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

If any point on a hyperbola has the coordinates $\displaystyle \left ( 5 \tan \phi , : 4 \sec \phi \right )$ then the ecentricity of the hyperbola is

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

We have   $x=5\tan\phi, y=4\sec\phi$
Eliminating $\phi$ we get
$\cfrac{y^2}{16}-\cfrac{x^2}{25}=\sec^2\phi-\tan^2\phi=1$
This is a hyperbola. Therefore, its eccentricity $= \sqrt{1+\cfrac{25}{16}}=\cfrac{\sqrt{41}}{4}$

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

Consider the hyoerbola ${ 3x^{2} }-{ y }^{ 2 }-{ 24x } + { 4y } { 4 } = 0$

  1. $its centre is \left(4,2\right)$
  2. $its centre is \left(2,4\right)$
  3. $length of latus rectum = 24$
  4. $length of latus rectum = 12$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Completing the square for 3x^2 - y^2 - 24x + 4y + 4 = 0: 3(x^2 - 8x + 16) - (y^2 - 4y + 4) = -4 + 48 - 4, resulting in 3(x-4)^2 - (y-2)^2 = 40. The center is (4, 2).

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

$y=mx+c$ is tangent to hyperbola find $c$ if hyperbola eqn is

  1. $\dfrac{{x}^{2}}{{a}^{2}}-\dfrac{{y}^{2}}{{b}^{2}}=1$$a>b$
  2. $\dfrac{{x}^{2}}{{a}^{2}}-\dfrac{{y}^{2}}{{b}^{2}}=1$$b>a$
  3. $\dfrac{{-x}^{2}}{{a}^{2}}-\dfrac{{y}^{2}}{{b}^{2}}=1$$a>b$
  4. $\dfrac{{-x}^{2}}{{a}^{2}}-\dfrac{{y}^{2}}{{b}^{2}}=1$$b>a$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For line y=mx+c to be tangent to hyperbola x^2/a^2 - y^2/b^2 = 1, we need c^2 = a^2m^2 - b^2. For real c, we need a^2m^2 > b^2. When a > b, we can find m values satisfying this condition. The question is poorly phrased but option A correctly identifies the standard hyperbola condition.

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 whose directrix is $2x+y=1$ and focus is at $(1,2)$ with $e=\sqrt{3}$ is :

  1. $7x^2+12xy+2y^2-2x+14y-22=0$
  2. $7x^2+12xy-2y^2-2x+14y-22=0$
  3. $7x^2+12xy-2y^2-2x-14y-22=0$
  4. $7x^2+12xy+2y^2+2x+14y-22=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The definition of a hyperbola is PF = e * PM, where PM is the perpendicular distance to the directrix. Squaring both sides: (x-1)^2 + (y-2)^2 = 3 * ((2x+y-1)/sqrt(2^2+1^2))^2. Expanding this leads to 5(x^2 - 2x + 1 + y^2 - 4y + 4) = 3(4x^2 + y^2 + 1 + 4xy - 4x - 2y). Simplifying results in 7x^2 + 12xy + 2y^2 - 2x + 14y - 22 = 0.

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

The vertices of a hyperbola are at $(0, 0)$ and $(10,0)$ and one of its focus is at $(18,0)$. The possible equation of the hyperbola is

  1. $\displaystyle \frac{x^2}{25}\, -\, \frac{y^2}{144}\, =\, 1$
  2. $\displaystyle \frac{(x\, -\, 5)^2}{25}\, -\, \frac{y^2}{144}\, =\, 1$
  3. $\displaystyle \frac{x^2}{25}\, -\, \frac{(y\, -\, 5)^2}{144}\, =\, 1$
  4. $\displaystyle \frac{(x\, -\, 5)^2}{25}\, -\, \frac{(y\, -\, 5)^2}{144}\, =\, 1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Centre of hyperbola is $(5, 0)$, so equation is
$\displaystyle \frac{(x\, -\, 5)^2}{a^2}\, -\, \frac{y^2}{b^2}\, =\, 1$
$a\, =\, 5,\, ae\, -\, a\, =\, 8\, \Rightarrow\, e\, =\, \displaystyle \frac{13}{5}$

$b^2\, =a^2(e^2-1)=\, 144$
So required equation is,  $\displaystyle \frac{(x\, -\, 5)^2}{25}\, -\, \frac{y^2}{144}\, =\, 1$
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 centre, vertex and focus of a hyperbola be $(0,0), (4, 0)$ and $(6,0)$ respectively, then the equation of the hyperbola is

  1. $4x^2\, -\, 5y^2\, =\, 8$
  2. $4x^2\, -\, 5y^2\, =\, 80$
  3. $5x^2\, -\, 4y^2\, =\, 80$
  4. $5x^2\, -\, 4y^2\, =\, 8$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given $a=4, ae=6\Rightarrow e=\cfrac{3}{2}\Rightarrow \cfrac{b^2}{a^2}=e^2-1=\cfrac{9}{4}-1=\cfrac{5}{4}\Rightarrow b^2=20$
Therefore, required hyperbola is, $\cfrac{x^2}{16}-\cfrac{y^2}{20}=1$
$\Rightarrow 5x^2-4y^2=80$
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

The hyperbola $\dfrac{x^2}{a^2}\, -\, \dfrac{y^2}{b^2}\, =\, 1\, (a,\, b\, >\, 0)$ passes through the point of intersection of the lines $7x + 13y - 87 = 0$ & $5x - 8y + 7 = 0$ and the latus rectum is $\dfrac{32 \sqrt{2}}5$. The values of $a$ and $b$ are:

  1. $\displaystyle a =\frac{5}{\sqrt{2}},\, b=3$.
  2. $\displaystyle a =\frac{5}{\sqrt{2}},\, b=4$.
  3. $\displaystyle a =\frac{7}{\sqrt{2}},\, b=3$.

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

Point of intersection of lines
$7x + 13y - 87 = 0$ & $5x - 8y + 7 = 0$ is $(5, 4)$.
Also this point lies on the given hyperbola
$\therefore \displaystyle \frac{25}{a^2}\, -\, \frac{16}{b^2}\, =\, 1$ ......(1)
Also latus rectum $\displaystyle LR\, =\, \frac{2b^2}{a}\, =\, \frac{32 \sqrt{2}}{5}$
$\displaystyle \Rightarrow\, b^2\, =\, \frac{16 \sqrt{2}a}{5}$ .....(ii)
From (i) & (ii) $\displaystyle a^2\, =\, \frac{25}{2},\, b^2\, =\, 16$.

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

For the hyperbola $16x^2\, -\, 9y^2\, +\, 32x\, +\, 36y\,-\, 164\, =\, 0$, find $2(a+b)$.

  1. $8$
  2. $6$
  3. $14$
  4. $12$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given hyperbola can be written as


$16x^2+32x+16-9y^2+36y-36=144$

$\displaystyle\, \frac{(x\, +\, 1)^2}{9}\, -\, \frac{(y\, -\, 2)^2}{16}\, =\, 1$

$\Rightarrow a =3, b = 4$

Length of major axis $=2\times 4=8$ and length of minor axis $=2\times 3 = 6$. 

Hence sum is $2(a+b)=2(3+4)=14.$

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 $\sqrt{2}$ is confocal with $3x^2 + 4y^2 = 12$, then its equation is:

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

Given ellipse may be written as, $\displaystyle \frac {x^2}{4}+\frac {y^2}{3}=1$
$\displaystyle \Rightarrow a^2 = 4, b^2 = 3\therefore e = \sqrt{1-\frac{3}{4}}=\frac{1}{2}$
Thus foci of ellipse $(\pm 1, 0)$
Hence foci of the hyperbola is $(\pm 1, 0)$
And semi-major axis $a =\cfrac{1}{\sqrt{2}}$
$\Rightarrow \pm 1=\sqrt {a^2+b^2}\Rightarrow b^2=\frac {1}{2}$
or $b^2=1-a^2=1-\frac {1}{2}=\frac {1}{2}$
The required equation of hyperbola is,
$\displaystyle \frac {x^2}{1/2}-\frac {y^2}{1/2}=1\Rightarrow 2x^2-2y^2=1$

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

A parabola is drawn with its vertex at $(0,-3)$, the axis of symmetry along the conjugate axis of the hyperbola $\displaystyle \frac { { x }^{ 2 } }{ 49 } -\frac { { y }^{ 2 } }{ 9 } =1$ and passing through the two foci of the hyperbola. The coordinates of the focus of the parabola are :

  1. $\displaystyle \left( 0,\frac { 11 }{ 6 } \right) $
  2. $\displaystyle \left( 0,-\frac { 11 }{ 6 } \right) $
  3. $\displaystyle \left( 0,\frac { 11 }{ 12 } \right) $
  4. $\displaystyle \left( 0,-\frac { 11 }{ 12 } \right) $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Equation of hyperbola is $\displaystyle \frac { { x }^{ 2 } }{ 49 } -\frac { { y }^{ 2 } }{ 9 } =1$

Its conjugate axis is y-axis
Also, $\displaystyle e=\sqrt { 1+\frac { { b }^{ 2 } }{ { a }^{ 2 } }  } =\sqrt { 1+\frac { 9 }{ 49 }  } =\frac { \sqrt { 58 }  }{ 7 } $
$\therefore$ Foci of hyperbola is $\left( \pm ae,0 \right) \Rightarrow \left( \pm \sqrt { 58 } ,0 \right) $
Now equation of parabola with vertex at $(0,-3)$ and axis along y-axis is ${ x }^{ 2 }=l\left( y+3 \right) $
It passes through $\left( \pm \sqrt { 58 } ,0 \right) $
$\displaystyle \therefore 58=l\left( 0+3 \right) \Rightarrow l=\frac { 58 }{ 3 } $
$\therefore$ parabola is $\displaystyle { x }^{ 2 }=\frac { 58 }{ 3 } \left( y+3 \right) $
Its focus is $\displaystyle \left( 0,-3+\frac { 58 }{ 4.3 }  \right) \equiv \left( 0,\frac { 11 }{ 6 }  \right) $

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

Which of the following is true for the hyperbola $9x^2\, -\, 16y^2\, -\, 18x\, +\, 32y\, -\, 151\, =\, 0$?

  1. The length of the transverse axes is $4$
  2. Length of latus rectum is $9$
  3. Equation of directrix is $x\, =\, \displaystyle \frac{21}{5}$ and $x\, =\, - \displaystyle \frac{11}{5}$

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

$9x^2\, -\, 16y^2\, -\, 18x\, +\, 32y\, -\, 151\, =\, 0$
$9(x^2-2x)-16(y^2-2y)=151$
$9(x^2-2x+1)-16(y^2-2y+1)=151-7=144$
$9(x-1)^2-16(y-1)^2=144$
$\cfrac{(x-1)^2}{16}-\cfrac{(y-1)^2}{9}=1$
$\Rightarrow a^2=16, b^2=9$
$\therefore e=\sqrt{1+\dfrac{b^2}{a^2}}=\cfrac{5}{4}$

$\therefore$ The length of the transverse axis is $=2a=8$
Length of latus rectum is $=2\cfrac{b^2}{a}=\cfrac{9}{2}$
Equation of directrix is, $x=1\pm \cfrac{a}{e}=1 \pm \cfrac{16}{5}=\cfrac{21}{5}$ or $-\cfrac{11}{5}$
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

The equations of the transverse and conjugate axes of a hyperbola are respectively $x + 2y - 3 = 0, 2x - y + 4 = 0$ and their respective lengths are $\displaystyle \sqrt{2}$ 2/$\displaystyle \sqrt{2}$. The equation of the hyperbola is 

  1. $\displaystyle \frac{2}{5}(x+2y-3)^{2}-\frac{3}{5}(2x-y+4)^{2}=1$
  2. $\displaystyle \frac{2}{5}(2x+y-4)^{2}-\frac{3}{5}(x+2y3-4)^{2}=1$
  3. $\displaystyle 2(2x-y+4)^{2}-3(x+2y-3)^{2}=1$
  4. $\displaystyle 2(2x+2y-3)^{2}-3(2x-y+4)^{2}=1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

It is given that $2a=\sqrt {2}$ which implies that $a=\dfrac { 1 }{ \sqrt { 2 }  }$.


Also, $2b=\dfrac { 2 }{ \sqrt { 3 }  } \Rightarrow b=\dfrac { 1 }{ \sqrt { 3 }  }$ 

If we take the two axes as the new coordinate system and the point of intersection of the axes of the new origin, then in the new coordinate system, equation of the hyperbola will be:

$\dfrac { { X }^{ 2 } }{ a^{ 2 } } -\dfrac { { Y }^{ 2 } }{ b^{ 2 } } =1\ \Rightarrow \dfrac { { X }^{ 2 } }{ \left( \dfrac { 1 }{ \sqrt { 2 }  }  \right) ^{ 2 } } -\dfrac { { Y }^{ 2 } }{ \left( \dfrac { 1 }{ \sqrt { 3 }  }  \right) ^{ 2 } } =1\ \Rightarrow \dfrac { { X }^{ 2 } }{ \dfrac { 1 }{ 2 }  } -\dfrac { { Y }^{ 2 } }{ \dfrac { 1 }{ 3 }  } =1\ \Rightarrow 2{ X }^{ 2 }-3{ Y }^{ 2 }=1\quad ....(1)$

Let $P(x,y)$ be the coordinates of a point on the hyperbola in original x-y system, then 

$X=\dfrac { \left| 2x-y+4 \right|  }{ \sqrt { 5 }  } ,\quad Y=\dfrac { \left| x+2y-3 \right|  }{ \sqrt { 5 }  } $

($\because$ $X$ is the distance of a point on hyperbola from $2x-y+4=0$ and $Y$ is the distance of a point on hyperbola from $x+2y-3=0$)

Therefore, equation 1 becomes:

$2{ \left( \dfrac { \left| 2x-y+4 \right|  }{ \sqrt { 5 }  }  \right)  }^{ 2 }-3{ \left( \dfrac { \left| x+2y-3 \right|  }{ \sqrt { 5 }  }  \right)  }^{ 2 }=1\ \Rightarrow \dfrac { 2 }{ 5 } { \left( 2x-y+4 \right)  }^{ 2 }-\dfrac { 3 }{ 5 } { \left( x+2y-3 \right)  }^{ 2 }=1$

Hence, the equation of the hyperbola is $\dfrac { 2 }{ 5 } { \left( 2x-y+4 \right)  }^{ 2 }-\dfrac { 3 }{ 5 } { \left( x+2y-3 \right)  }^{ 2 }=1$.