-2, 3)$
  3. $(2, -3)$
  4. $(5, 4)$
  5. $(25, 16)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
  • the equation of hyperbola is $\dfrac { { x }^{ 2 }+4x+4 }{ 25 } -\dfrac { { y }^{ 2 }-6x+9 }{ 16 } =1$
  • $\dfrac { { (x+2) }^{ 2 } }{ 25 } -\dfrac { { (y-3) }^{ 2 } }{ 16 } =1$
  • Therefore the center is $(-2,3)$
Multiple choice mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

For hyperbola  $-\dfrac{(x-1)^2}{3}+\dfrac{(y+2)^2}{16}=1$ vertices are

  1. $(\pm\sqrt3,0)$
  2. $(\pm\sqrt3+1,-2)$
  3. $(\pm1,-2)$
  4. $(0,0)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given, hyperbola is conjugate hyperbola of $\dfrac { { (x-1) }^{ 2 } }{ 3 } -\dfrac { { (y+2) }^{ 2 } }{ 16 } =-1$

So the vertices of given hyperbola are
${ (x-1) }^{ 2 }=3,{ (y+2) }^{ 2 }=0\ \Rightarrow x-1=\pm \sqrt { 3 } ,y+2=0\ \Rightarrow x=1\pm \sqrt { 3 } ,y=-2\ \Rightarrow \left( 1\pm \sqrt { 3 } ,-2 \right) $
So, option B is correct.

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

Find the equation to the hyperbola, referred to its axes as axes of coordinates, whose transverse axis is $7$ and which passes through the point $\left( 3,-2 \right) $.

  1. $65y^2-16x^2=196$
  2. $65y^2-14x^2=196$
  3. $85y^2-16x^2=196$
  4. $85y^2-16x^2=147$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

General equation of hyperbola is $\dfrac{y^2}{b^2}-\dfrac{x^2}{a^2}=1$

Length of transverse axis is $2a$.
So, $2a=7$
$\Rightarrow a=\dfrac{7}{2}$
Equation becomes,
$\dfrac{y^2}{b^2}-\dfrac{4x^2}{49}=1$
It passes through $(3,-2)$, so it should satisfy the parabola,
$\dfrac{4}{b^2}-\dfrac{36}{49}=1$
On solving, we get 
$85y^2-16x^2=196$

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

Equation of the hyperbola with vertices at $(\pm 5, 0)$ and foci at $(\pm 7, 0)$ is

  1. $24x^2-25y^2=600$
  2. $25x^2-24y^2=600$
  3. $\displaystyle \frac{x^2}{25}-\frac{y^2}{24}=1$
  4. $\displaystyle \frac{x^2}{24}-\frac{y^2}{25}=1$
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

$Vertices(\pm 5,0)\quad Foci(\pm 7,0)$ 

$a=\pm 5$ and $ae=\pm 7$ 
And $e=\dfrac { 7 }{ 5 } $ 
We know $e=\sqrt { 1+\frac { { b }^{ 2 } }{ { a }^{ 2 } }  }$ 
On squaring both sides we get:
${ e }^{ 2 }=\quad 1+\frac { { b }^{ 2 } }{ { a }^{ 2 } }$ 
Or $\dfrac { 49 }{ 25 } =1+\frac { { b }^{ 2 } }{ 25 }$ 
Or $\dfrac { { b }^{ 2 } }{ 25 } =\frac { 24 }{ 25 } $
${ b }^{ 2 }=24$ 
The equation of hyperbola is 
$\dfrac { { x }^{ 2 } }{ 25 } -\dfrac { y^{ 2 } }{ 24 } =1$ 
$24{ x }^{ 2 }-25y^{ 2 }=600$

Hence, Option [A] and [C] are correct.

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

The equation of a hyperbola is given in its standard form as $16x^2-9y^2=144$.Equations of directrices is

  1. $5x \pm 16=0$
  2. $5y \pm 16=0$
  3. $5x \pm 12=0$
  4. $5y \pm 12=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$Given\quad :\quad 16{ x }^{ 2 }−9{ y }^{ 2 }=144\quad \quad \quad \ Or,\quad \frac { { x }^{ 2 } }{ 9 } -\frac { { y }^{ 2 } }{ 16 } =1\ We\quad know,\ be=\sqrt { { a }^{ 2 }+{ b }^{ 2 } } \quad \quad \quad \quad \quad \because (b>a)\ Or,\quad be=\sqrt { 9+16 } \ Or,\quad be=\pm 5\ Or,\quad \frac { b }{ e } \quad =\frac { 16 }{ \pm 5 } \ We\quad know\quad equation\quad of\quad directrix\quad is\quad y=\frac { b }{ e } \ \therefore \quad 5y\pm 16=0$


Option [B]

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

Equation of the transverse and conjugate axis of a hyperbola are respectively $x+2y-3=0$, $2x-y+4=0$ and their respectively lengths are $\sqrt {2}$ and $\cfrac { 2 }{ \sqrt { 3 }  } $ then answer the following 
Equation of one of the directrix is

  1. $2x-y+4+\sqrt { \cfrac { 3 }{ 2 } } =0\quad $
  2. $x+2y+4-\sqrt { \cfrac { 2 }{ 3 } } =0$
  3. $2x-y=\sqrt { \cfrac { 3 }{ 2 } } $
  4. $2x-y+4+\sqrt { \cfrac { 3 }{ 2 } } =\sqrt { 3 } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The directrix of a hyperbola is given by the line parallel to the conjugate axis at a distance a/e from the center. Given the axes and lengths, one can determine the specific equation for the directrix.

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

The equation of a hyperbola is given in its standard form as $16x^2-9y^2=144$.Coordinates of foci is

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

$Given\quad :\quad 16{ x }^{ 2 }−9{ y }^{ 2 }=144\quad \quad \quad \ Or,\quad \frac { { x }^{ 2 } }{ 9 } -\frac { { y }^{ 2 } }{ 16 } =1\ We\quad know,\ be=\sqrt { { a }^{ 2 }+{ b }^{ 2 } } \quad \quad \quad \quad \quad \because (b>a)\ Or,\quad be=\sqrt { 9+16 } \ Or,\quad be=\pm 5\ \therefore \quad Focii\quad is\quad (0,\pm 5)\quad $


Option [D]

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

The foci of hyperbola $9x^2-16y^2+18x+32y=151$ are 

  1. $(-4,1),(6,1)$
  2. $(-11,2),(-6,1)$
  3. $(4,1),(-6,1)$
  4. $(2,1),(1,-6)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$9{x}^{2}−16{y}^{2}+18x+32y=151$
$\left(9{x}^{2}+18x\right)-\left(16{y}^{2}-32y\right)=151$
$\Rightarrow \left({\left(3x\right)}^{2}+2\times 3x\times 3+{3}^{2}-{3}^{2}\right)-\left({\left(4y\right)}^{2}-2\times 4y\times 4+{4}^{2}-{4}^{2}\right)=151$ by completing the square method
$\Rightarrow {\left(3x+3\right)}^{2}-9-{\left(4y-4\right)}^{2}+16=151$
$\Rightarrow {\left(3x+3\right)}^{2}-{\left(4y-4\right)}^{2}=151-7$
$\Rightarrow {\left(3x+3\right)}^{2}-{\left(4y-4\right)}^{2}=144$
$\Rightarrow 9{\left(x+1\right)}^{2}-16{\left(y-1\right)}^{2}=144$
$\Rightarrow \dfrac{9{\left(x+1\right)}^{2}}{144}-\dfrac{16{\left(y-1\right)}^{2}}{144}=1$ by dividing both sides by $144$
$\Rightarrow \dfrac{{\left(x+1\right)}^{2}}{16}-\dfrac{{\left(y-1\right)}^{2}}{9}=1$ is the equation of the horizontal hyperbola.
center$=\left(-1,1\right)$
We have $a=4$ and $b=3$
${c}^{2}={a}^{2}+{b}^{2}={4}^{2}+{3}^{2}=16+9=25$
$\therefore c=\sqrt{25}=\pm 5$
Foci$=\left(-1\pm 5, 1\right)$
$\therefore$Foci$=\left(-1+5,1\right)$ and $\left(-1-5,1\right)$
Hence Foci$=\left(4,1\right)$ and $\left(-6,1\right)$
Multiple choice mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

The vertices and the foci of a hyperbola are the points $\displaystyle \left ( \pm 5, 0 \right )$ and $\displaystyle \left ( \pm 7, 0 \right )$.Which of the following holds true?

  1. $\displaystyle a^{2}\neq b^{2}$
  2. $a^2=b^2$
  3. $\dfrac{a^2}{b^2}=2$

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

$\displaystyle a= 5, ae = 7\Rightarrow a^{2}e^{2}= 49$
or $\displaystyle a^{2}\left ( 1+\frac{b^{2}}{a^{2}} \right )= 49$ 

Or 
$\displaystyle a^{2}+b^{2}=49$ 
$\displaystyle b^{2}= 24$ 
Since, $\displaystyle a^{2}\neq b^{2},$ hence hyperbola is not rectangular.

Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

If focus of the parabola is $(3,0)$ and length of latus rectum is $8$, then its vertex is

  1. $(2,0)$
  2. $(1,0)$
  3. $(0,0)$
  4. $(-1,0)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given, focus $=(3,0)$ and Length of latus rectum $= 8$

$\Rightarrow 4a=8$ $\Rightarrow a=2$

$\Rightarrow$ Vertex = $(3-a,0)$ $=(1,0)$

$\therefore $ Option B is correct
Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

If $(0,0)$ be the vertex and $3x-4y+2=0$ be the directrix of a parabola, then the length of its latus rectum is

  1. $4/5$
  2. $2/5$
  3. $8/5$
  4. $1/5$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Distance of vertex from directrix = $\dfrac{\left | 3(0)-4(0)+2 \right |}{\sqrt{3^{2}+4^{2}}}= \dfrac{2}{5}=a$

Length of latus rectum = $4a= \dfrac{8}{5}$

$\therefore $ Option C is correct
Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

The length of the latusrectum of the parabola $169\left{ { \left( x-1 \right)  }^{ 2 }+{ \left( y-3 \right)  }^{ 2 } \right} ={ \left( 5x-12y+17 \right)  }^{ 2 }$

  1. $\cfrac { 14 }{ 13 } $
  2. $\cfrac { 28 }{ 13 } $
  3. $\cfrac { 12 }{ 13 } $

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

Here ${ \left( x-1 \right)  }^{ 2 }+{ \left( y-3 \right)  }^{ 2 }={ \left{ \cfrac { 5x-12y+17 }{ \sqrt { { 5 }^{ 2 }+{ \left( -12 \right)  }^{ 2 } }  }  \right}  }^{ 2 }$


$\therefore$ The focus is $(1,3)$ and the directrix is $5x-12y+17=0$

The distance of the focus from the directrix

$=\left| \cfrac { 5\times 1-12\times 3+17 }{ \sqrt { { 5 }^{ 2 }+{ \left( -12 \right)  }^{ 2 } }  }  \right| =\cfrac { 14 }{ 13 } $

$\therefore$ Length of latusrectum $=2\times \cfrac { 14 }{ 13 } =\cfrac { 28 }{ 13 } $

Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

The equation to the locus of the point of intersection of any two perpendicular tangents to $x^{2}+ y^{2} = 4$ is

  1. $\mathrm{x}^{2}+\mathrm{y}^{2}=8$
  2. $\mathrm{x}^{2}+\mathrm{y}^{2}=12$
  3. $\mathrm{x}^{2}+\mathrm{y}^{2}=16$
  4. $\mathrm{x}^{2}+\mathrm{y}^{2}=4\sqrt{3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equation of the tangent to the circle $x^2+y^2=4$ is

$y=mx+2\sqrt{1+m^2}$
$P(h,k)$ lies on the tangent, then
$k-mh=2\sqrt{1+m^2}$
or, $(k-mh)^2=4(1+m^2)$
or, $m^2(h^2-4)-2mhk+k^2-4=0$
This is the quadratic equation in $m.$ Let $m _1$ and $m _2$ be roots
$m _1m _2=\cfrac{k^2-4}{h^2-4}=-1$
or, $k^2-4=-h^2+4$
or, $h^2+k^2=8$
Therefore, Equation to the locus of the intersection of any two perpendicular tangents is
$x^2+y^2=8$
Hence, A is the correct option.