Conic Sections Class 11 Notes: Formulas, Worked Examples and Exam Tips

These conic sections class 11 notes compress Chapter 10 of the NCERT Mathematics textbook into a page you can revise from alone — the definitions, standard equations, eccentricity, latus rectum, and axis relations for the circle, parabola, ellipse and hyperbola.

You also get four stepwise worked examples with fresh numbers, an error-correction table for mistakes that actually cost marks, and a one-glance comparison of all four curves. Every formula here traces to the rationalised NCERT text, which you can cross-check against the official PDF linked in the recap.

Chapter 10 builds directly on coordinate geometry: the distance formula you used in our straight lines notes derives every equation in this chapter. Once you can write \((x-h)^2+(y-k)^2=r^2\) from the centre–radius form, the rest is just applying three more distance definitions — and one number, eccentricity, decides which curve you get.

How a Plane Cuts the Cone: Circle, Ellipse, Parabola, Hyperbola

A fixed vertical line \(l\) is the axis of the cone. A second line \(m\), the generator, pivots at a fixed point \(V\) — the vertex — and keeps a constant angle \(\alpha\) with the axis. Rotating \(m\) around \(l\) sweeps out a double-napped right circular cone; each half of the cone is a nappe (NCERT, p. 177).

Now let an intersecting plane make an angle \(\beta\) with the vertical axis of the cone. Compare \(\beta\) with the fixed angle \(\alpha\), and the shape of the cut is decided before a single equation is written (NCERT, p. 178):

When the plane passes exactly through the vertex, you get the degenerated conic sections (NCERT, pp. 178–179): a single point (when \(\alpha \lt \beta \leq 90^\circ\)), a straight line (when \(\beta = \alpha\), the degenerate parabola), or a pair of intersecting straight lines (when \(0 \leq \beta \lt \alpha\), the degenerate hyperbola).

The same angle conditions, but cutting at the vertex, produce these collapsed curves.

The chapter then rebuilds each curve from a distance definition, and one number — the eccentricity \(e\) — separates all four: \(e = 0\) for the circle, \(0 \lt e \lt 1\) for the ellipse, \(e = 1\) for the parabola, and \(e \gt 1\) for the hyperbola. Keep that ladder in mind; the formula table stamps it in.

Circle: Centre–Radius Equation and Completing the Square

Definition 1 (circle): all points in a plane equidistant from a fixed point — the centre — with that common distance called the radius (NCERT, p. 180). It is the simplest conic: one fixed point, one distance, no focus–directrix pair.

Take centre \(C(h,k)\) and radius \(r\). For any point \(P(x,y)\) on the circle, \(|CP| = r\). The distance formula gives \(\sqrt{(x-h)^2+(y-k)^2} = r\), so the required equation is (NCERT, p. 180):

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

When the centre is the origin, \(h=k=0\), and this collapses to \(x^2+y^2 = r^2\) — a form worth recognising instantly.

The reverse skill — reading the centre and radius. A question gives a general quadratic like \(x^2+y^2+6x-8y-11=0\) and asks for centre and radius. Group the \(x\) and \(y\) terms and complete the square, adding the same number to both sides:

1. Group: \((x^2+6x) + (y^2-8y) = 11\).
2. Complete each square: \((x^2+6x+9) + (y^2-8y+16) = 11+9+16\).
3. Factor: \((x+3)^2 + (y-4)^2 = 36\).
4. Read off: centre \((-3, 4)\), radius \(6\).

You add \(9 = (6/2)^2\) to complete the \(x\)-square and \(16 = (-8/2)^2\) for the \(y\)-square, and balance the right side the same way — that is what “add the same number to both sides” means.

Where does a point sit? Compare the point’s distance from the centre with the radius (as in Exercise 10.1 Q15): distance \(\lt r\) means inside, \(\gt r\) outside, \(= r\) on the circle. For \((-2.5, 3.5)\) against \(x^2+y^2=25\), distance \(= \sqrt{18.5} \approx 4.3 \lt 5\), so the point lies inside.

Parabola: Four Standard Forms and the Latus Rectum

Definition 2 (parabola): all points equidistant from a fixed line — the directrix — and a fixed point not on that line — the focus (NCERT, p. 182). If the focus lies on the directrix, the set collapses to the straight line through the focus perpendicular to the directrix: the degenerate parabola.

The line through the focus perpendicular to the directrix is the axis; the point where the parabola meets its axis is its vertex.

With the vertex at the origin and the axis along a coordinate axis, only four standard equations exist (NCERT, pp. 183–185):

The quick-reading rule you will use on every parabola question (NCERT, p. 185): the squared variable names the axis, and the sign of the linear coefficient names the direction it opens. So \(y^2 = -24x\) has a \(y^2\), hence the axis is the \(x\)-axis; the coefficient \(-24\) is negative, so it opens left.

Latus rectum (Definition 3, NCERT, p. 186): the chord through the focus perpendicular to the axis, with endpoints on the parabola. Because a point on the parabola stays equidistant from focus and directrix, each half is \(2a\) long, and symmetry doubles it — so the length is always \(4a\).

Watch the factor of 4. In \(y^2 = 4ax\) the coefficient is \(4a\), so \(a\) is a quarter of the coefficient, not the coefficient itself. For \(y^2 = 20x\), \(4a = 20\), so \(a = 5\), focus \((5,0)\), directrix \(x = -5\), latus rectum 20.

Ellipse: Eccentricity and Finding the Major Axis

Definition 4 (ellipse): all points whose sum of distances from two fixed points — the foci — is a constant (NCERT, p. 187). That constant is taken as \(2a\), always greater than the gap \(2c\) between the foci.

The centre is the midpoint of the segment joining the foci. The line through the foci is the major axis (length \(2a\)); the perpendicular through the centre is the minor axis (length \(2b\)); the ends of the major axis are the vertices (NCERT, p. 188).

Why \(a^2 = b^2 + c^2\)? Put a point at the end of the major axis: the sum of distances to the foci is \(F_1P + F_2P = 2a\) directly. Put a point at the end of the minor axis: each distance is \(\sqrt{b^2+c^2}\), so the sum is \(2\sqrt{b^2+c^2}\).

Both points lie on the ellipse, so both sums equal \(2a\), giving \(2\sqrt{b^2+c^2}=2a\), i.e. \(a^2 = b^2 + c^2\) (NCERT, p. 189). Picture it as a right triangle: centre–focus–vertex with hypotenuse \(a\).

Eccentricity: \(e = \frac{c}{a}\) (NCERT, p. 189). Since \(0 \lt c \lt a\), an ellipse always has \(0 \lt e \lt 1\), and the focus sits at distance \(ae\) from the centre.

Two standard forms. Foci on the \(x\)-axis: \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\). Foci on the \(y\)-axis: \(\frac{x^2}{b^2}+\frac{y^2}{a^2}=1\) (NCERT, pp. 190, 192).

The denominator test decides which is which (NCERT, p. 192): the larger denominator belongs to \(a^2\), and that variable names the major axis. In \(\frac{x^2}{4}+\frac{y^2}{9}=1\) the larger denominator 9 sits under \(y^2\), so the major axis is the \(y\)-axis and \(a=3\), \(b=2\). The trap is reading \(a=2\) from the \(x^2\) denominator — \(a\) is always the semi-major axis.

Renormalise before you read. An equation like \(9x^2+4y^2=36\) must first be divided through by 36 to reach \(\frac{x^2}{4}+\frac{y^2}{9}=1\) before you compare denominators.

Latus rectum: the chord through a focus perpendicular to the major axis; its length is \(\frac{2b^2}{a}\) (NCERT, p. 193).

Hyperbola: Transverse Axis and the \(c^2 = a^2 + b^2\) Relation

Definition 7 (hyperbola): all points whose difference of distances from two fixed points — the foci — is a constant (NCERT, pp. 195–196). “Difference” here means the distance to the farther point minus the distance to the closer point, taken as the constant \(2a\).

The line through the foci is the transverse axis (length \(2a\)); the perpendicular through the centre is the conjugate axis (length \(2b\)); the curve meets the transverse axis at the vertices (NCERT, p. 196). The gap between the foci is \(2c\), with \(b = \sqrt{c^2 – a^2}\).

The relation that flips sign compared with the ellipse. For an ellipse \(c^2 = a^2 – b^2\) because \(a\) is the largest length (the semi-major axis). For a hyperbola \(a\) is not the largest — the branches never touch the conjugate axis, and the focus sits farther out — so the relation becomes \(c^2 = a^2 + b^2\), i.e. \(b^2 = c^2 – a^2\) (NCERT, p. 199).

The eccentricity \(e = \frac{c}{a}\) is therefore never less than one (NCERT, p. 198).

Standard forms. Transverse axis on the \(x\)-axis: \(\frac{x^2}{a^2} – \frac{y^2}{b^2} = 1\). On the \(y\)-axis: \(\frac{y^2}{a^2} – \frac{x^2}{b^2} = 1\) (NCERT, pp. 199–200).

Which term names the transverse axis? The positive term’s denominator. For \(\frac{x^2}{9} – \frac{y^2}{16} = 1\) the transverse axis is the \(x\)-axis with length \(2a = 6\); for \(\frac{y^2}{25} – \frac{x^2}{16} = 1\) it is the \(y\)-axis with length 10 (NCERT, p. 200). When \(a = b\) the hyperbola is called equilateral (NCERT, p. 199).

The latus rectum — a chord through a focus perpendicular to the transverse axis — has length \(\frac{2b^2}{a}\), the same formula as the ellipse (NCERT, p. 201).

Conics Formula Table: All Four at a Glance

One table carries the chapter’s whole summary (NCERT, p. 206). Every entry below matches the textbook’s Summary table:

That eccentricity row is the memory device: circle zero, ellipse between zero and one, parabola one, hyperbola greater than one. The \(ae\) focus distance for ellipse and hyperbola comes directly from \(e = c/a\) (NCERT, pp. 189, 198).

Key Terms in Conic Sections — Definitions Table

Quick-reference glossary of the terms Definitions 1–9 in the chapter introduce. Meanings are in plain words you can recall under exam pressure:

Worked Examples: Stepwise Solutions with Original Numbers

Example 1 (Circle): Find the equation of the circle with centre \((-2, 5)\) and radius 3

Example 2 (Parabola): For \(y^2 = -24x\), find the focus, axis, directrix and latus rectum

Example 3 (Ellipse): For \(\frac{x^2}{49}+\frac{y^2}{16}=1\), find the foci, vertices, eccentricity and latus rectum

Example 4 (Hyperbola): Find the equation with foci \((\pm 10, 0)\) and vertices \((\pm 6, 0)\)

Common Mistakes: Error Pairs That Cost Marks

Five error–correction pairs specific to this chapter, each with the reason behind the rule:

Exam Notes: Patterns That Recur in Question Papers

Based on the exercise sets in this chapter (Exercises 10.1–10.4 and the Miscellaneous Exercise), four question families keep coming back. Practise these specifically:

• Write the equation from given data. Centre and radius (Ex 10.1 Q1–5); focus and directrix, or vertex and focus (Ex 10.2 Q7–12); vertices and foci, or axis and point (Ex 10.3 Q10–20); vertices and foci, or foci and latus rectum (Ex 10.4 Q7–15).
• Read off the parts from a given equation. Focus, axis, directrix and latus rectum of a parabola (Ex 10.2 Q1–6); foci, vertices, axes, eccentricity and latus rectum of ellipse and hyperbola (Ex 10.3 Q1–9, Ex 10.4 Q1–6).
• Renormalise first. Questions deliberately hide the standard form: \(2x^2+2y^2-x=0\) (circle, Ex 10.1 Q9), \(9x^2+4y^2=36\) (ellipse, Ex 10.3 Q9), \(9y^2-4x^2=36\) and \(16x^2-9y^2=576\) (hyperbolas, Ex 10.4 Q3, Q4).
• Model an application. The Miscellaneous Examples and Miscellaneous Exercise use a parabolic reflector, a deflected beam, a suspension cable and a sliding rod to force you to set up the curve yourself.

A full-marks answer shows the standard form written out before substitution — that single line earns the method mark even if a later step slips.

Where Conics Show Up in Real Life

The chapter opens by naming the real-world pull of these curves — planetary motion, the design of telescopes and antennas, reflectors in flashlights and automobile headlights (NCERT, p. 176). The geometry, not history, is the point: a reflector shaped as a parabola sends every incoming ray through the focus.

Parabolic mirror (Example 17, p. 202): the focus lies 5 cm from the vertex and the mirror is 45 cm deep, so the section obeys \(y^2 = 20x\); at \(x = 45\), \(y^2 = 900\) gives \(y = \pm 30\), so the opening \(AB\) is 60 cm. Deflected beam (Example 18, p. 203): a beam supported 12 m apart sags 3 cm at the centre into a parabola \(x^2 = 4ay\).

Sliding rod (Example 19, p. 204): a 15 cm rod resting across the axes, with a point 6 cm from one end, traces the ellipse \(\frac{x^2}{81}+\frac{y^2}{36}=1\) — the sum-of-distances definition made concrete.

The same curves return in later chapters of the course, including Introduction to Three Dimensional Geometry notes, where circles and spheres reappear in space. Understand the plane-curve version here, and the 3D extension is mostly new notation.

Quick Revision Recap for the Night Before

• Circle: points equidistant from one fixed point (centre); equation \((x-h)^2+(y-k)^2=r^2\).
• Parabola: points equidistant from a fixed line (directrix) and a fixed point (focus); \(e = 1\).
• Parabola reading rule: squared variable = axis; sign of coefficient = direction it opens.
• Parabola latus rectum: \(4a\). Remember \(a\) = one quarter of the linear coefficient.
• Ellipse: sum of distances from two foci \(= 2a\); \(0 \lt e \lt 1\).
• Ellipse relation: \(a^2 = b^2 + c^2\) (\(a\) largest); the larger denominator names the major axis.
• Hyperbola: difference of distances (farther minus closer) \(= 2a\); \(e \gt 1\).
• Hyperbola relation: \(b^2 = c^2 – a^2\) (\(c\) largest); the positive term’s denominator names the transverse axis.
• Ellipse and hyperbola latus rectum: both \(\frac{2b^2}{a}\).
• Always renormalise equations like \(9x^2+4y^2=36\) to RHS = 1 before reading parts.

If any formula feels shaky, open the official NCERT Chapter 10 PDF and check it against the Summary table on p. 206. For the full picture, browse the Class 11 Mathematics notes hub, the Class 11 notes index, or the complete CBSE notes collection.

FAQs: Conic Sections Questions Students Ask

Q1. How do I tell whether the major axis of an ellipse is on the x-axis or the y-axis?

Look at the larger denominator. Whatever variable has the bigger denominator in \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) carries the major axis (NCERT, p. 192). For \(\frac{x^2}{25}+\frac{y^2}{9}=1\), the major axis is the \(x\)-axis; for \(\frac{x^2}{4}+\frac{y^2}{9}=1\), the \(y\)-axis.

Q2. What is the quickest way to read the focus and directrix from \(y^2 = 4ax\)?

Divide the coefficient of \(x\) by 4 to get \(a\). The focus is \((a, 0)\), the directrix is \(x = -a\), and the parabola opens right. Example: \(y^2 = 12x\) gives \(a = 3\), focus \((3, 0)\), directrix \(x = -3\).

Q3. Why is the relation \(c^2 = a^2 + b^2\) for a hyperbola but \(c^2 = a^2 – b^2\) for an ellipse?

In an ellipse the foci sit inside the curve, so \(a\) (the semi-major axis) is the largest length and \(c^2 = a^2 – b^2\). In a hyperbola the branches never meet the conjugate axis and the foci lie farther out than the vertices, so \(c\) is largest and the relation becomes \(c^2 = a^2 + b^2\), i.e. \(b^2 = c^2 – a^2\) (NCERT, pp. 189, 199).

Q4. How do I convert \(4x^2 + 9y^2 = 36\) into the standard form of an ellipse?

Divide every term by 36 to make the right side 1: \(\frac{x^2}{9}+\frac{y^2}{4}=1\). Now the larger denominator 9 is under \(x^2\), so \(a = 3\), \(b = 2\), and the major axis is the \(x\)-axis.

Q5. What does eccentricity mean, and how do I use it to identify which conic I am looking at?

Eccentricity is the ratio \(e = \frac{c}{a}\) of the focus distance from the centre to the vertex distance (NCERT, pp. 189, 198) — loosely, how “stretched” the curve is. The values separate the conics: \(e = 0\) circle, \(0 \lt e \lt 1\) ellipse, \(e = 1\) parabola, \(e \gt 1\) hyperbola.

Reference: NCERT Class 11 Mathematics textbook, Chapter 10 Conic Sections.