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 general knowledge math & puzzles

x² /a² - y²/b² = 1 is the equation of a hyperbola. When |x| is very high the graph for the above equation resembles –

  1. cycloid

  2. parabola

  3. ellipse

  4. straight line
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For the hyperbola x²/a² - y²/b² = 1, as |x| approaches infinity, the curve approaches its asymptotes y = ±(b/a)x, which are straight lines passing through the origin. This asymptotic behavior is fundamental to hyperbola geometry.

Multiple choice general knowledge math & puzzles

First derivative of a parabola is a -

  1. straight line inclined at 45 deg to x axis

  2. straight line parallel to y axis

  3. straight line parallel to x axis

  4. straight line
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

A parabola is defined by a quadratic equation (y = ax^2 + bx + c). Its first derivative is a linear equation (y' = 2ax + b). A linear equation graphically represents a straight line.

Multiple choice general knowledge math & puzzles

A parabola has the origin as its focus and the line x = 2 as the directrix. Then the vertex of the parabola is at

  1. (0,2)

  2. (0,1)

  3. (1,0)

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

For a parabola with focus at (0,0) and directrix x=2, every point on the parabola is equidistant from the focus and directrix. The vertex is the midpoint along the axis of symmetry (the x-axis, since focus and directrix are horizontal). Distance from focus to directrix is 2 units, so the vertex is at (1,0), exactly halfway between x=0 and x=2.

Multiple choice

For a scalar function f(x, y, z) = x2 + 3y2 + 2z2, the gradient at the point P (1, 2, - 1) is
  1. $2i + 6 \bar{j} + 4 \bar{k}$
  2. $2i + 12 \bar{j} - 4 \bar{k}$
  3. $2i + 12 \bar{j} + 4 \bar{k}$
  4. $\sqrt{56}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The gradient of a scalar function f(x, y, z) is the vector (df/dx, df/dy, df/dz). Calculating these partial derivatives gives 2x, 6y, and 4z. Evaluating at (1, 2, -1) yields 2(1)i + 6(2)j + 4(-1)k, which is 2i + 12j - 4k.

Multiple choice maths functions and graphs different forms of equation of a line

A line passing through the points of intersection of $x+y=4$ and $x-y=2$ makes an angle $\tan^{-1}(3/4)$ with the x-axis. It intersects the parabola $y^2=4(x-3)$ at points $(x _1, y _1)$ and $(x _2, y _2)$ respectively. Then $|x _1-x _2|$ is equal to?

  1. $\dfrac{16}{9}$
  2. $\dfrac{32}{9}$
  3. $\dfrac{40}{9}$
  4. $\dfrac{80}{9}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Solving x+y=4 and x-y=2 gives the intersection point (3,1). A line through this point with slope 3/4 has equation: y-1 = (3/4)(x-3), or 4y-4 = 3x-9, or 3x-4y=5. Substituting into y²=4(x-3) gives a quadratic. The x-coordinates of the intersection points are found by solving for the parameter along the line. The difference |x₁ - x₂| = 16/9.

Multiple choice intersection of a line and a parabola conic section maths

The length of the chord of the parabola $y^2 = 4x$ which passes through the vertex and makes $30^o$ angle with x-axis is

  1. $\dfrac{\sqrt{3}}{2}$
  2. $\dfrac{3}{2}$
  3. $8\sqrt{3}$
  4. $\sqrt{3}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The chord passes through the vertex (0,0) and makes 30 degrees with the x-axis, so its equation is y = tan(30)x = x/sqrt(3). Substituting into y^2 = 4x gives x^2/3 = 4x, so x = 12. The length is sqrt(x^2 + y^2) = sqrt(144 + 144/3) = sqrt(144 * 4/3) = 12 * 2/sqrt(3) = 8*sqrt(3).

Multiple choice intersection of a line and a parabola conic section maths

The length of normal chord to the parabola $y^{2} = 4x$ which subtends a right angle at the vertex is

  1. $6\sqrt {3}$
  2. $6\sqrt {2}$
  3. $7\sqrt {2}$
  4. $7\sqrt {3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A chord is a line segment that passes through any two points on the parabola. A normal chord is a chord that is perpendicular to a tangent of the parabola at the point of intersection of the chord with the parabola. 


Let $y=mx+c$   is the tangent of parabola.


As given parabola is $y^2=4x$

So any point on the parabola is $(t^2,2t)$

As $y=mx+\dfrac{a}{m}$ , is the tangent to the parabola $y^2=4ax$

then the tangent equation to this given parabola is $y=mx+\dfrac{1}{m}$

Let the tangent passes through point $P(t _1^{2},2t _1)$

On substituting P in parabola equation we get $m=\dfrac{1}{t _1}$

So the slope of the normal is $-t _1$

Let the chord joins $P(t _1^2,2t _1)$ and $Q(t _2^2,2t _2)$ 

On solving we will get slope of line PQ as $\dfrac{2}{t _1+t _2}$

So , $\dfrac{2}{{t _1}+{t _2}}= \dfrac{1}{t _1}$

$\Rightarrow t _1^{2}+{t _1}{t _2}=-2$

As from properties of a normal chord which subtends a right angle at the vertex, ${t _1}{t _2}=-4$

On solving above two equations we get $ t _1=\sqrt{2} , t _2=-2\sqrt{2} $

Hence the points are $P(2,2\sqrt{2})$ and $Q(8,-4\sqrt{2}) $

By applying distance formula we get the distance between P and Q as

$\Rightarrow PQ=\sqrt{(8-2)^2+(-4\sqrt{2}-2\sqrt{2})^2}$

$\Rightarrow PQ=\sqrt{108}=6\sqrt{3} units $

Multiple choice intersection of a line and a parabola conic section maths

The number of focal chord(s) of length $\dfrac{4}{7}$ in the parabola $7y^2 = 8x$ is

  1. $1$
  2. $0$

  3. infinite

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

Parabola: ${ y }^{ 2 }=4(\cfrac { 2 }{ 7 } )x\quad ...........(1)\quad (a=\cfrac { 2 }{ 7 } )$

Let there be a point $P(a{ t }^2,2at)$ on parabola lying on focal chord so it will intersect parabola at $Q(\cfrac { a }{ t }^2,\cfrac { -2a }{ t })$(as ${ t } _{ 1 }{ t } _{ 2 }=-1)$ 
So, length of $PQ:\sqrt { (\cfrac { a }{ { t }^{ 2 } } -{ { at }^{ 2 }) }^{ 2 }+4(\cfrac { a }{ t } +{ at) }^{ 2 } } $
Given $PQ=\cfrac{ 4 }{ 7 }=\cfrac{ 2 }{ 7 }\sqrt { { (\cfrac { 1 }{ { t }^{ 2 } } -{ t }^{ 2 }) }^{ 2 }+4({ \cfrac { 1 }{ t } +t) }^{ 2 } } $
$=>4=(\cfrac { 1 }{ { t }^{ 2 } } -{ { t }^{ 2 }) }^{ 2 }+4(\cfrac { 1 }{ t } +{ t })^{ 2 }$
$=>4=[(\cfrac { 1 }{ t } +t)(\cfrac { 1 }{ t } -{ t)] }^{ 2 }+4(\cfrac { 1 }{ t } +{ t })^{ 2 }$
$=>4={ (\cfrac { 1 }{ t } +t) }^{ 2 }[({ \cfrac { 1 }{ t } -2) }^{ 2 }+4]$
$=>4=({ \cfrac { 1 }{ t } +t) }^{ 2 }(\cfrac { 1 }{ { t }^{ 2 } } +{ t }^{ 2 }+4-2)$
$=>4=({ \cfrac { 1 }{ t } +t) }^{ 2 }({ \cfrac { 1 }{ t } +t) }^{ 2 }$
$=>4=(\cfrac { 1 }{ t } +{ t) }^{ 4 }$
$=>0=(\cfrac { 1 }{ t } +{ t })^{ 2 }(\cfrac { 1 }{ t } +{ t })^{ 2 }-({ 2 })^{ 2 }$
$=>[(\cfrac { 1 }{ t } +{ t })^{ 2 }+2][(\cfrac { 1 }{ t } +{ t })^{ 2 }-2]=0$
We know, $[(\cfrac { 1 }{ t } +{ t })^{ 2 }+2]$ can never be zero so $[(\cfrac { 1 }{ t } +{ t })^{ 2 }-2]=0$
$=>(\cfrac{1}{t}+t-\sqrt2)(\cfrac{1}{t}+t+\sqrt2)=0$
$=>({t}^2-\sqrt2t+1)(t^2+\sqrt2t+1)=0$
$=> $either$ I:(t^2-\sqrt2t+1)=0\quad $or$\quad II:(t^2+\sqrt2t+1)=0$
$D _1=(-\sqrt2)^2-4(1)=-2 <0, $ No solution.
$D _2=(\sqrt2)^2-4(1)=-2 <0,$ No solution.
No such focal chord is possible.

Multiple choice intersection of a line and a parabola conic section maths

Find the length of the chord of the parabola $y^2\, =\, 8x$, whose equation is $x + y = 1$.

  1. $8 \sqrt{3}$
  2. $4 \sqrt {3}$
  3. $2 \sqrt {3}$
  4. $\sqrt {3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Substitute $x=1-y$ in $y^2=8x$

Then $y^2=8(1-y)$
$y=\dfrac{8\pm\sqrt{96}}{2}=4\pm 2\sqrt6$
Therefore, $x=3 \mp 2\sqrt6$
Using distance formula length of chord $=8\sqrt3$

Multiple choice intersection of a line and a parabola conic section maths

The length of the chord of the parabola $x^2 = 4y $ passing through the vertex and having slope $cot \alpha $ is 

  1. $4 \cos \alpha . cosec^2\alpha$
  2. $ 4a \tan \alpha \sec \alpha $
  3. $4 \sin \alpha . \sec^2 \alpha $

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Equation of parabola $x^2=4ay$
Vertex of parabola $(0)=(0, 0)$
Equation of chord passing through vertex $y-y _1=m(x-x _1)$

Here, $x _1=0$ and $y _1=0$
$y-0=m(x-0)$
$\therefore y=mx$

Now, slope$=m=\tan\alpha$
Let the chord intersect the parabola at $P(h, k)$
Since, point P lies on chord
$k=(\tan\alpha)h$ ………..$(1)$

Point P lies on parabola
$h^2=4ak$ ………$(2)$

Substituting equation $(1)$ in $(2)$
$\therefore h^2=4a(\tan\alpha \times h)$
$\therefore h^2(h-4a\tan\alpha)=0$
$\therefore h=0$
$h=4a\tan\alpha$

Substituting $h=4a\tan\alpha$ in equation $(2)$, we get
$\therefore (4a)^2\tan^2\alpha=4ak$
$\therefore k=4a\tan^2\alpha$
$\therefore p=(h, k)=(4a\tan\alpha, 4a\tan^2\alpha)$

Distance $OP=\sqrt{(4a\tan\alpha -0)^2+(4a\tan^2\alpha -0)}$
$=\sqrt{16a^2\tan^2\alpha +16a^2\tan^4\alpha}$
$=4a\tan\alpha\sqrt{1+\tan^2\alpha}$
$=4a\tan\alpha\sqrt{\sec^2\alpha}$
$=4a\tan\alpha \sec\alpha$.
Multiple choice intersection of a line and a parabola conic section maths

Let $AB$ be a chord of the parabola $y^{2}=4ax$.If the pole of $AB$ with respect to the parabola be $\left ( 2a,3a \right )$ then the length of $AB$ is 

  1. $\sqrt{13}a$
  2. $4a$
  3. $5a$
  4. $2\sqrt{3}a$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given pole of AB w.r.t the parabola $y^2=4ax$ is $(2a,3a)$
Thus the equation of chord $AB$ is given by $T=0$
$\Rightarrow y(2a)-2a(x+3a)=0$
$\Rightarrow y =\dfrac{2}{3}x+\dfrac{4}{3}a$
Clearly here $m = \dfrac{2}{3}, c = \dfrac{4}{3}a$
Therefore length of chord AB is $=\cfrac{4}{m^2}\sqrt{a(1+m^2)(a-cm)}=\sqrt{13}a$