Mathematics

Conic Sections

285 Questions

Conic sections deal with the geometry of curves like ellipses, hyperbolas, and parabolas formed by the intersection of a plane with a cone. This is an important topic in advanced mathematics sections of competitive exams. Practice these questions to understand eccentricity, focal distances, and equations of tangents and normals.

Ellipse equationsHyperbola eccentricityFocal distance calculationsTangent and normal equationsConic standard forms

Conic Sections Questions

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

In the hyperbola $4x^2\, -\, 9y^2\, =\, 36$, find lengths of the axes, the co-ordinates of the foci, the eccentricity, and the latus rectum.

  1. $6, 4;\, (\pm\, \sqrt{13},\, 0);\, \dfrac{\sqrt{13}}3;\, \dfrac8 3$
  2. $9, 4;\, (\pm\, \sqrt{8},\, 0);\, \dfrac{\sqrt{8}}3;\, \dfrac83$
  3. $9, 4;\, (\pm\, \sqrt{13},\, 0);\, \dfrac{\sqrt{13}}3;\, \dfrac83$
  4. $6, 4;\, (\pm\, \sqrt{8},\, 0);\, \dfrac{\sqrt{8}}3;\, \dfrac83$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given hyperbolas may be written as,  $\displaystyle \frac{x^2}{9}-\frac{y^2}{4}=1$
$\Rightarrow a^2 =9, b^2=4$
$\therefore$ Eccentricity is, $\displaystyle e= \sqrt{1+\frac{b^2}{a^2}}=\frac{\sqrt{13}}{3}$
Thus, length of axes are $2a$ and $2b \Rightarrow 6 $ and $4$
Focus $\equiv (\pm ae, 0) =(\pm \sqrt{13},0)$
And length of latus rectum $=\cfrac{2b^2}{a}=\cfrac{8}{3}$

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, whose eccentricity is $\displaystyle \frac{5}{4}$, focus is $(a, 0)$ and whose directrix is $4x - 3y = a$.

  1. $7y^2\, +\, 24xy\, -\, 12ax\, -\, 3ay\, +\, 15a^2\, =\, 0$
  2. $7y^2\, +\, 24xy\, +\, 12ax\, +\, 3ay\, +\, 15a^2\, =\, 0$
  3. $7y^2\, +\, 24xy\, +\, 24ax\, +\, 6ay\, +\, 15a^2\, =\, 0$
  4. $7y^2\, +\, 24xy\, -\, 24ax\, -\, 6ay\, +\, 15a^2\, =\, 0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using Hyperbola definition, $PS^2=e^2.PM^2$
$\displaystyle (x-a)^2+(y-0)^2=\frac{25}{16}\left| \frac{4x-3y-a}{\sqrt{3^2+4^2}} \right|^2$
$\Rightarrow 16(x^2+y^2-2ax+a^2)=16x^2+9y^2+a^2-24xy+6ya-8ax$
$\Rightarrow 7y^2\, +\, 24xy\, -\, 24ax\, -\, 6ay\, +\, 15a^2\, =\, 0$

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

The foci of the ellipse $\displaystyle \frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1$ and the hyperbola $\displaystyle \frac { { x }^{ 2 } }{ 144 } -\frac { { y }^{ 2 } }{ 81 } =\frac { 1 }{ 25 } $ coincide. The value of ${ b }^{ 2 }$ is

  1. $9$
  2. $1$
  3. $5$
  4. $7$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The equation of hyperbola is $\displaystyle \frac { { x }^{ 2 } }{ 144 } -\frac { { y }^{ 2 } }{ 81 } =\frac { 1 }{ 25 } $


Here, $\displaystyle a=\sqrt { \frac { 144 }{ 25 }  } ,b=\sqrt { \frac { 81 }{ 25 }  } ,e=\sqrt { 1+\frac { 81 }{ 144 }  } =\frac { 15 }{ 12 } =\frac { 5 }{ 4 } $
$\therefore$ Foci $=\left( \pm 3,0 \right) $
Also, focus of ellipse $\displaystyle =\left( 3,0 \right) \Rightarrow e=\frac { 3 }{ 4 } $
$\displaystyle \therefore { b }^{ 2 }=16\left( 1-\frac { 9 }{ 16 }  \right) =7$

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, the distance between whose foci is $16$ and whose eccentricity is $\sqrt{2}$.

  1. $x^2\, -\, y^2\, =\, 32$
  2. $x^2\, -\, y^2\, =\, 18$
  3. $x^2\, -\, y^2\, =\, 64$
  4. $x^2\, -\, y^2\, =\, 48$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given eccentricity of the hyperbola is $\sqrt{2}$
Hence hyperbola is rectangular $a=b$
Also given distance between focii is 16. $\Rightarrow 2ae = 16\Rightarrow a =4\sqrt{2}$
Hence required hyperbola is given by, $x^2-y^2=a^2=32$

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

An ellipse intersects the hyperbola $\displaystyle 2x^{2}-2y^{2}=1$ orthogonally at point $P$. The eccentricity of the ellipse is reciprocal to that of the hyperbola. If the axes of the ellipse are along the co-ordinate axes and product of focal distances of $P$ is $x$ then $2x$ is:

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

As ellipse and hyperbola intersect orthogonally 
$\displaystyle \Rightarrow $ their foci are coincident
Now we have $\displaystyle SP+S,P=2a\Rightarrow $ length of major axis ellipse ..........(1)
And $\displaystyle \left | SP-S,P \right |=2a'\Rightarrow $ length of transverse axis of hyperbola ....... (2)
$\displaystyle (1)^{2}-(2)^{2}\Rightarrow 4PS'PS=4(a^{2}-a'^{2})=4\left [ \left ( \frac{a'e}{e} \right )^{2} -a'^{2}\right ]=4\left ( 2-\frac{1}{2} \right )=6$
$\displaystyle \Rightarrow 2x = 3$

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

MATCH THE FOLLOWING
Hyperbola                                                   Length of latusrectum
A}$x^{2}-4y^{2}=4$                                               1. 1
B}$25x^{2}-16y^{2}=400$                                     2.12
C}$ 2x^{2}-y^{2}-4x-4y-20=0$                   3.9/2
D)$9x^{2}-16y^{2}+72x-32y-16=0$           4. 25/2

The correct match is

  1. I II III IV

    1 2 3 4

  2. 1 4 2 3

  3. 3 1 2 4

  4. 2 3 4 1
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

(A) $\dfrac{x^{2}}{4}-y^{2}=1$


$LR=\dfrac{2b^{2}}{a}$

$=\dfrac{2}{2}$

$=1$

(B)$\dfrac{x^{2}}{16}-\frac{y^{2}}{25}=1$

$b=5, a=4$

$LR=\dfrac{2.25}{4}$

$=\dfrac{25}{2}$

(C) $2(x-1)^{2}-(y-2)^{2}=18$

$\dfrac{(x-1)^{2}}{9}-\frac{(y-2)^{2}}{18}=1$

$b^{2}=18, a=3$

$LR=\dfrac{2\times 18}{3}$

$=12$

(D) $\dfrac{(x+4)^{2}}{16}-\frac{(y+1)^{2}}{9}=1$

$b^{2}=9$

$a=4$

$LR=2\times \dfrac{9}{4}$

$=\dfrac {9}{2}$

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

The equation to the hyperbola having its eccentricity $2$ and the distance between its foci is $8$, is

  1. $\dfrac {x^{2}}{12} - \dfrac {y^{2}}{4} = 1$
  2. $\dfrac {x^{2}}{4} - \dfrac {y^{2}}{12} = 1$
  3. $\dfrac {x^{2}}{8} - \dfrac {y^{2}}{2} = 1$
  4. $\dfrac {x^{2}}{16} - \dfrac {y^{2}}{9} = 1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the equation of hyperbola is $\dfrac {x^{2}}{a^{2}} - \dfrac {y^{2}}{b^{2}} = 1$
Given, $e = 2, 2ae = 8$
$\Rightarrow ae = 4\Rightarrow a = 2$
Now, $b^{2} = a^{2} (e^{2} - 1)$
$\Rightarrow b^{2} = 4(4- 1)$
$\Rightarrow b^{2} = 12$
$\therefore$ Equation of hyperbola is
$\dfrac {x^{2}}{4} - \dfrac {y^{2}}{12} = 1$

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

Hyperbola $\dfrac{{x}^{2}}{{a}^{2}}-\dfrac{{y}^{2}}{3}=1$ of eccentricity $e$ is confocal with the ellipse $\dfrac{{x}^{2}}{8}+\dfrac{{y}^{2}}{4}=1$. Let $A$, $B$, $C$ & $D$ are points of intersection of hyperbola & ellipse, then-

  1. $e=\dfrac{5}{2}$
  2. $e=2$
  3. $A$, $B$, $C$, $D$ are concyclic points
  4. Number of common tangents of hyperbola & ellipse is $2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

If foci of $\dfrac {x^2}{a^2}-\dfrac {y^2}{b^2}=1$ coincide with the foci of $\dfrac {x^2}{25}+\dfrac {y^2}{9}=1$ and eccentricity of the hyperbola is 2, then :

  1. $a^2+b^2=16$

  2. there is no director circle to the hyperbola
  3. centre of the director circle is $(0, 0)$
  4. length of latus rectum of the hyperbola $=12$
Reveal answer Fill a bubble to check yourself
A,B,D Correct answer
Explanation

$\displaystyle \frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 9 } =1$
which is an ellipse.
For ellipse, $a=5$ , $b=3$ 
$\Rightarrow \displaystyle e=\sqrt {\frac {25-9}{25}}=\frac {4}{5}$
$\therefore ae=4$
Hence, the foci are $(-4, 0)$ and $(4, 0)$.
For the hyperbola,
$ae=4, e=2$
$\Rightarrow a=2$
$b^2=4(4-1)=12$
$\Rightarrow b=\sqrt {12}$
$\Rightarrow a^2+b^2=16$
Since $b^2>a^2$, so there is no director circle to the hyperbola.
Length of latus rectum $\displaystyle=\frac{2b^2}{a}=12$

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

An ellipse intersects the hyperbola $2x^{2}-2y^{2}=1$ orthogonally. The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinates axes, then

  1. equation of ellipse is $x^{2}+2y^{2}=2$
  2. the foci of ellipse are $\left ( \pm 1, 0 \right )$
  3. equation of ellipse is $x^{2}+2y^{2}=4$
  4. the foci of ellipse are $\left ( \pm \sqrt{2}, 0 \right )$
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation

Eccentricity of the hyperbola is $\sqrt{2}$ as it is a rectangular hyperbola. 

So eccentricity $e$ of the ellipse is $\dfrac1{\sqrt{2}}$
Let the equation of the ellipse be $\displaystyle \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ where $b^{2}=a^{2}\left ( 1-e^{2} \right )=\dfrac{a^{2}}2\Rightarrow a^{2}=2b^{2}$
So equation of the ellipse is $x^{2}+2y^{2}=a^{2}$
Let $\left ( x _{1}, y _{1} \right )$ be a point of intersection of the ellipse and the hyperbola
Then $2x _{1}^{2}-2y _{1}^{2}=1$ and $x _{1}^{2}+2y _{1}^{2}=a^{2}$          (1)
Equations of the tangents at $\left ( x _{1}, y _{1} \right )$ to the two conics are
   $2xx _{1}-2yy _{1}=1$ and $xx _{1}+2yy _{1}=a^{2}$
Since the two conics intersect orthogonally
$\displaystyle \left ( \frac{x _{1}}{y _{1}} \right )\left ( -\frac{x _{1}}{2y _{1}} \right )=-1\Rightarrow x _{1}^{2}=2y _{1}^{2}$
And from (1) we get $x _{1}^{2}=1$, $a^{2}=2$.
Hence the equation of the ellipse is $x^{2}+2y^{2}=2$ and its focus is
   $\displaystyle \left ( \pm ae, 0 \right )=\left ( \pm \sqrt{2}\times \frac{1}{\sqrt{2}}, 0 \right )=\left ( \pm 1, 0 \right )$

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

The equation of the ellipse whose equation of directrix is $3x+4y-5=0$, coordinates of the focus are $(1,2)$ and the eccentricity is $\dfrac{1}{2}$ is $91x^2+84y^2-24xy-170x-360y+475=0$

  1. True

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

Let $P(x,y)$ be any point on the ellipse and PM be the perpendicular from P upon the directrix $3x+4y-5=0$.

Then by the definition,
$\dfrac{SP}{PM}=e$

$SP=e.PM$
$\sqrt{(x-1)^2+(y-2)^2}=\dfrac{1}{2}|\dfrac{3x+4y-5}{\sqrt{3^2+4^2}}|$

$(x-1)^2+(y-2)^2=\dfrac{1}{4}. \dfrac{(3x+4y-5)^2}{25}$

$100(x^2+y^2-2x-4y+5)=9x^2+16y^2+24xy-30x-40y+25$
$91x^2+84y^2-24xy-170x-360y+475=0$ is the equation of the ellipse.

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

The equation of the ellipse whose foci are $(\pm5,0)$ and of the directrix is $5x=36$, is

  1. $\dfrac{x^2}{36}+\dfrac{y^2}{11}=1$
  2. $\dfrac{x^2}{6}+\dfrac{y^2}{\sqrt{11}}=1$
  3. $\dfrac{x^2}{6}+\dfrac{y^2}{11}=1$

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

Given foci $(\pm 5,0)$ and directrix $x=\cfrac{36}{5}$

Then $ae=5$ (focus coordinates ($\pm ae,0)]$....(1)
$\cfrac{a}{e}=\cfrac{36}{5}$ (directrix equation $x=\cfrac{a}{e}$]....(2)
From (1) and (2) ${a}^{2}=36\Rightarrow$ $a=6$
$e=\cfrac{5}{6}\Rightarrow $ $\sqrt { 1-\cfrac { { b }^{ 2 } }{ { a }^{ 2 } }  } =\cfrac { 5 }{ 6 } $
$1-\cfrac { { b }^{ 2 } }{ 36 } =\cfrac{25}{36}$
$b=\sqrt 11$
required equation $\cfrac{{x}^{2}}{36}+\cfrac{{y}^{2}}{11}=1$

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

If the eccentricity of the ellipse $\dfrac{x^2}{a^2 + 1} + \dfrac{y^2}{a^2 + 2 } = 1$ is $\dfrac{1}{\sqrt{6}}$, then the length of latusrectum is

  1. $\dfrac{5}{\sqrt{6}}$
  2. $\dfrac{10}{\sqrt{6}}$
  3. $\dfrac{8}{\sqrt{6}}$

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

Given the ellipse equation x^2/(a^2+1) + y^2/(a^2+2) = 1, we identify the semi-axes. Since a^2+2 > a^2+1, the ellipse is vertical. The eccentricity e = 1/sqrt(6). Using e^2 = 1 - (a^2+1)/(a^2+2) = 1/(a^2+2), we find 1/6 = 1/(a^2+2), so a^2+2 = 6, a^2 = 4. The semi-axes are b^2 = 5 and a^2 = 6. Latus rectum = 2 * (minor^2) / major = 2 * 5 / sqrt(6) = 10/sqrt(6).

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

Which of the following can be the equation of an ellipse?

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

An ellipse equation in standard form is x^2/a^2 + y^2/b^2 = 1. Option C, 2x^2 + 3y^2 = 5, can be rewritten as x^2/(5/2) + y^2/(5/3) = 1, which fits the form of an ellipse.