Mathematics

Conic Sections

278 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 introduction to hyperbola standard equation of hyperbola conic sections

If area of quadrilateral formed by tangents drawn at ends of latus rectum of hyperbola $\dfrac { { x }^{ 2 } }{ { a }^{ 2 } } -\dfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1$ is equal to square of distance between centre and one focus of hyperbola,then ${ e }^{ 3 }$ is (e is eccentricity of hyperbola)

  1. $2\sqrt { 2 } $

  2. 2

  3. 3

  4. 8
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The area of the quadrilateral formed by tangents at the ends of the latus rectum is 2*b^2. Setting this equal to (ae)^2 and using b^2 = a^2(e^2 - 1) leads to e^3 = 2*sqrt(2).

Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

Eccentricity of a hyperbola is always less than 1.

  1. True

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

Standard equation of the hyperbola is $\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}$

The eccentricity of the hyperbola is given by
$e=\sqrt { 1+\dfrac { { b }^{ 2 } }{ { a }^{ 2 } }  }$ which is always greater than $1$.
Thus, the given statement is false.
Hence, option B is correct.

Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

The equation $\frac{x^2}{1-k}-\frac{y^2}{1+k}=1$, $k<1$ represents 

  1. $circle$
  2. $ellipse$
  3. $hyperbola$
  4. $none$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Equating the above equation with the second-degree equation
$A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0$ with $\dfrac{{x}^{2}}{1-k}-\dfrac{{y}^{2}}{1+k}=1$
we get $A=\dfrac{1}{1-k}, B=0, C=\dfrac{1}{1+k},D=0,E=0$ and $F=-1$
$(i)$For the second degree equation to represent a circle , the coefficients must satisfy the discriminant condition ${B}^{2}-4AC=0$ and also $A=C$
$\Rightarrow -4\times \dfrac{1}{1-k}\times \dfrac{1}{1+k}=0$
$\Rightarrow \dfrac{1}{1-{k}^{2}}=0$
This case does not exist
$(ii)$For the second degree equation to represent a ellipse , the coefficients must satisfy the discriminant condition ${B}^{2}-4AC<0$ and also $A\neq C$
$\Rightarrow -4\times \dfrac{1}{1-k}\times \dfrac{1}{1+k}<0$
$\Rightarrow \dfrac{1}{1-{k}^{2}}>0$
$\Rightarrow 1-{k}^{2}<0$
$\Rightarrow -{k}^{2}<-1$
$\Rightarrow {k}^{2}>1$ does not exist since it is given that $k<1$
$(iii)$For the second degree equation to represent a hyperbola, the coefficients must satisfy the discriminant condition ${B}^{2}-4AC>0$ and also $A\neq C$
$\Rightarrow -4\times \dfrac{1}{1-k}\times \dfrac{1}{1+k}>0$
$\Rightarrow \dfrac{1}{1-{k}^{2}}<0$
$\Rightarrow 1-{k}^{2}>0$
$\Rightarrow -{k}^{2}>-1$
$\Rightarrow {k}^{2}<1$ 
$\therefore k<1$
Hence the above equation represents a hyperbola.
Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

The equation $\displaystyle\frac{x^2}{10-\lambda}+\frac{y^2}{6-\lambda}=1$ represents

  1. a hyperbola if $\lambda < 6$
  2. an ellipse if $\lambda>6$
  3. a hyperbola if $6 < \lambda < 10$
  4. an ellipse if $0 < \lambda < 6$
Reveal answer Fill a bubble to check yourself
C,D Correct answer
Explanation

The general equation of an ellipse is $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ and that of a hyperbola is $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$
Using this, we get that the above equation is an ellipse if $10 - \lambda > 0$ and $6 - \lambda > 0$. The combined solution gives $\lambda < 6.$
For a hyperbola, the coefficient of $x^2$ and $y^2$ must be of opposite sign. Hence, 
 $10 - \lambda > 0$ and $6 - \lambda < 0$ which gives $6 < \lambda < 10$

Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

The point to which the axes are to be translated to eliminate $x$ and $y$ terms in the equation $3x^{2}-4xy-2y^{2}-3x-2y-1=0$ is 

  1. $\left(\dfrac{5}{2},3\right)$
  2. $(-4,\dfrac{3}{2})$
  3. $ (-2,3)$
  4. $ (2,3)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given equation is $3x^{2}-4xy-2y^{2}-3x-2y-1=0$

Let $\left({x} _{1},{y} _{1}\right)$ be a point to which the origin is shifted by translation

Let $\left(X,Y\right)$ be the new coordinates of the point $\left(x,y\right)$

$\therefore\,$ the equations of the transformation are $x=X+{x} _{1},\,y=Y+{y} _{1}$

Now the transformed equation is 
$3{\left(X+{x} _{1}\right)}^{2}-4\left(X+{x} _{1}\right)\left(Y+{y} _{1}\right)-2{\left(Y+{y} _{1}\right)}^{2}-3\left(X+{x} _{1}\right)-2\left(Y+{y} _{1}\right)-1=0$

$\Rightarrow\,3\left({X}^{2}+2X{x} _{1}+{{x} _{1}}^{2}\right)-4\left(XY+X{y} _{1}+Y{x} _{1}+{x} _{1}{y} _{1}\right)-2\left({Y}^{2}+{{y} _{1}}^{2}+2Y{Y} _{1}\right)-3X-{x} _{1}-2Y-2{y} _{1}-1=0$

$\Rightarrow\,3{X}^{2}+3{{x} _{1}}^{2}+6X{x} _{1}-4X{y} _{1}-4{x} _{1}Y-4{x} _{1}{y} _{1}-2{Y}^{2}-2{{y} _{1}}^{2}+4Y{y} _{1}-3X-3{x} _{1}-2Y-2{y} _{1}-1=0$

$\Rightarrow\,\left(3{X}^{2}-4XY-2{Y}^{2}\right)+\left(3{{x} _{1}}^{2}-2{{y} _{1}}^{2}-4{x} _{1}{y} _{1}-3{x} _{1}-2{y} _{1}-1\right)+2X\left(3{x} _{1}-2{y} _{1}-\dfrac{3}{2}\right)+2Y\left(-2{x} _{1}+2{y} _{1}-1\right)=0$

Solving the first degree terms,we have
$3{x} _{1}-2{y} _{1}=\dfrac{3}{2}$

$-2{x} _{1}+2{y} _{1}=1$

Adding the above equations, we get
$3{x} _{1}-2{y} _{1}-2{x} _{1}+2{y} _{1}=\dfrac{3}{2}+1$

$\Rightarrow\,{x} _{1}=\dfrac{5}{2}$

From equation ,$-2{x} _{1}+2{y} _{1}=1$

$\Rightarrow\,2{y} _{1}=1+2{x} _{1}=1+2\times\dfrac{5}{2}=1+5=6$

$\Rightarrow\,{y} _{1}=\dfrac{6}{2}=3$

$\therefore\,\left({x} _{1},{y} _{1}\right)=\left(\dfrac{5}{2},3\right)$


Hence the point is $\left(\dfrac{5}{2},3\right)$

Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

Eccentricity of hyperbola$ \dfrac { { x }^{ 2 } }{ k } -\dfrac { { y }^{ 2 } }{ k } =1$

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

$Standard\, hyperbola\, : $


$\dfrac { { { x^{ 2 } } } }{ k } -\dfrac { { { y^{ 2 } } } }{ k } =1$

$Now, \ \dfrac { { { { \left( { x-h } \right)  }^{ 2 } } } }{ { { a^{ 2 } } } } -\dfrac { { { { \left( { y-k } \right)  }^{ 2 } } } }{ { { b^{ 2 } } } } =1$

$Therefore\, Hyperbola\, properties\, are \ (h,k)=\left( { 0,0 } \right) ,\, \, a=\sqrt { k } ,\, b=\sqrt { k }  $

$=\dfrac { { \sqrt { { { \left( \sqrt k \right)  }^{ 2 } }+{ { \left( { \sqrt { k }  } \right)  }^{ 2 } } }  } }{ { \sqrt { k }  } } $

$=\sqrt { 2 }  \ Hence,\, the\, option\, C\, is\, the\, correct\, answer$

Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

A hyperbola passes through the focus of the ellipse $\dfrac{x^2}{25}+\dfrac{y^2}{16}=1,$ and its transverses and conjugate axes coincide with the major and minor axes of the ellipse. If the product of the eccentricites of the two curve is $1$, then the focus of the hyperbola is

  1. $(5\sqrt3,0)$
  2. $(5,0)$
  3. $\left(\dfrac{5}{3},0\right)$

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

The focus of the ellipse x^2/25 + y^2/16 = 1 is (3, 0). If the hyperbola passes through (3, 0) and shares axes, its equation is x^2/a^2 - y^2/b^2 = 1. Using the eccentricity product condition, one can solve for the focus.

Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

If eccentricity of the hyperbola $\dfrac {x^{2}}{\cos^{2}\theta}-\dfrac {y^{2}}{\sin^{2}\theta}=1$ is more then $2$ when $\theta\ \in \ \left(0,\dfrac {\pi}{2}\right)$. Find the possible values of length of latus rectum 

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

The eccentricity e = sqrt(1 + tan^2(theta)) = sec(theta). If e > 2, then sec(theta) > 2, so cos(theta) < 1/2. The latus rectum length is 2*sin^2(theta)/cos(theta). Solving this interval gives (3, infinity).

Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

The latus rectum of the hyperbola $16{x^2} - 9{y^2} = 144$ is-

  1. $\dfrac{13}{6}$
  2. $\dfrac{32}{3}$
  3. $\dfrac{8}{3}$
  4. $\dfrac{4}{3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We have,


$16{x^2} - 9{y^2} = 144$

$\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$

Here, $a=3,  b=4$

We know that the latus rectum 

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

Therefore,

$=\dfrac{2\times 16}{3}$

$=\dfrac{32}{3}$

Hence, this is the answer.

Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

Find the locus of a point which moves so that the difference of its distances from the points, $(5, 0)$ and $(-5, 0)$ is $2$ is:

  1. $\dfrac{x^2}{1}+\dfrac{y^2}{24}=1$
  2. $\dfrac{x^2}{24}+\dfrac{y^2}{1}=1$
  3. $\dfrac{x^2}{24}-\dfrac{y^2}{2}=1$
  4. $\dfrac{x^2}{1}-\dfrac{y^2}{24}=1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The locus is nothing but hyperbola.
Difference of distance of a point from foci $=2a$ 

Given distance is $2 \Rightarrow a=1$
Distance between foci $=2ae=2\sqrt{a^2+b^2}=\sqrt{(5+5)^2}$
                                                 $\Rightarrow a^2+b^2 =25$
                                                  $\Rightarrow b^2=24$
Therefore, locus is $\dfrac{x^2}{1}-\dfrac{y^2}{24}=1$ 

Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

If $e$ and $e'$ be the eccentricities of two conics $S$ and $S'$ such that $\displaystyle e^{2}+(e')^{2}= 3,$  then both $S$ and $S'$ are

  1. Ellipses

  2. Parabolas

  3. Hyperbolas

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

For a parabola the eccentricity is $1$

$\therefore e^2 + e'^2 = 1 + 1 = 2$
For an ellipse the eccerntricity is less than $1$
$\therefore$ for a hyperbola the eccentricity is greater than $1$
So, the conics can be hyperbolas
Hence, hyperbola correct.

Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

The eccentricity the hyperbola $x=\left( t+\dfrac { 1 }{ t }  \right) ,y=\dfrac { a }{ 2 } \left( t-\dfrac { 1 }{ t }  \right) $ is ____________.

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

The parametric equations x = a*sec(t) and y = b*tan(t) define a hyperbola. The given equations are a variation of this, leading to an eccentricity of sqrt(2).

Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

The equation $ \displaystyle 3x^{2}-2xy+y^{2}=0 $ represents:

  1. a circle

  2. hyperbola

  3. a pair of lines

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

Given expression,$\displaystyle 3{ x }^{ 2 }-2xy+{ y }^{ 2 }=0$ 
As Coefficient of $\displaystyle xy$ is not zero,It will not be a circle and hyperbola.
Let $\displaystyle \frac { y }{ x } =m$

We get $\displaystyle { m }^{ 2 }-2m+3=0$ will not have any real solutions as discriminant is less than zero.
$\displaystyle \therefore $ They will not be pair of lines too.

Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections
The difference between the length $2a$ of the transverse axis of a hyperbola of eccentricity $e$ and the length of its latus rectum is :
  1. $2a\left| 3-{ e }^{ 2 } \right| $
  2. $2a\left| 2-{ e }^{ 2 } \right| $
  3. $2a\left( { e }^{ 2 }-1 \right) $
  4. $a\left( 2{ e }^{ 2 }-1 \right) $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the equation of hyperbola be $\displaystyle \frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1$

Length of transverse axis is $2a$ and 
Length of latus rectum is $\displaystyle \frac { 2{ b }^{ 2 } }{ a } $
Now, difference $\displaystyle =\left| 2a-\frac { 2{ b }^{ 2 } }{ a }  \right| =\frac { 2 }{ a } \left| 2{ a }^{ 2 }-{ a }^{ 2 }{ e }^{ 2 } \right| $
$\therefore$ Difference $=2a\left| 2-{ e }^{ 2 } \right| $

Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

The equation of hyperbola whose coordinates of the foci are $(\pm8,0)$ and the lenght of latus rectum is $24$ units, is

  1. $3{ x }^{ 2 }-{ y }^{ 2 }=48$
  2. $4{ x }^{ 2 }-{ y }^{ 2 }=48$
  3. ${ x }^{ 2 }-3{ y }^{ 2 }=48$
  4. ${ x }^{ 2 }-4{ y }^{ 2 }=48$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The Foci of hyperbola are $(\pm8,0)$, hence Foci lie on $x$ - axis. 


We know that foci of hyperbola lie at $(\pm ae,0)$, So $ae = 8$ ...$(1)$

squaring both sides of equation $(1)$, we get,

$\Rightarrow a^2e^2 = 64$

Eccentricity of hyperbola $e^2 =1 + \dfrac {b^2}{a^2}$  

$\Rightarrow a^2(1+\dfrac{b^2}{a^2}) = 64$

$\Rightarrow a^2 + b^2 = 64$ ...$(2)$

Now the length of latus rectum is given as 24 units.

length of latusrectum of hyperbola $ = \dfrac{2b^2}{a} = 24$

$\Rightarrow b^2 = 12a$ ...$(3)$

putting value of $b^2$ in eq. $(2)$, we get,

$\Rightarrow a^2 +12a -64 = 0$

Hence $a  = 4, -16$

As $a$ is always taken as positive value so $a =4$ 

from eq. $(3)$,  $b = \sqrt{48}$

Hence equation of hyperbola is $\dfrac{x^2}{16} -\dfrac {y^2}{48} = 1$

Or $3x^2 -y^2 = 48$, So correct option is $A$.