Home / Mathematics / Class XI / Conic Sections
Conic Sections — CUET Mathematics hero
Class XI 📐 Mathematics Ch 10 of 14

Conic Sections

CUET unit: Conic Sections

📌 Snapshot

  • Circle, parabola, ellipse and hyperbola: standard equations, foci, eccentricity and latus rectum for CUET.
  • How slicing a cone with a plane produces all four curves, and what happens when the plane hits the vertex.
  • The sign rules, the two c squared relations and the focus placement rules that decide most conic questions.
  • Six worked examples with common wrong steps, plus a quick table for naming a conic from its equation.

📖 Detailed Notes

Written and reviewed by Devendra Dubey, Mathematics faculty at UniDrill (M.Sc. Mathematics, Osmania University · 10 years teaching Mathematics). Based on NCERT Mathematics Class XI, Chapter 10.

What shape does a torch make on the wall?

Hold a torch straight at a wall and the lit patch is a circle. Tilt it and the patch stretches into an oval. Tilt it further, and at one exact angle the oval springs open into a U-shaped curve that never closes. Tilt it more and the patch becomes the open curve of a hyperbola. The beam of a torch is a cone, the wall is a plane, and all four curves are what you get when a plane cuts a cone. These are the conic sections, and this chapter gives you the equation and the key numbers of each.

Slicing a cone

Rotate a line about a vertical axis, keeping a fixed angle $\alpha$ (the semi-vertical angle) between the line and the axis. This traces a double-napped cone: two cones joined at the vertex. Cut it with a plane that makes angle $\beta$ with the axis, not passing through the vertex.

Cutting plane Condition Curve
Perpendicular to the axis $\beta = 90^\circ$ Circle
Tilted, cuts one nappe all the way round $\alpha < \beta < 90^\circ$ Ellipse
Parallel to the slant side of the cone $\beta = \alpha$ Parabola
Steep, cuts both nappes $0 \le \beta < \alpha$ Hyperbola

If the plane passes through the vertex, the curves collapse. These are the degenerate conics.

Plane through the vertex Condition Result
Steeper than the slant side $\alpha < \beta \le 90^\circ$ A single point
Along the slant side $\beta = \alpha$ A straight line (a degenerate parabola)
Flatter than the slant side $0 \le \beta < \alpha$ A pair of intersecting straight lines (a degenerate hyperbola)

Circle: all points at one distance from a centre

A circle with centre $(h,k)$ and radius $r$ is the set of points at distance $r$ from the centre. Writing the distance formula and squaring:

$$(x-h)^2 + (y-k)^2 = r^2$$

With the centre at the origin this becomes $x^2 + y^2 = r^2$. An equation with equal coefficients of $x^2$ and $y^2$ and no $xy$ term is a circle, and you find its centre and radius by completing the square.

Worked example: centre and radius. Find the centre and radius of $x^2 + y^2 + 8x + 10y - 8 = 0$.

Step What we do Result
1 Group $x$ terms and $y$ terms $(x^2 + 8x) + (y^2 + 10y) = 8$
2 Add $\left(\tfrac82\right)^2 = 16$ and $\left(\tfrac{10}{2}\right)^2 = 25$ to both sides $(x+4)^2 + (y+5)^2 = 8 + 16 + 25$
3 Compare with $(x-h)^2+(y-k)^2=r^2$ $(x+4)^2 + (y+5)^2 = 49$
4 Read off Centre $(-4,-5)$, radius $7$

Common wrong step: reading the centre as $(4,5)$, or the radius as $\sqrt 8$. The centre has the opposite signs of the numbers inside the brackets, and the radius is the square root of the right-hand side after completing the squares.

Parabola: equidistant from a point and a line

A parabola is the set of points equidistant from a fixed point (the focus) and a fixed line (the directrix) that does not contain the focus. The line through the focus perpendicular to the directrix is the axis; the parabola meets the axis at its vertex.

Why the equation is $y^2 = 4ax$. Put the vertex at the origin, the focus at $(a,0)$ and the directrix at $x = -a$. A point $(x,y)$ is on the parabola when its distance to the focus equals its distance to the directrix:

$$\sqrt{(x-a)^2 + y^2} = x + a \;\Longrightarrow\; (x-a)^2 + y^2 = (x+a)^2 \;\Longrightarrow\; y^2 = 4ax$$

Flip signs or swap $x$ and $y$ and you get the other three standard forms. The squared variable tells you the axis, and the sign of the other term tells you the direction.

Equation Axis Opens Focus Directrix Length of latus rectum
$y^2 = 4ax$ x-axis Right $(a,0)$ $x=-a$ $4a$
$y^2 = -4ax$ x-axis Left $(-a,0)$ $x=a$ $4a$
$x^2 = 4ay$ y-axis Up $(0,a)$ $y=-a$ $4a$
$x^2 = -4ay$ y-axis Down $(0,-a)$ $y=a$ $4a$

The latus rectum is the chord through the focus perpendicular to the axis. For $y^2 = 4ax$, put $x=a$: $y^2 = 4a^2$, so $y=\pm 2a$ and the chord has length $4a$.

Worked example: reading a parabola. Find the focus, directrix and latus rectum of $x^2 = -12y$.

Step What we do Result
1 $x$ is squared, so the axis is the y-axis; the right side is negative, so it opens downward Form $x^2 = -4ay$
2 Compare $4a$ with 12 $a = 3$
3 Focus $(0,-a) = (0,-3)$
4 Directrix $y = a$, that is $y=3$
5 Latus rectum $4a = 12$

Common wrong step: taking $a = 12$. The coefficient is $4a$, so $a$ is one quarter of it.

Worked example: building a parabola. Find the equation of the parabola with vertex at the origin and directrix $y = 2$.

Step What we do Result
1 The directrix is above the vertex, so the focus is below it and the parabola opens downward Form $x^2 = -4ay$
2 Distance from vertex to directrix is $a$ $a = 2$
3 Substitute $x^2 = -8y$

Common wrong step: writing $x^2 = 8y$. The curve always opens away from the directrix, so a directrix above the vertex means a downward parabola.

Ellipse: constant sum of distances

An ellipse is the set of points whose distances from two fixed points (the foci) add up to a constant $2a$, which must be larger than the distance $2c$ between the foci. A string pinned at two points and pulled tight by a pencil draws one.

The line through the foci is the major axis (length $2a$); the perpendicular bisector is the minor axis (length $2b$). The numbers $a$, $b$, $c$ are linked by a right triangle. Take the end $B$ of the minor axis. It is the same distance from both foci, and those two distances add to $2a$, so each equals $a$. The triangle with sides $b$, $c$ and hypotenuse $a$ then gives

$$c^2 = a^2 - b^2, \qquad e = \dfrac{c}{a} < 1$$

The eccentricity $e$ measures how stretched the ellipse is: near 0 is almost a circle, near 1 is very flat. It is below 1 because $c<a$.

Standard form (with $a>b$) Major axis Foci Vertices
$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ Along the x-axis $(\pm c, 0)$ $(\pm a, 0)$
$\dfrac{x^2}{b^2} + \dfrac{y^2}{a^2} = 1$ Along the y-axis $(0, \pm c)$ $(0, \pm a)$

The rule: the foci lie on the axis under the larger denominator. The latus rectum is the focal chord perpendicular to the major axis. Putting $x=c$ in the first form gives $y^2 = b^2\left(1 - \dfrac{c^2}{a^2}\right) = \dfrac{b^4}{a^2}$, so each half is $\dfrac{b^2}{a}$ and

$$\text{length of latus rectum} = \dfrac{2b^2}{a}$$

Worked example: a tilted-looking ellipse. For $9x^2 + 4y^2 = 36$, find the foci, eccentricity and latus rectum.

Step What we do Result
1 Divide by 36 to get 1 on the right $\dfrac{x^2}{4} + \dfrac{y^2}{9} = 1$
2 The larger denominator is under $y^2$, so the major axis is along the y-axis $a^2 = 9,\ b^2 = 4$
3 Find $c$ $c = \sqrt{9-4} = \sqrt5$
4 Foci $(0,\pm\sqrt5)$
5 Eccentricity $e = \dfrac{\sqrt5}{3}$
6 Latus rectum $\dfrac{2b^2}{a} = \dfrac{2\cdot4}{3} = \dfrac83$

Common wrong step: taking $a^2 = 4$ because $x^2/4$ comes first. The letter $a$ always belongs to the larger denominator, wherever it sits.

Pause and predict: on which axis are the foci of $\dfrac{x^2}{16} + \dfrac{y^2}{25} = 1$, and how far from the centre?

Check your answer: 25 is the larger denominator and sits under $y^2$, so the foci are on the y-axis. Here $c = \sqrt{25-16} = 3$, so the foci are $(0,\pm3)$.

Pause and predict: can an ellipse have eccentricity exactly 1? What about exactly 0?

Check your answer: no to 1, and yes to 0. If $e=1$ then $c=a$, which makes $b=0$: the ellipse flattens into a line segment between the foci. If $e=0$ then $c=0$, the foci coincide at the centre and $a=b$, which is a circle.

Hyperbola: constant difference of distances

A hyperbola is the set of points whose distances from two fixed foci differ by a constant $2a$. The line through the foci is the transverse axis (length $2a$), the perpendicular through the centre is the conjugate axis (length $2b$), and the curve cuts the transverse axis at its two vertices.

The key relation looks like the ellipse's but with the opposite sign:

$$c^2 = a^2 + b^2, \qquad e = \dfrac{c}{a} > 1$$

Why $e>1$? Two sides of a triangle always differ by less than the third, so the constant difference $2a$ is less than the focal distance $2c$, which gives $c>a$.

Standard form Transverse axis Foci Vertices
$\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ Along the x-axis $(\pm c, 0)$ $(\pm a, 0)$
$\dfrac{y^2}{a^2} - \dfrac{x^2}{b^2} = 1$ Along the y-axis $(0, \pm c)$ $(0, \pm a)$

The rule: the foci lie on the axis of the positive term. The latus rectum has the same formula as for the ellipse, $\dfrac{2b^2}{a}$ (put $x=c$ in the first form and $y^2 = \dfrac{b^4}{a^2}$ again). For $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ the lines $y = \pm\dfrac{b}{a}x$ are the asymptotes, the lines the branches approach but never meet. They are the diagonals of the rectangle of sides $2a$ and $2b$ drawn about the centre. A hyperbola with $a = b$ is equilateral and has $e = \sqrt2$.

Worked example: reading a hyperbola. Find the foci, eccentricity, latus rectum and asymptotes of $\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$.

Step What we do Result
1 The positive term is the $x^2$ term $a^2 = 9,\ b^2 = 16$
2 Use the hyperbola relation $c = \sqrt{9+16} = 5$
3 Foci $(\pm5, 0)$
4 Eccentricity $e = \dfrac53$
5 Latus rectum $\dfrac{2b^2}{a} = \dfrac{32}{3}$
6 Asymptotes $y = \pm\dfrac43 x$

Common wrong step: using the ellipse relation $c^2 = a^2 - b^2 = 9 - 16 = -7$. A negative value of $c^2$ is impossible and signals the wrong formula.

Worked example: finding the equation. A hyperbola has foci $(0,\pm13)$ and a conjugate axis of length 24. Find its equation.

Step What we do Result
1 Foci are on the y-axis, so the transverse axis is vertical Form $\dfrac{y^2}{a^2} - \dfrac{x^2}{b^2} = 1$
2 Read the data $c = 13,\ 2b = 24 \Rightarrow b = 12$
3 Use $c^2 = a^2 + b^2$ $a^2 = 169 - 144 = 25$
4 Write the equation $\dfrac{y^2}{25} - \dfrac{x^2}{144} = 1$

Common wrong step: computing $a^2 = c^2 + b^2$. For a hyperbola, $c$ is the largest of the three and $a^2 = c^2 - b^2$.

All four conics side by side

Circle Parabola Ellipse Hyperbola
Defined by Constant distance from a centre Equal distance from focus and directrix Constant sum of distances to two foci Constant difference of distances to two foci
Standard equation $x^2+y^2=r^2$ $y^2=4ax$ $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$
Key relation Radius $r$ $a$ = vertex to focus $c^2=a^2-b^2$ $c^2=a^2+b^2$
Eccentricity $0$ $1$ $0<e<1$ $e>1$
Latus rectum none $4a$ $\dfrac{2b^2}{a}$ $\dfrac{2b^2}{a}$

A quick way to name a conic from its equation: look at the coefficients of $x^2$ and $y^2$.

Equation Observation Conic
$x^2 + y^2 = 16$ Equal, same sign Circle
$y^2 = -6x$ Only one squared term Parabola, opens left
$4x^2 + 9y^2 = 36$ Same sign, unequal Ellipse
$9x^2 - 4y^2 = 36$ Opposite signs Hyperbola
$x^2 - y^2 = 9$ Opposite signs, equal size Equilateral hyperbola

Beyond the textbook. For the exam: each conic has its own definition and its own formulas. In reality: there is one definition that covers the three non-circular conics. A conic is the set of points whose distance from a fixed point (the focus) is a constant multiple $e$ of its distance from a fixed line (the directrix). The value of $e$ alone decides the shape: less than 1 is an ellipse, exactly 1 is a parabola, more than 1 is a hyperbola. This is why the eccentricity column in the table above runs smoothly from 0 to 1 to beyond.

Beyond the textbook. For the exam: learn the formulas and the focus rules. In reality: the planets move around the Sun along ellipses with the Sun at one focus. The Earth's orbit has $e \approx 0.017$, so it is nearly a circle, while Halley's comet has $e \approx 0.97$, an extremely stretched ellipse. A body that passes the Sun once at high speed and never returns follows a hyperbola. A ball thrown in air, when air resistance is ignored, follows a parabola.

Beyond the textbook. For the exam: the focus of a parabola is a point on the axis at distance $a$ from the vertex. In reality: the focus is where a parabolic mirror collects all rays that arrive parallel to its axis. That is why satellite dishes and solar cookers place the receiver or pot at the focus, and why a headlamp with the bulb at the focus sends out a parallel beam. In an ellipse, a sound or light ray leaving one focus passes through the other focus, which is the principle of whispering galleries.

Exam traps

The trap Why it's tempting The truth
Using $c^2 = a^2 - b^2$ for a hyperbola The ellipse formula is learned first Hyperbola: $c^2 = a^2 + b^2$. Ellipse: $c^2 = a^2 - b^2$
Ellipse foci always on the x-axis $x^2$ is written first Foci are under the larger denominator, so $\dfrac{x^2}{4}+\dfrac{y^2}{9}=1$ has foci on the y-axis
Hyperbola foci under the larger denominator Same rule as the ellipse Foci are on the axis of the positive term
$a = 8$ for $y^2 = 8x$ The number is right there $4a = 8$, so $a = 2$ and the focus is $(2,0)$
Latus rectum $4a$ for every conic Parabola formula learned first Parabola: $4a$. Ellipse and hyperbola: $\dfrac{2b^2}{a}$
Major axis has length $a$ The letter looks like the length Major axis is $2a$, minor axis $2b$, transverse axis $2a$, conjugate axis $2b$
Eccentricity of an ellipse can be 1 $e=1$ is the boundary value Ellipse $e<1$. Parabola is the case $e=1$. Circle has $e=0$
$y^2=-4ax$ opens right Positive-looking $x$ The negative sign sends it left; focus $(-a,0)$

Formula sheet

Result Formula When to use
Circle $(x-h)^2+(y-k)^2=r^2$ Centre and radius; complete the square from the general form
Parabola, right $y^2=4ax$; focus $(a,0)$; directrix $x=-a$ Axis along the x-axis, opens right
Parabola, left $y^2=-4ax$; focus $(-a,0)$; directrix $x=a$ Opens left
Parabola, up and down $x^2=\pm4ay$; focus $(0,\pm a)$; directrix $y=\mp a$ Axis along the y-axis
Latus rectum, parabola $4a$ Focal chord perpendicular to the axis
Ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1,\ c^2=a^2-b^2,\ e=\dfrac ca$ Foci under the larger denominator
Hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1,\ c^2=a^2+b^2,\ e=\dfrac ca$ Foci on the positive term's axis
Latus rectum, ellipse and hyperbola $\dfrac{2b^2}{a}$ Both curves
Asymptotes $y=\pm\dfrac ba x$ Hyperbola with transverse axis along x
Equilateral hyperbola $a=b,\ e=\sqrt2$ Special case
Eccentricity ranges Circle 0, ellipse $<1$, parabola 1, hyperbola $>1$ Classifying a conic

60-second revision

Topic Lock this in
Slicing $\beta=90^\circ$ circle, $\alpha<\beta<90^\circ$ ellipse, $\beta=\alpha$ parabola, $\beta<\alpha$ hyperbola
Degenerate Point, line, pair of intersecting lines
Circle $(x-h)^2+(y-k)^2=r^2$; complete the square
Parabola $y^2=4ax$ right; sign and squared variable give direction; latus rectum $4a$
Ellipse Sum of distances; $c^2=a^2-b^2$; foci on larger denominator; $e<1$
Hyperbola Difference of distances; $c^2=a^2+b^2$; foci on positive term; $e>1$
Latus rectum $4a$ for the parabola, $\dfrac{2b^2}{a}$ for ellipse and hyperbola
Axes Major $2a$, minor $2b$, transverse $2a$, conjugate $2b$
Equilateral $a=b$, $e=\sqrt2$
Identify Equal coefficients: circle. One squared: parabola. Same sign unequal: ellipse. Opposite signs: hyperbola

Where this chapter connects

🎯 Practice MCQs

First 3 questions free · create a free account to unlock the rest — answers & explanations included, no payment needed

Q1. A plane passes through the vertex of a double napped right circular cone and makes an angle $\beta = \alpha$ with the axis, where $\alpha$ is the semi vertical angle. The section obtained is

▸ Show answer & explanation

Answer: B

For a plane through the vertex, $\beta = \alpha$ gives a straight line, which is the degenerate parabola. A point comes from $\alpha < \beta \le 90^\circ$ and a pair of intersecting lines from $0 \le \beta < \alpha$. A parabola needs a plane that does not pass through the vertex.

Q2. The centre and radius of the circle $x^2 + y^2 + 8x + 10y - 8 = 0$ are

▸ Show answer & explanation

Answer: C

Completing the squares gives $(x+4)^2 + (y+5)^2 = 8 + 16 + 25 = 49$, so the centre is $(-4,-5)$ and the radius is $7$. Option (B) forgets to add the 16 and 25.

Q3. The focus and the length of the latus rectum of the parabola $x^2 = -12y$ are

▸ Show answer & explanation

Answer: A

The curve has the form $x^2 = -4ay$ with $4a = 12$, so $a=3$. It opens downward with focus $(0,-3)$, and the latus rectum is $4a = 12$.

🔒 9 more practice MCQs

Create a free account to unlock every MCQ in this chapter — answers and explanations included. No payment needed.

Already registered? Just log in and they'll all appear here.

📊 Previous-Year Questions

Practise with real CUET Mathematics previous-year papers — every question solved, with the correct answer and a step-by-step explanation.

View solved CUET PYQ papers →

Ready to drill Mathematics?

Unlock all MCQs, chapter tests, mocks & PYQs for ₹199/year.

Get UniDrill Pro