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

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

  1. x-axis

  2. y-axis

  3. z-axis

  4. none
Reveal answer Fill a bubble to check yourself
B 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 foci lie on its transverse axis. $i.e.$ $y$ - axis. one each side of the hyperbola, at $(0, be)$ and $(0,-be)$ respectively.

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

The foci of the hyperbola $4{ x }^{ 2 }-9{ y }^{ 2 }-1=0$ are

  1. $\left( \pm \sqrt { 13 } ,0 \right) $
  2. $\left( \pm \dfrac { \sqrt { 13 } }{ 6 } ,0 \right) $
  3. $\left( 0,\pm \dfrac { \sqrt { 13 } }{ 6 } \right) $

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

The given equation of hyperbola can be re-written as,

$\dfrac { x^{ 2 } }{ \left( \dfrac { 1 }{ 2 }  \right) ^{ 2 } } -\dfrac { { y }^{ 2 } }{ \left( \dfrac { 1 }{ 3 }  \right) ^{ 2 } } =1$
So, the hyperbola has x-axis as the transverse axis, with $a=\dfrac { 1 }{ 2 } $
and $b=\dfrac { 1 }{ 3 }$ 

And, $e=\displaystyle \sqrt { 1+\dfrac { b^{ 2 } }{ a^{ 2 } }  }$

 $ =\displaystyle \sqrt { 1+\dfrac { 4 }{ 9 }  } =\dfrac { \sqrt { 13 }  }{ 3 }$ 

$\therefore$ Focii $=(\pm ae,0)=\left(\pm \dfrac { \sqrt { 13 }  }{ 6 } ,0\right)$
Hence, B is correct.

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. $(2,3)$
  2. $(\pm\sqrt3,3)$
  3. $(2,\pm3)$
  4. $(1,2)$,$(1,-6)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

It has center at $(0,0)$ and vertex at $(0,\pm a)$.
For hyperbola $\dfrac{(y+2)^{2}}{16}-\dfrac{(x-1)^{2}}{3}=1$
Its center shifted to $(1,-2)$.
So, vertex will also shift by $(1,-2)$. 
So, vertex is $(0,\pm a)+(1,-2)=(0,\pm 4)+(1,-2) $
$\Rightarrow  (1,2)$ and $(1,-6)$

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

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

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

The given hyperbola is $-\dfrac{(x-1)^2}{3} + \dfrac{(y+2)^2}{16} = 1$ is a conjugate hyperbola.


It can be written as $\dfrac{(x-1)^2}{3} - \dfrac{(y+2)^2}{16} = -1$  ......$(1)$

Now let $x-1 = X$ and $y+2 = Y$

Putting values of $x-1$ and $y+2$ in eq. $(2)$ we get,

$\rightarrow  \ \dfrac{(X)^2}{3} - \dfrac{(Y)^2}{16} = -1$  ....$(2)$

We can see the eq$(2)$ is a standard form of conjugate hyperbola and it's center lies at origin $(0,0)$

So $X =0 , Y = 0$ is the center of the hyperbola given in eq.$(2)$

$\rightarrow$ $ X = x-1 = 0$ or $x = 1$

$\rightarrow$ $ Y = y+2 = 0$ or $y = -2$

Hence center of the given hyperbola in eq. $(1)$ is $(1,-2)$. So correct option is $A$.

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 of given length of transverse axis $6$ and the join of centre and focus is bisected by vertex.

  1. $3x^{2} - y^{2} = 27$.
  2. $3x^{2} + y^{2} = 27$.
  3. $x^{2} - y^{2} = 27$.

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

Transverse axis 2a = 6, so a = 3. Vertex bisects center and focus means a = ae/2, so e = 2. Then b^2 = a^2(e^2 - 1) = 9(4 - 1) = 27. Equation is x^2/9 - y^2/27 = 1, or 3x^2 - y^2 = 27.

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

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

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

The given hyperbola is $-\dfrac{(x-1)^2}{3} + \dfrac{(y+2)^2}{16} = 1$ is a conjugate hyperbola.


It can be written as $\dfrac{(x-1)^2}{3} - \dfrac{(y+2)^2}{16} = -1$  ......$(1)$

Now let $x-1 = X$ and $y+2 = Y$

Putting values of $x-1$ and $y+2$ in eq. $(2)$ we get,

$\rightarrow  \ \dfrac{(X)^2}{3} - \dfrac{(Y)^2}{16} = -1$  ....$(2)$

We can see the eq$(2)$ is a standard form of conjugate hyperbola and it's center lies at origin $(0,0)$

So $X =0 , Y = 0$ is the center of the hyperbola given in eq.$(2)$

$\rightarrow$ $ X = x-1 = 0$ or $x = 1$

$\rightarrow$ $ Y = y+2 = 0$ or $y = -2$

Hence center of the given hyperbola in eq. $(1)$ is $(1,-2)$. So correct option is $C$.

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

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

  1. $y = 2$
  2. $y = 6$
  3. $y = 8$
  4. $y = 3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equation at the conjugate axis of the hyperbola

$\frac{{{{\left( {y - 2} \right)}^2}}}{9} - \frac{{{{\left( {x + 3} \right)}^2}}}{9} = 1$ 
There for,
      $y-2=0$
      $y=2$
 option (A) is correct answer

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

The transverse axis of a hyperbola is of length $2a$ and a vertex divides the segment of the axis between the centre and the corresponding focus in the ratio $2:1$. The equation of the hyperbola is

  1. $4x^2-5y^2=4a^2$
  2. $4x^2-5y^2=5a^2$
  3. $5x^2-4y^2=4a^2$
  4. $5x^2-4y^2=5a^2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We have given  $\cfrac{a}{ae-a}=2\Rightarrow e=\cfrac{3}{2}$
Using $e^2=1+\cfrac{b^2}{a^2}$
$\cfrac{9}{4}=1+\cfrac{b^2}{a^2}\Rightarrow b^2=\cfrac{5}{4}a^2$
Hence required hyperbola is $\cfrac{x^2}{a^2}-\cfrac{y^2}{b^2}=1$
$\cfrac{x^2}{a^2}-\cfrac{y^2}{\frac{5}{4}a^2}=1$
$\Rightarrow 5x^2-4y^2=5a^2$
Hence option 'D' is correct.

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

The equation of the hyperbola whose directrix is $x + 2y = 1$, focus is $(2, 1)$ and eccentricity $2$ is

  1. $x^2 + 16 xy - 11y^2 - 12 x + 6y + 21 = 0$
  2. $x^2 - 16 xy - 11y^2 - 12 x + 6y + 21 = 0$
  3. $x^2 - 4 xy - y^2 - 12 x + 6y + 21 = 0$

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

Using Definition of hyperbola
$PS^2=e^2\cdot PM^2$
$(x-2)^2+(y-1)^2=2^2\left(\cfrac{x+2y-1}{\sqrt{5}}\right)^2$
$5(x^2+y^2-4x-2y+5)=4(x^2+4y^2+1+4xy-2x-4y)$
$\Rightarrow x^2 - 16 xy - 11y^2 - 12 x + 6y + 21 = 0$
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

Equation of the hyperbola whose vertices are at ($\pm3, 0$) and focii at ($\pm5, 0$) is

  1. $16x^2 - 9y^2 = 144$
  2. $9x^2 - 16y^2 = 144$
  3. $25x^2 - 9y^2 = 255$
  4. $9x^2 - 25y^2 = 81$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given $a=3$ and $ae=5\Rightarrow e=\cfrac{5}{3}$
Using $e^2=1+\cfrac{b^2}{a^2}$
$\cfrac{25}{9}=1+\cfrac{b^2}{9}\Rightarrow b^2=16$
Therefore required hyperbola is, $\cfrac{x^2}{9}-\cfrac{y^2}{16}=1$
$\Rightarrow 16x^2-9y^2=144$
Hence, option 'A' is correct.

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

The tangent of a point $P$ on the hyperbola $\dfrac {x^{2}}{a^{2}}-\dfrac {y^{2}}{b^{2}}=1$ passes through the point $(0,\ -b)$ and the normal at $P$ pases through the point $(2a\sqrt {2},\ 0)$. Then the eccentricity of the hyperbola is   

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

Using the condition that the tangent passes through (0, -b) and the normal through (2a*sqrt(2), 0) for a point P on the hyperbola, one can derive the eccentricity e = 2.

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

Find the equation of the hyperbola whose directrix is $2x+y=1$, focus $(1,2)$ and eccentricity $\sqrt{3}$

  1. $7x^2-2y^2 +12xy-2x+14y-22=0$
  2. $7x^2-2y^2 +2xy-2x+14y-22=0$
  3. $7x^2-2y^2 +xy-14x+2y-22=0$

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

Using the definition of a hyperbola (distance from focus = e * distance from directrix), the equation is (x-1)^2 + (y-2)^2 = 3 * (2x+y-1)^2 / (2^2 + 1^2). Expanding this yields the provided equation.

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

The ecentricity of the hyperbola passing through the origin and whose asymptotes are given by straight lines $y=3x-1$ and $x+3y=3$, is

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

The asymptotes are 3x - y - 1 = 0 and x + 3y - 3 = 0. The angle between them is 90 degrees because the product of their slopes is -1 (3 * -1/3 = -1). A hyperbola with perpendicular asymptotes is a rectangular hyperbola, which has an eccentricity of sqrt(2).