Chapter 10 of 14

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Karnataka (KSEEB)Class 11 Mathematics★ Unit III · part of the 12-mark Coordinate Geometry block shared with Straight Lines — CBSE states no chapter-wise split

Conic Sections

Slice a cone at different angles and the same surface gives you a circle, an ellipse, a parabola, or a hyperbola. This chapter derives the standard equation of each from its own defining distance property, covering circles, parabolas, ellipses, and hyperbolas in full — with no formative-only carve-out this time, unlike Straight Lines next door. CBSE Class 11 Mathematics Chapter 10.

⏱ 9 min readMedium difficulty✓ Free · NCERT-aligned

Worksheet

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Mind Map

The chapter at a glance

Every concept and exam-favourite branch — revisable in 5 minutes.

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Study Guide

Night-before notes

Key points, glossary and the questions most likely to show up.

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By the end of this chapter you'll be able to…

1Identify which conic section results from a given angle of intersection between a plane and a cone, including the three degenerate cases (point, line, pair of intersecting lines)
2Write and derive the standard equation of a circle, and recover centre and radius from an expanded equation by completing the square
3Write and derive the standard equations of a parabola in all four orientations, and find its focus, directrix, axis, and latus rectum
4Write and derive the standard equations of an ellipse and a hyperbola, and find foci, vertices, eccentricity, and latus rectum from either the equation or from given conditions
5Determine which axis is major (for an ellipse) or transverse (for a hyperbola) by comparing denominators or signs, not by assumption
6Construct the equation of any of the four conics from partial information such as focus and directrix, vertices and foci, or points the curve passes through

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Why this chapter matters

This chapter derives the standard equation of every curve you get by slicing a cone: circle, parabola, ellipse, hyperbola, plus their degenerate cases. Unlike its neighbouring chapters, nothing here is formative-only — the entire syllabus line is examinable. It also sets up Class 12's applications of conics (area under curves, tangents and normals) and is a heavily tested JEE topic in its own right.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Class 10 coordinate geometry: the distance formula, reused directly in every derivation in this chapter to turn a geometric distance condition into an algebraic equation

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Completing the square, needed to recover a circle's centre and radius from its expanded form, and used again to convert non-standard-looking equations into recognisable conic forms

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Straight Lines (Class 11, Chapter 9): the general equation of a line Ax+By+C=0, used whenever a conic's centre or vertex is constrained to lie on a given line

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Comfort solving simultaneous equations, needed throughout 'find the equation given two points or two conditions' problems

Conic Sections

1. Check this before you revise anything

Everything in this chapter's syllabus line is fully examinable — there is no formative-only carve-out here. CBSE's formative-only block lists specific dropped topics for Straight Lines (Normal Form, general equation of a line) and Introduction to Three-Dimensional Geometry (section formula) under Unit III, but Conic Sections has no entry there at all.

Every phrase in its syllabus line — sections of a cone including the degenerate cases, and the standard equations and simple properties of the circle, parabola, ellipse, and hyperbola — is summative.

Asymptotes are not part of the current book. Coaching material commonly teaches the hyperbola's asymptotes ( $y = \pm \frac{b}{a} x$ ) alongside its standard equation. The word "asymptote" does not appear anywhere in this book, and none of Exercise 10.4's 15 questions ask for one.

The book's own figure of a hyperbola does sketch the diagonal lines the branches approach, but never names or derives them. This chapter follows the book and does not teach them as a core topic — a brief flag appears in the JEE section since they show up often at that level.

"General equation of a circle" ( $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ , centre $(- g, - f)$ , radius $g^{2} + f^{2} - c$ ) is not how the book teaches expanded-form circles either. The book's own Example 3 recovers the centre and radius of $x^{2} + y^{2} + 8 x + 10 y - 8 = 0$ by completing the square directly, never introducing $g$ , $f$ , $c$ as named constants.

Since Exercise 10.1's own Q6–9 require exactly this skill, this chapter teaches completing the square as the primary method — matching the book — and mentions the $g, f, c$ shortcut only as a quick equivalent, not a formula to memorise in place of the technique.

2. What this chapter covers

Textbook section	Topic
10.2	Sections of a cone: circle, ellipse, parabola, hyperbola, and their degenerate cases (a point, a line, a pair of intersecting lines)
10.3	Circle: definition, standard equation, completing the square
10.4	Parabola: definition, four standard equations, latus rectum
10.5	Ellipse: definition, standard equation, eccentricity, latus rectum
10.6	Hyperbola: definition, standard equation, eccentricity, latus rectum

3. How the four curves come from one cone

Take a fixed vertical line $l$ and another line $m$ crossing it at a point $V$ , held at a constant angle $α$ to $l$ . Rotating $m$ around $l$ sweeps out a double-napped cone with vertex $V$ ; $m$ in any position is a generator, and $V$ splits the cone into two nappes.

Slice this cone with a plane at angle $β$ to the cone's axis. When the plane misses the vertex and cuts across one nappe:

Angle $β$	Curve
$β = 90°$	Circle
$α < β < 90°$	Ellipse
$β = α$	Parabola
$0 \leq β < α$ (cuts both nappes)	Hyperbola

When the plane passes through the vertex instead, the section degenerates: $α < β \leq 90°$ gives a single point; $β = α$ gives a straight line (the degenerate parabola); $0 \leq β < α$ gives a pair of intersecting straight lines (the degenerate hyperbola). These three degenerate cases are exactly what the syllabus names alongside the four curves themselves.

4. Circle

Definition. The set of all points in a plane equidistant from a fixed point (the centre); that fixed distance is the radius.

Standard equation. For centre $C (h, k)$ and radius $r$ , any point $P (x, y)$ on the circle satisfies $∣ C P ∣ = r$ . By the distance formula:

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

Centred at the origin, this reduces to $x^{2} + y^{2} = r^{2}$ .

Worked, mirroring the textbook's own Example 3 — recovering centre and radius from an expanded equation. Find the centre and radius of $x^{2} + y^{2} + 8 x + 10 y - 8 = 0$ . Group and complete the square on each variable: $(x^{2} + 8 x + 16) + (y^{2} + 10 y + 25) = 8 + 16 + 25$ , i.e. $(x + 4)^{2} + (y + 5)^{2} = 49 = 7^{2}$ . Reading this against $(x - h)^{2} + (y - k)^{2} = r^{2}$ gives centre $(- 4, - 5)$ and radius $7$ .

The same technique, run in general on $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ , always completes to $(x + g)^{2} + (y + f)^{2} = g^{2} + f^{2} - c$ — so centre $(- g, - f)$ and radius $g^{2} + f^{2} - c$ is a fast shortcut for the identical result, valid whenever $g^{2} + f^{2} - c > 0$ .

5. Parabola

Definition. The set of all points in a plane equidistant from a fixed line (the directrix) and a fixed point not on that line (the focus). The line through the focus perpendicular to the directrix is the axis; where the parabola meets its axis is the vertex.

Deriving the standard equation, mirroring the book's own derivation. Take the vertex at the origin, focus at $(a, 0)$ with $a > 0$ , and directrix $x = - a$ . For $P (x, y)$ on the parabola, $P F = P B$ where $B = (- a, y)$ is the foot of the perpendicular to the directrix. By the distance formula, $(x - a)^{2} + y^{2} = (x + a)^{2}$ ; squaring both sides and simplifying:

$y^{2} = 4 a x$

The other three orientations follow the same way:

Vertex-origin form	Focus	Directrix	Opens
$y^{2} = 4 a x$	$(a, 0)$	$x = - a$	Right
$y^{2} = - 4 a x$	$(- a, 0)$	$x = a$	Left
$x^{2} = 4 a y$	$(0, a)$	$y = - a$	Up
$x^{2} = - 4 a y$	$(0, - a)$	$y = a$	Down

A $y^{2}$ term means the axis of symmetry is the x-axis; an $x^{2}$ term means it's the y-axis. The sign of the linear term then fixes the direction it opens.

Latus rectum: the chord through the focus, perpendicular to the axis, with both endpoints on the curve. For $y^{2} = 4 a x$ , the book's own argument (using the definition directly on the latus rectum's endpoints) gives length $4 a$ .

Worked, mirroring the textbook's own Example 5. Find the focus, directrix, and latus rectum of $y^{2} = 8 x$ . Comparing with $y^{2} = 4 a x$ gives $a = 2$ . Focus $(2, 0)$ , directrix $x = - 2$ , latus rectum $4 a = 8$ .

6. Ellipse

Definition. The set of all points in a plane whose distances from two fixed points (the foci) sum to a constant. For $P$ on the ellipse, $P F_{1} + P F_{2} = 2 a$ , where $2 a$ (the constant sum) is necessarily greater than the distance between the foci.

The midpoint of the foci is the centre; the segment through the foci is the major axis (length $2 a$ ); the segment through the centre perpendicular to it is the minor axis (length $2 b$ ).

Relating $a$ , $b$ , and $c$ (the centre-to-focus distance). Taking $P$ at the far end of the major axis gives $P F_{1} + P F_{2} = 2 a$ automatically. Taking $P$ at the end of the minor axis gives $P F_{1} + P F_{2} = 2 b^{2} + c^{2}$ (both distances equal by symmetry). Equating these two expressions for the same constant sum: $2 b^{2} + c^{2} = 2 a$ , so

$a^{2} = b^{2} + c^{2}, c = a^{2} - b^{2}$

Eccentricity $e = \frac{c}{a}$ , the ratio of the centre-to-focus distance to the centre-to-vertex distance; since $0 < c < a$ for an ellipse, $0 < e < 1$ .

Standard equation (centre at origin, foci on the x-axis, derived the same way as the circle and parabola — using the distance formula on the defining sum and simplifying):

$\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1 (a > b > 0, major axis along the x-axis)$

If the foci sit on the y-axis instead, the roles swap: $\frac{x ^{2}}{b ^{2}} + \frac{y ^{2}}{a ^{2}} = 1$ , still with $a > b > 0$ — here $a$ sits under $y^{2}$ , since the major axis is now vertical. Whichever denominator is larger tells you which axis is major; don't assume it's always under $x^{2}$ .

Latus rectum: perpendicular to the major axis through a focus, endpoints on the curve. Using the point $(a e, ℓ)$ on the ellipse where $ℓ$ is the half-length, substituting into the standard equation and using $e^{2} = 1 - \frac{b ^{2}}{a ^{2}}$ gives length $\frac{2 b ^{2}}{a}$ .

Worked, mirroring the textbook's own Example 9. Find the foci, vertices, eccentricity, and latus rectum of $\frac{x ^{2}}{25} + \frac{y ^{2}}{9} = 1$ . Here $25 > 9$ , so the major axis is along the x-axis, $a = 5$ , $b = 3$ . Then $c = 25 - 9 = 4$ : foci $(\pm 4, 0)$ , vertices $(\pm 5, 0)$ , eccentricity $e = 4/5$ , latus rectum $\frac{2 ( 9 )}{5} = \frac{18}{5}$ .

7. Hyperbola

Definition. The set of all points in a plane whose distances from two fixed points (the foci) have a constant difference (farther distance minus nearer distance). The midpoint of the foci is the centre; the line through the foci is the transverse axis (length $2 a$ between the two vertices where the curve meets it); the line through the centre perpendicular to it is the conjugate axis (length $2 b$ , where $b = c^{2} - a^{2}$ and $2 c$ is the distance between the foci).

Eccentricity $e = \frac{c}{a}$ ; since $c \geq a$ always for a hyperbola, $e \geq 1$ .

Standard equation (centre at origin, foci on the x-axis, derived by the same distance-formula-and-simplify method as the ellipse, using the difference instead of the sum):

$\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$

With foci on the y-axis instead: $\frac{y ^{2}}{a ^{2}} - \frac{x ^{2}}{b ^{2}} = 1$ . Whichever variable is positive tells you which axis is the transverse one.

Latus rectum has the same form as the ellipse's: $\frac{2 b ^{2}}{a}$ .

A hyperbola with $a = b$ is called an equilateral hyperbola — the one named special case the book itself calls out.

Worked, mirroring the textbook's own Example 14(i). Find the foci, vertices, eccentricity, and latus rectum of $\frac{x ^{2}}{9} - \frac{y ^{2}}{16} = 1$ . Here $a = 3$ , $b = 4$ , so $c = 9 + 16 = 5$ : foci $(\pm 5, 0)$ , vertices $(\pm 3, 0)$ , eccentricity $e = 5/3$ , latus rectum $\frac{2 ( 16 )}{3} = \frac{32}{3}$ .

Summary

A conic section is what you get slicing a double-napped cone with a plane; the angle of the cut relative to the cone's own angle determines circle, ellipse, parabola, or hyperbola. Cutting through the vertex instead gives the degenerate cases: a point, a line, or a pair of intersecting lines.
Circle: $(x - h)^{2} + (y - k)^{2} = r^{2}$ . Expanded-form equations are solved by completing the square, as the book's own Example 3 does — not by a memorised $g, f, c$ formula.
Parabola $y^{2} = 4 a x$ : focus $(a, 0)$ , directrix $x = - a$ , latus rectum $4 a$ ; the other three orientations follow by symmetry.
Ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ ( $a > b$ , major axis along x): $c = a^{2} - b^{2}$ , foci $(\pm c, 0)$ , $e = c / a \in (0, 1)$ , latus rectum $\frac{2 b ^{2}}{a}$ .
Hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ : $c = a^{2} + b^{2}$ , foci $(\pm c, 0)$ , $e = c / a \geq 1$ , latus rectum $\frac{2 b ^{2}}{a}$ . Asymptotes are not part of the current book.
Unlike Straight Lines and Introduction to 3D Geometry next door, this chapter has no formative-only carve-out — everything in its syllabus line is summative.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Circle (standard)

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

Centre (h,k), radius r. Centred at the origin: x^2+y^2=r^2. For an expanded equation, recover h, k, r by completing the square, as the book's own Example 3 does.

Parabola (all four orientations)

y^2=4ax (right) | y^2=-4ax (left) | x^2=4ay (up) | x^2=-4ay (down)

Vertex at origin, a > 0. A y^2 term means the axis is the x-axis; an x^2 term means it's the y-axis. Focus is at distance a from the vertex along the axis; directrix is the same distance on the opposite side. Latus rectum = 4a.

Ellipse (major axis along x)

x^2/a^2 + y^2/b^2 = 1, a > b > 0

c^2 = a^2 - b^2, foci (+-c, 0), e = c/a with 0 < e < 1, latus rectum = 2b^2/a

Ellipse (major axis along y)

x^2/b^2 + y^2/a^2 = 1, a > b > 0

Same relations as above, foci (0, +-c). The larger denominator always marks the major axis — don't assume it sits under x^2.

Hyperbola (transverse axis along x)

x^2/a^2 - y^2/b^2 = 1

c^2 = a^2 + b^2, foci (+-c, 0), e = c/a with e >= 1, latus rectum = 2b^2/a. a=b is called an equilateral hyperbola.

Hyperbola (transverse axis along y)

y^2/a^2 - x^2/b^2 = 1

Same relations, foci (0, +-c). Whichever variable is positive marks the transverse axis.

Eccentricity, as a classification device

Circle: e=0 | Ellipse: 0<e<1 | Parabola: e=1 | Hyperbola: e>=1

Only the ellipse and hyperbola definitions of eccentricity (e=c/a) are actually derived in this book. Circle e=0 and parabola e=1 are standard classification facts that follow from the definitions, not separately derived formulas to memorise.

Latus rectum, by conic

Parabola: 4a | Ellipse and Hyperbola: 2b^2/a

These are different formulas for different curves — a common slip is applying 4a to an ellipse or hyperbola, or 2b^2/a to a parabola.

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Confusing c^2=a^2-b^2 (ellipse) with c^2=a^2+b^2 (hyperbola)

✓ For an ellipse, c is always less than a (the foci sit inside the curve), so c^2 = a^2 MINUS b^2. For a hyperbola, c is always greater than a, so c^2 = a^2 PLUS b^2. The relation flips because the geometric picture flips.

WATCH OUT

✗ Assuming the major axis of an ellipse is always along the x-axis

✓ Compare the two denominators directly: whichever is larger marks the major axis, regardless of which variable it sits under. x^2/9+y^2/25=1 has its major axis along the y-axis, since 25 > 9.

WATCH OUT

✗ Mixing up which sign or which squared variable makes a parabola open left/right versus up/down

✓ A y^2 term means the axis of symmetry is the x-axis (opens left or right); an x^2 term means the axis is the y-axis (opens up or down). The sign of the linear term then decides the direction: positive opens toward the positive axis.

WATCH OUT

✗ Using latus rectum = 4a for an ellipse or hyperbola, or 2b^2/a for a parabola

✓ These are two different formulas for two different families of curves. Parabola: 4a. Ellipse and hyperbola: 2b^2/a. Check which conic you're working with before reaching for either one.

WATCH OUT

✗ Expecting a named 'general equation of a circle' formula (with g, f, c) as this chapter's method for expanded-form equations

✓ The book recovers centre and radius from an expanded equation by completing the square directly (its own Example 3), never introducing g, f, c as named constants. The g,f,c shortcut is a fast equivalent, but the completing-the-square technique is what Exercise 10.1's own questions expect.

WATCH OUT

✗ Expecting asymptotes to be taught or tested from this chapter

✓ The word 'asymptote' never appears in the current book, and none of Exercise 10.4's questions ask for one. It's genuinely useful JEE-level knowledge, but it's beyond what this chapter itself covers — don't expect it in board exam questions drawn from this chapter.

WATCH OUT

✗ Forgetting that a hyperbola's eccentricity can equal 1 only in the limiting a=b equilateral case, never assuming e is close to 1 for hyperbolas in general

✓ Since c >= a always for a hyperbola, e = c/a >= 1 always, with equality only when c=a, which isn't possible for a genuine hyperbola (c>a strictly). e is often well above 1 — don't expect it to behave like an ellipse's e.

WATCH OUT

✗ Forgetting to check g^2+f^2-c > 0 before calling an expanded equation a real circle

✓ If g^2+f^2-c is zero, the 'circle' is a single point; if negative, no real curve exists at all. Completing the square makes this visible immediately, since the right-hand side of (x+g)^2+(y+f)^2 = g^2+f^2-c must be positive.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Ex 10.1

Exercise 10.1

Circles — find the equation given centre and radius; find centre and radius by completing the square; applied 'find the equation' problems

Questions

Ex 10.2

Exercise 10.2

Parabolas — find focus, axis, directrix, and latus rectum from a standard equation; find the equation from given conditions

Questions

Ex 10.3

Exercise 10.3

Ellipses — find foci, vertices, major/minor axis lengths, eccentricity, and latus rectum; find the equation from given conditions

Questions

Ex 10.4

Exercise 10.4

Hyperbolas — find foci, vertices, eccentricity, and latus rectum; find the equation from given conditions

Questions

Miscellaneous Ex

Miscellaneous Exercise

Applied problems on parabolic reflectors, arches, suspension-bridge cables, a rod-locus ellipse, and a parabola-inscribed triangle

Questions

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Conic Sections?

10 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

10 questions~7 min worth ~12 marks in Karnataka (KSEEB) exams

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•A conic section comes from slicing a cone; the angle of the cut determines circle, ellipse, parabola, or hyperbola, or one of three degenerate cases (point, line, pair of intersecting lines) when the cut passes through the vertex
•Circle (x-h)^2+(y-k)^2=r^2: recover centre/radius from an expanded equation by completing the square
•Parabola y^2=4ax: focus (a,0), directrix x=-a, latus rectum 4a; y^2 term means axis is the x-axis, x^2 term means axis is the y-axis
•Ellipse x^2/a^2+y^2/b^2=1 (a>b): c^2=a^2-b^2, foci (+-c,0), e=c/a in (0,1), latus rectum 2b^2/a; the larger denominator marks the major axis
•Hyperbola x^2/a^2-y^2/b^2=1: c^2=a^2+b^2, foci (+-c,0), e=c/a>=1, latus rectum 2b^2/a; a=b is an equilateral hyperbola
•Asymptotes and the g,f,c general-circle formula are both absent from the current book — useful elsewhere, but not this chapter's own content
•Unlike Straight Lines and Introduction to 3D Geometry, Conic Sections has no formative-only carve-out — everything in its syllabus line is summative

Karnataka (KSEEB) marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: Part of Unit III's 12-mark Coordinate Geometry block (shared with Straight Lines; no chapter-wise split, per CBSE)

Question type	Marks each	Typical count	What it tests
Circle: Standard Equation and Completing the Square	2-3	1	Writing the equation from centre and radius; recovering centre and radius from an expanded equation
Parabola: Focus, Directrix, Axis, and Latus Rectum	3-4	1	Reading off focus/directrix/latus rectum from a standard equation; finding the equation from given conditions
Ellipse and Hyperbola: Foci, Vertices, Eccentricity, and Latus Rectum	4-6	1-2	Identifying the major/transverse axis correctly, then computing all standard parameters
Equation from Given Conditions	3-5	1	Constructing the equation of any of the four conics from partial information such as foci, vertices, or points on the curve

Prep strategy

Identify the conic first (circle, parabola, ellipse, or hyperbola) before reaching for any formula — the four families don't share formulas
For an ellipse or hyperbola, compare the two denominators explicitly rather than assuming which axis is major or transverse
For expanded circle equations, complete the square directly rather than trying to recall a g,f,c shortcut under exam pressure

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Planetary and satellite orbits

Kepler's First Law states that planets orbit the Sun in ellipses with the Sun at one focus. Earth's orbit has eccentricity around 0.017, close enough to 0 that it looks nearly circular.

Parabolic reflectors

Satellite dishes, car headlights, and telescope mirrors are shaped as parabolas because a parabola reflects every ray parallel to its axis through a single point — the focus — which is exactly the defining property derived in this chapter.

Suspension bridge cables

A uniformly loaded suspension bridge cable hangs in the shape of a parabola, exactly as worked in the book's own Miscellaneous Example on a 100 m roadway — the same standard equation used throughout this chapter models the cable's curve directly.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

Name the conic first, then pick its formula — don't start substituting into a formula before confirming which of the four families the question is about

For 'find the equation' problems, list exactly what's given (focus? directrix? foci? vertices? a point on the curve?) before choosing which standard form to set up

For circles given in expanded form, complete the square as a first step even before the question asks for centre and radius — it clarifies what you're working with

Always double check which denominator is larger for an ellipse, or which term is positive for a hyperbola, before naming the major or transverse axis

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

STRETCH

The general second-degree equation Ax^2+Bxy+Cy^2+Dx+Ey+F=0 represents a conic whose type is determined by the discriminant B^2-4AC: negative for an ellipse, zero for a parabola, positive for a hyperbola — a natural generalisation of everything derived in this chapter for axis-aligned conics

STRETCH

Parametric forms are a fast route through many JEE problems: circle (r cos(theta), r sin(theta)), ellipse (a cos(theta), b sin(theta)), parabola (at^2, 2at) — none of these appear in the current book, but all follow directly from substituting into the standard equations derived here

STRETCH

Asymptotes of a hyperbola (y = +-(b/a)x) are absent from the current book but standard for JEE — see the Advanced problem in the JEE section for the derivation

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JEE Main & Advanced practice

Competitive-level problems on this chapter, above the board pattern. Try each one on paper before opening the solution.

JEE MainCircle through the origin with given axis interceptsGeneral-form circle, three-condition system

Find the equation of the circle which passes through the origin and has intercepts $a$ and $b$ on the x-axis and y-axis respectively.

Stuck? Show the approach

Use the general form $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ . Passing through the origin fixes $c$ . Passing through $(a, 0)$ and $(0, b)$ (the two intercept points) fixes $g$ and $f$ .

Show the full solution

Through $(0, 0)$ : $c = 0$ . Through $(a, 0)$ : $a^{2} + 2 g a = 0 \Rightarrow g = - a /2$ (for $a \neq = 0$ ). Through $(0, b)$ : $b^{2} + 2 f b = 0 \Rightarrow f = - b /2$ . Substituting back: $x^{2} + y^{2} - a x - b y = 0$ .

Answer: x^2 + y^2 - ax - by = 0

The trap

Treating $(a, 0)$ and $(0, b)$ as the centre or radius information instead of as two more points the circle passes through is the most common misread of this problem.

JEE MainParabola from a focal-chord-style latus rectum conditionWorking backward from the latus rectum

A parabola has its vertex at the origin, axis along the positive x-axis, and its latus rectum passes through the point $(2, 4)$ and $(2, - 4)$ . Find its equation.

Stuck? Show the approach

The latus rectum of $y^{2} = 4 a x$ is the vertical chord $x = a$ with endpoints $(a, \pm 2 a)$ . Match these endpoints to the given points.

Show the full solution

Endpoints $(a, 2 a)$ and $(a, - 2 a)$ must match $(2, 4)$ and $(2, - 4)$ : so $a = 2$ and $2 a = 4$ , consistent. Hence the equation is $y^{2} = 4 (2) x$ .

Answer: y^2 = 8x

The trap

Assuming the latus rectum's half-length is $a$ instead of $2 a$ (it's the full latus rectum that has length $4 a$ , so each endpoint sits $2 a$ above or below the axis) throws off the answer by a factor of 2.

JEE MainEllipse from eccentricity and one axis lengthTwo-equation system in a and b

Find the equation of the ellipse whose length of the major axis is 20 and eccentricity is $\frac{2}{5}$ , with the major axis along the x-axis.

Stuck? Show the approach

Use $2 a = 20$ to get $a$ , then $e = c / a$ to get $c$ , then $b^{2} = a^{2} - c^{2}$ .

Show the full solution

$2 a = 20 \Rightarrow a = 10$ . $e = c / a = 2/5 \Rightarrow c = 4$ . $b^{2} = a^{2} - c^{2} = 100 - 16 = 84$ .

Answer: x^2/100 + y^2/84 = 1

The trap

Using the given length (20) directly as $a$ instead of recognising it as $2 a$ is the single most common slip in 'length of major/minor axis' problems.

JEE MainHyperbola from eccentricity and latus rectumTwo-equation system in a and b

Find the equation of the hyperbola with transverse axis along the x-axis, eccentricity $\frac{5}{4}$ , and latus rectum of length 9.

Stuck? Show the approach

Write $e = c / a$ and latus rectum $= 2 b^{2} / a$ as two equations, and use $c^{2} = a^{2} + b^{2}$ to eliminate one variable.

Show the full solution

From $e = 5/4$ : $c = 5 a /4$ , so $c^{2} = 25 a^{2} /16$ . From $c^{2} = a^{2} + b^{2}$ : $b^{2} = 25 a^{2} /16 - a^{2} = 9 a^{2} /16$ . From the latus rectum: $2 b^{2} / a = 9 \Rightarrow b^{2} = 9 a /2$ . Equating the two expressions for $b^{2}$ : $9 a^{2} /16 = 9 a /2 \Rightarrow a = 8$ . Then $b^{2} = 9 (8) /2 = 36$ .

Answer: x^2/64 - y^2/36 = 1

The trap

Solving for $b^{2}$ from only one of the two given conditions and forgetting to check it against the other (eccentricity vs. latus rectum) leaves an equation that satisfies just one of the two stated conditions.

JEE AdvancedAsymptotes of a hyperbola — beyond this chapter's own scopeLimiting-behaviour derivation

Find the equations of the asymptotes of the hyperbola $\frac{x ^{2}}{9} - \frac{y ^{2}}{16} = 1$ , and state what an asymptote means for this curve.

Stuck? Show the approach

Asymptotes are not covered anywhere in this NCERT book — no formula, no worked example, no exercise question — but they're standard JEE Main and Advanced material for hyperbolas. Rewrite the equation for $y$ and examine what happens as $x \to \infty$ .

Show the full solution

From $\frac{x ^{2}}{9} - \frac{y ^{2}}{16} = 1$ : $y^{2} = 16 (\frac{x ^{2}}{9} - 1) = \frac{16}{9} x^{2} (1 - \frac{9}{x ^{2}})$ . As $x \to \infty$ , the $9/ x^{2}$ term vanishes, so $y^{2} \to \frac{16}{9} x^{2}$ , i.e. $y \to \pm \frac{4}{3} x$ . These two lines are the asymptotes: the hyperbola gets arbitrarily close to them but never touches them.

Answer: y = 4x/3 and y = -4x/3

The trap

Assuming this technique is something the current book expects you to reproduce — it isn't part of the chapter at all, so treat it as extension knowledge for competitive exams, not board-exam content from this chapter.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 11 BoardVery High

JEE MainVery High

JEE AdvancedHigh

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Questions students ask

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Not as a named formula. The book's own Example 3 recovers centre and radius from an expanded equation by completing the square directly. The g,f,c shortcut gives the same result faster, but completing the square is the method Exercise 10.1's own questions expect.

No. The word 'asymptote' doesn't appear anywhere in the current book, and none of Exercise 10.4's questions ask for one. They're useful to know for JEE, but they aren't part of what this specific chapter covers.

For an ellipse, compare the two denominators directly — the larger one marks the major axis, regardless of which variable it sits under. For a hyperbola, whichever variable is positive (not subtracted) marks the transverse axis.

No. CBSE's formative-only list has entries for Straight Lines (Normal Form, general equation of a line) and Introduction to Three-Dimensional Geometry (section formula) under Unit III, but nothing for Conic Sections. Everything in this chapter's syllabus line is summative.

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