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 asymptote asymptotes of a curve introduction to asymptotes

The angle between the asymptotes of a hyperbola is $30^{o}$. The eccentricity of the hyperbola may be

  1. $\sqrt{3}\pm 1$
  2. $\sqrt{3}+1$
  3. $\pm\sqrt{2}$
  4. $none\ of\ these$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The angle between asymptotes is 2*sec^-1(e). If the angle is 30 degrees, then sec^-1(e) = 15 degrees. e = sec(15 degrees) = 1/cos(15 degrees) = 1/cos(45-30) = 1/(cos45cos30 + sin45sin30) = 1/((sqrt(2)/2 * sqrt(3)/2) + (sqrt(2)/2 * 1/2)) = 4/(sqrt(6)+sqrt(2)) = sqrt(6)-sqrt(2). None of the options match.

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

If $e$ is the eccentricity of $\dfrac {x^{2}}{a^{2}}-\dfrac {y^{2}}{b^{2}}=1$ and '$\theta $' be the angle between its asymptotes then $\cos (\theta /2)$ is equal to.

  1. $1/ 2e$
  2. $1/ e$
  3. $2/e^{2}$
  4. $none\ of\ these$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The angle between asymptotes is 2*sec^-1(e). Thus theta/2 = sec^-1(e), which means sec(theta/2) = e. Therefore, cos(theta/2) = 1/e.

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

If $\theta$ is the angle between the asymptotes of the hyperbola $\displaystyle \frac{x^2}{a^2}\, -\, \displaystyle \frac{y^2}{b^2}\, =\, 1$ with eccentricity $e$, then $\sec \displaystyle  \frac{\theta}{2}$can be

  1. $e$
  2. $\dfrac{e}2$
  3. $\dfrac{e}3$
  4. $\displaystyle \frac{e}{\sqrt{e^2\, -\, 1}}$
Reveal answer Fill a bubble to check yourself
A,D Correct answer
Explanation

$\tan\, \displaystyle \frac{\theta}{2}\, =\, \displaystyle \frac{b}{a}\,

\Rightarrow\, e^2\, -\, 1\, =\, \tan^2\, \displaystyle

\frac{\theta}{2}\, \Rightarrow\, \sec \displaystyle \frac{\theta}{2}\,

=\, e$
or $e^2\, -\, 1\, =\, \cot^2\, \displaystyle \frac{\theta}{2}\, \Rightarrow\, co\sec\, \displaystyle \frac{\theta}{2}\, =\, e$
$\Rightarrow\, \sec\, \displaystyle \frac{\theta}{2}\, =\, \displaystyle \frac{e}{\sqrt{e^2\, -\, 1}}$.

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

If $e$ is the eccentricity of $\displaystyle \frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1$ and $\theta$ be the angle between the asymptotes then $\displaystyle \sec { \frac { \theta  }{ 2 }  } $ equals :

  1. ${ e }^{ 2 }$
  2. $\displaystyle \frac { 1 }{ e } $
  3. $2e$
  4. $e$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Equation of asymptotes to $\displaystyle \frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1$ are given by 

$\displaystyle y=-\frac { b }{ a } x$
$\displaystyle \therefore { m } _{ 1 }=-\frac { b }{ a } $
Similarly $\displaystyle y=\frac { bx }{ a } $
$\therefore \displaystyle { m } _{ 2 }=\frac { b }{ a } $
Now $\displaystyle \theta =2\tan ^{ -1 }{ \frac { b }{ a }  } $
$\displaystyle \Rightarrow \tan { \frac { \theta  }{ 2 }  } =\frac { b }{ a } \Rightarrow \tan ^{ 2 }{ \frac { \theta  }{ 2 }  } =\frac { { b }^{ 2 } }{ { a }^{ 2 } } ={ e }^{ 2 }-1$
$\displaystyle \Rightarrow \sec ^{ 2 }{ \frac { \theta  }{ 2 }  } ={ e }^{ 2 }\Rightarrow \sec { \frac { \theta  }{ 2 }  } =e$

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

If e is the eccentricity of the hyperbola and $\theta$ is angle between the asymptotes, then $\dfrac{cos\theta}{2}$ = 

  1. $\dfrac{(1-e)}{e}$
  2. $\dfrac{1}{e}-1$
  3. $\dfrac{1}{e}$

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

Let $\dfrac {x^{2}}{a^{2}}-\dfrac {y^{2}}{b^{2}}=1$  be the hyperbola


It has asymptotes $y=\pm \dfrac {b} {a} x$
Angle between the asymptotes $= 2tan^-1 (\dfrac{b}{a})=\theta$
$\Rightarrow \tan \dfrac {\theta} {2}=\pm \dfrac {b} {a} $

$\Rightarrow sec^{2}\dfrac {\theta} {2}=1+\tan ^{2}\dfrac {\theta} {2}=1+\dfrac {b^{2}}{a^{2}}$

$\Rightarrow \sec ^{2}\dfrac {\theta} {2}=\sqrt {1+\dfrac{b^{2}}{a^{2}}} $

$\Rightarrow \sec ^{2}\dfrac {\theta} {2} =e^{2}$

$\Rightarrow \cos \dfrac {\theta} {2}=\dfrac {1}{e}$

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

From a point $P (1, 2)$ two tangents are drawn to a hyperbola $H$ in which one tangent is drawn to each arm of the hyperbola. If the equations of asymptotes of hyperbola $H$ are $\sqrt 3x-y+5=0$ and $\sqrt 3x+y-1=0$, then eccentricity of $H$ is :

  1. $2$
  2. $\dfrac {2}{\sqrt 3}$
  3. $\sqrt 2$
  4. $\sqrt 3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Since ${c} _{1}{c} _{2}\left({a} _{1}{a} _{2}+{b} _{1}{b} _{2}\right)<0$

$\therefore$ origin lies in acute angle. 

$P\left(1,2\right)$ lies in obtuse angle

Slope of asymptotes${m} _{1}=\sqrt{3},\,{m} _{2}=-\sqrt{3}$

$\tan{\theta}=\left|\dfrac{{m} _{1}-{m} _{2}}{1+{m} _{1}{m} _{2}}\right|$

$=\left|\dfrac{\sqrt{3}-\left(-\sqrt{3}\right)}{1+\sqrt{3}\times-\sqrt{3}}\right|$

$=\left|\dfrac{2\sqrt{3}}{1-3}\right|$

$=\left|\dfrac{2\sqrt{3}}{-2}\right|$

$\Rightarrow\,\tan{\theta}=\sqrt{3}$

Acute angle between the asymptotes is $\dfrac{\pi}{3}$

Hence eccentricity $e=\sec{\dfrac{\theta}{2}}=\sec{\dfrac{\pi}{6}}=\dfrac{2}{\sqrt{3}}$
Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

The asymptotes of the hyperbola $\dfrac {x^2}{a^2}-\dfrac {y^2}{b^2}=1$ form with any tangent to the hyperbola a triangle whose area is $a^2 \tan\lambda$ in magnitude, then its eccentricity is :

  1. $\sec \lambda$
  2. $\cos ec \lambda$
  3. $\sec^2\lambda$
  4. $\cos ec^2\lambda$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Any tangent to hyperbola forms a triangle with the asymptotes which has constant area $ab$.

$\Rightarrow ab=a^2 \tan\lambda$

$\displaystyle \Rightarrow \frac {b}{a}=\tan \lambda$

$\displaystyle e=\sqrt{1+\frac{b^2}{a^2}} $

$\Rightarrow e = \sqrt{1+\tan^2{\lambda}} =\sec{\lambda}$

Multiple choice maths logarithms more about logarithms a relation between logarithmic functions properties of logarithms

If the eccentricity of the ellipse $\cfrac { { x }^{ 2 } }{ { \left( \log { a }  \right)  }^{ 2 } } +\cfrac { { y }^{ 2 } }{ { \left( \log { b }  \right)  }^{ 2 } } =1\left( a>b>0,a\neq 1 \right) $ is $\cfrac { 1 }{ \sqrt { 2 }  } $ and $c$ be the eccentricity of the hyperbola $\cfrac { { x }^{ 2 } }{ { \left( \log _{ b }{ a }  \right)  }^{ 2 } } -{ y }^{ 2 }=1\quad $ then ${e}^{2}$ is greater than (where $\log{x}-\ln{x}$)

  1. $\dfrac{3}{2}$
  2. $\dfrac{1}{2}$
  3. $\dfrac{2}{3}$
  4. $\dfrac{5}{4}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Eqn of ellipse
$\dfrac{x^2}{(\log a)^2}+\dfrac{y^2}{(\log b)^2}=1\left(a>b>0, a\neq 1\right)$
if $a>b>0$ then $\log a> \log b$
For ellipse
$e=\dfrac{1}{\sqrt{2}}$
$e=\sqrt{1-\left(\dfrac{\log b}{\log a}\right)^2}=\dfrac{1}{\sqrt{2}}\quad \begin{cases} as\  & \dfrac { x^{ 2 } }{ a^{ 2 } } +\dfrac { y^{ 2 } }{ b^{ 2 } } =1 \\ e=\sqrt { 1-\dfrac { b^{ 2 } }{ a^{ 2 } }  }  & if\quad a>b \end{cases}$
$1-\left(\dfrac{\log b}{\log a}\right)^2=\dfrac{1}{2}$
$\dfrac{1}{2}=\left(\dfrac{\log b}{\log a}\right)^2\quad ----(1)$
Eqn of hyperbola
$\dfrac{x^2}{(\log _b a)^2}-y^2=1$
$e=\sqrt{1+\dfrac{b^2}{a^2}}\quad \left\{\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1\right\}$
So
$e=\sqrt{1+\dfrac{1}{(\log _b a)^2}}=\sqrt{1+\left(\dfrac{\log b}{\log a}\right)}\quad \left\{ as\ \log _{ b }{ a=\dfrac { \log { a }  }{ \log { b }  }  }  \right. $
$=\sqrt{1+\dfrac{1}{2}}=\sqrt{\dfrac{3}{2}}$
$e^2=\dfrac{3}{2}$
Multiple choice mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

The eccentricity of the hyperbola $16x^2-9y^2=1$ is

  1. $\dfrac{3}{5}$
  2. $\dfrac{5}{3}$
  3. $\dfrac{4}{5}$
  4. $\dfrac{5}{4}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Equation of hyperbola: $16x^2-9y^2=1$
can be written as $\cfrac{x^2}{\frac{1}{16}}+\cfrac{y^2}{\frac{1}{9}}=1$
$e^2=1+\cfrac{b^2}{a^2}=1+\cfrac{16}{9}=\cfrac{25}9$
or, $e=\cfrac53$
Hence, B is the correct option.
Multiple choice mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

An ellipse and a hyperbola have the same principle axes. From a point on the ellipse, tangents are drawn to the hyperbola . then  the chord contact of these tangents touches the ellipse.

  1. True

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

This is a standard property of conics where the chord of contact of tangents from a point on one conic to another conic touches the original conic under specific conditions.

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

The eccentricity of the conic represented by$2{x}^{2}+5xy+2{y}^{2}+11x-7y-4=0$ is

  1. $\dfrac {\sqrt {10}}{3}$
  2. $\dfrac {\sqrt {10}}{4}$
  3. $\dfrac {5}{4}$
  4. $\dfrac {3}{5}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equation represents a hyperbola. Calculating the eccentricity involves finding the angle between the asymptotes or using the general conic eccentricity formula. The result sqrt(10)/3 is the standard derivation for this specific equation.

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

The equation $\dfrac{x^{2}}{29 -p} + \dfrac{y^{2}}{4 -p} =1(p\neq4, 29)$ represents - 

  1. an ellipse if $p$ is any constant greater than $4$
  2. hyperbola if $p$ is any constant between $4$ and $29$.
  3. a rectanglar hyperbola is $p$ is any constant greater than $29$.
  4. no real curve is $p$ is less than $29$.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Equation of Hyperbola is $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$

if p lies between $4$ and $29$ then coefficient of $y^2$  is negative and coefficient of $x^2$ is positive

Hence, it satisfies the equation of Hyperbola between $4$ and $29$
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 $2x + y = 1$,corresponding focus is $(1, 1)$ and eccentricity $\sqrt { 3 }$, is given by 

  1. $7 x ^ { 2 } + 12 x y - 2 y ^ { 2 } - 2 x + 4 y - 7 = 0$
  2. $2 x ^ { 2 } + 12 x y - 7 y ^ { 2 } - 2 x + 14 y - 7 = 0$
  3. $7 x ^ { 2 } - 12 x y + 2 y ^ { 2 } - 2 x + 14 y - 22 = 0$
  4. $7 x ^ { 2 } + 12 x y - 2 y ^ { 2 } - 2 x - 14 y - 22 = 0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Let$ P(x, y)$ is any point on the hyperbola.
given, focus of parabola is $S(1,1)$.
equation of directrix is $2x + y = 1$
From P draw PM perpendicular to the directrix then $PM = (2x + y – 1)/√(2² + 1²) = (2x + y – 1)/√5$
Also from the definition of the hyperbola, we have
$SP/PM = e ⇒ SP = ePM$
$⇒ √{(x–1)² + (y–1)²} = √3{(2x + y – 1)/√5}$
$⇒ (x – 1)² + (y – 1)² = 3 (2x + y – 1)²/5$
$⇒ 5[(x² – 2x + 1) + (y² –2y + 1)] = 3(4x² + y² + 1 + 4xy – 4x – 2y)$
$⇒5x² - 10x + 5 + 5y² - 10y + 5 = 12x² + 3y² + 3 + 12xy - 12x - 6y $
$⇒7x² + 2y² + 12xy - 2x + 4y - 7 = 0$
hence, equation of hyperbola is $7x² - 2y² + 12xy - 2x + 4y - 7 = 0$