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Conic Sections

NCERT Class 11 · Mathematics Based on NCERT Class 11 Mathematics textbook · Free CBSE study kit

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Chapter Notes

Overview
This chapter introduces conic sections—curves obtained by intersecting a plane with a double-napped right circular cone. These curves (circle, ellipse, parabola, hyperbola) are fundamental in fields like planetary motion, telescope design, and reflectors. The one big idea is that each conic has a unique geometric definition (locus) leading to a standard equation, and students must master deriving, identifying, and applying these equations to solve real-world and abstract problems.
Key Concepts (32)
Conic section
The curves obtained by intersecting a right circular cone by a plane.
e.g. When you cut a cone-shaped birthday hat with a flat sheet of paper, the shape of the cut edge is a conic section.
Double-napped right circular cone
The surface generated by rotating a line (generator) around a fixed vertical line (axis) at a constant angle, extending indefinitely in both directions.
e.g. Two ice cream cones placed tip-to-tip form a double-napped cone.
Vertex
The fixed point V where the axis and generator intersect; it separates the cone into two nappes.
e.g. The tip of a party cone is its vertex.
Generator
The rotating line m that generates the cone surface.
e.g. A string attached to the tip of a rotating toy top traces a generator.
Nappes
The two separate parts of the cone on either side of the vertex.
e.g. The upper and lower halves of an hourglass are like nappes.
Circle
The set of all points in a plane that are equidistant from a fixed point (centre).
e.g. A bicycle wheel's rim is a circle, with the hub as centre.
Centre of a circle
The fixed point from which all points on the circle are equidistant.
e.g. The centre of a round table is the point exactly halfway across.
Radius of a circle
The distance from the centre to any point on the circle.
e.g. The distance from the centre of a pizza to its edge is the radius.
Parabola
The set of all points in a plane that are equidistant from a fixed line (directrix) and a fixed point (focus) not on the line.
e.g. The path of a ball thrown in the air is a parabola.
Focus of a parabola
The fixed point used in the definition of a parabola.
e.g. In a satellite dish, signals reflect to the focus point.
Directrix of a parabola
The fixed line used in the definition of a parabola.
e.g. The line from which distances are measured in a parabolic reflector.
Axis of a parabola
The line through the focus and perpendicular to the directrix.
e.g. The vertical line through the centre of a parabolic mirror is its axis.
Vertex of a parabola
The point of intersection of the parabola with its axis.
e.g. The lowest point of a U-shaped curve is the vertex.
Latus rectum of a parabola
A line segment perpendicular to the axis of the parabola, through the focus, with endpoints on the parabola.
e.g. The width of a parabolic reflector at the focus level is the latus rectum.
Ellipse
The set of all points in a plane for which the sum of distances from two fixed points (foci) is constant.
e.g. The orbit of Earth around the Sun is an ellipse.
Foci of an ellipse
The two fixed points used in the definition of an ellipse.
e.g. The two points inside an elliptical pool table where balls reflect.
Centre of an ellipse
The midpoint of the line segment joining the foci.
e.g. The centre of an elliptical mirror is halfway between its two foci.
Major axis of an ellipse
The line segment through the foci, with endpoints on the ellipse; its length is 2a.
e.g. The longer diameter of an oval racetrack is the major axis.
Minor axis of an ellipse
The line segment through the centre perpendicular to the major axis; its length is 2b.
e.g. The shorter diameter of an oval is the minor axis.
Vertices of an ellipse
The endpoints of the major axis.
e.g. The two farthest points on an ellipse from its centre are vertices.
Eccentricity of an ellipse
The ratio e = c/a, where c is the distance from centre to focus and a is the semi-major axis (0 < e < 1).
e.g. Earth's orbit has eccentricity about 0.0167, nearly circular.
Latus rectum of an ellipse
A line segment perpendicular to the major axis through a focus, with endpoints on the ellipse.
e.g. The chord through a focus perpendicular to the major axis.
Hyperbola
The set of all points in a plane for which the absolute difference of distances from two fixed points (foci) is constant.
e.g. The shape of a cooling tower or the path of a comet escaping the solar system.
Foci of a hyperbola
The two fixed points used in the definition of a hyperbola.
e.g. Two points such that the difference of distances to any point on the hyperbola is constant.
Centre of a hyperbola
The midpoint of the line segment joining the foci.
e.g. The centre of a hyperbola is halfway between its two foci.
Transverse axis of a hyperbola
The line through the foci; its length between vertices is 2a.
e.g. The axis along which the hyperbola opens.
Conjugate axis of a hyperbola
The line through the centre perpendicular to the transverse axis; its length is 2b.
e.g. The axis perpendicular to the transverse axis.
Vertices of a hyperbola
The points where the hyperbola intersects the transverse axis.
e.g. The two closest points on the two branches of a hyperbola.
Eccentricity of a hyperbola
The ratio e = c/a, where c is distance from centre to focus and a is semi-transverse axis (e > 1).
e.g. For a hyperbola, e is always greater than 1.
Latus rectum of a hyperbola
A line segment perpendicular to the transverse axis through a focus, with endpoints on the hyperbola.
e.g. The chord through a focus perpendicular to the transverse axis.
Degenerated conic sections
Sections obtained when the plane cuts at the vertex of the cone: a point, a straight line, or a pair of intersecting lines.
e.g. When a plane cuts exactly at the tip of a cone, you get a point.
Equilateral hyperbola
A hyperbola in which a = b.
e.g. The hyperbola x² - y² = 1 is equilateral.
Formulas & Equations (17)
(x – h)² + (y – k)² = r²
Reads as: x minus h whole square plus y minus k whole square equals r square
Variables: (h, k) = centre of circle; r = radius
Use when: Finding equation of a circle given centre and radius, or finding centre/radius from general equation.
⚠ Students forget to change sign when centre is negative: e.g., for centre (-3,2), they write (x-3)² instead of (x+3)².
x² + y² = r²
Reads as: x square plus y square equals r square
Variables: r = radius; centre at origin (0,0)
Use when: Circle with centre at origin.
⚠ Forgetting that this is a special case of (x-h)²+(y-k)²=r² with h=k=0.
y² = 4ax
Reads as: y square equals 4 a x
Variables: a = distance from vertex to focus (a > 0); focus at (a,0); directrix x = -a
Use when: Parabola opening right, vertex at origin, axis along x-axis.
⚠ Writing y² = 4ax when focus is (-a,0) — that would be y² = -4ax.
y² = -4ax
Reads as: y square equals negative 4 a x
Variables: a > 0; focus at (-a,0); directrix x = a
Use when: Parabola opening left, vertex at origin.
⚠ Confusing sign: if focus is (-2,0), equation is y² = -8x, not y² = 8x.
x² = 4ay
Reads as: x square equals 4 a y
Variables: a > 0; focus at (0,a); directrix y = -a
Use when: Parabola opening upward, vertex at origin.
⚠ Writing x² = 4ay when parabola opens downward — that would be x² = -4ay.
x² = -4ay
Reads as: x square equals negative 4 a y
Variables: a > 0; focus at (0,-a); directrix y = a
Use when: Parabola opening downward, vertex at origin.
⚠ Mixing up x² and y² forms: for vertical axis, equation has x² term.
Length of latus rectum of parabola y² = 4ax is 4a
Reads as: Length of latus rectum of parabola y squared equals 4 a x is 4 a
Variables: a = distance from vertex to focus
Use when: Finding latus rectum length for any standard parabola.
⚠ Writing 2a instead of 4a — remember it's double the distance from focus to directrix.
x²/a² + y²/b² = 1 (a > b)
Reads as: x square over a square plus y square over b square equals 1
Variables: a = semi-major axis (along x-axis); b = semi-minor axis; foci at (±c,0) where c² = a² - b²
Use when: Ellipse with major axis along x-axis, centre at origin.
⚠ Assuming a is always the larger denominator — check which denominator is larger to identify major axis.
x²/b² + y²/a² = 1 (a > b)
Reads as: x square over b square plus y square over a square equals 1
Variables: a = semi-major axis (along y-axis); b = semi-minor axis; foci at (0,±c) where c² = a² - b²
Use when: Ellipse with major axis along y-axis, centre at origin.
⚠ Writing a under x² when major axis is vertical — a always goes under the larger denominator.
c² = a² - b² (for ellipse)
Reads as: c square equals a square minus b square
Variables: c = distance from centre to focus; a = semi-major axis; b = semi-minor axis
Use when: Finding foci or eccentricity of an ellipse.
⚠ Writing c² = a² + b² (that's for hyperbola) — ellipse uses subtraction.
e = c/a (for ellipse)
Reads as: e equals c over a
Variables: e = eccentricity (0 < e < 1); c = distance from centre to focus; a = semi-major axis
Use when: Calculating eccentricity of an ellipse.
⚠ Forgetting that e < 1 for ellipse; if e > 1, it's a hyperbola.
Length of latus rectum of ellipse = 2b²/a
Reads as: Length of latus rectum of ellipse equals 2 b square over a
Variables: a = semi-major axis; b = semi-minor axis
Use when: Finding latus rectum length for any ellipse.
⚠ Writing 2a²/b — the formula has b² in numerator, a in denominator.
x²/a² - y²/b² = 1
Reads as: x square over a square minus y square over b square equals 1
Variables: a = semi-transverse axis (along x-axis); b = semi-conjugate axis; foci at (±c,0) where c² = a² + b²
Use when: Hyperbola with transverse axis along x-axis, centre at origin.
⚠ Writing plus sign instead of minus — hyperbola equation has subtraction.
y²/a² - x²/b² = 1
Reads as: y square over a square minus x square over b square equals 1
Variables: a = semi-transverse axis (along y-axis); b = semi-conjugate axis; foci at (0,±c) where c² = a² + b²
Use when: Hyperbola with transverse axis along y-axis, centre at origin.
⚠ Confusing which variable has positive term — the positive term gives the transverse axis.
c² = a² + b² (for hyperbola)
Reads as: c square equals a square plus b square
Variables: c = distance from centre to focus; a = semi-transverse axis; b = semi-conjugate axis
Use when: Finding foci or eccentricity of a hyperbola.
⚠ Writing c² = a² - b² (that's for ellipse) — hyperbola uses addition.
e = c/a (for hyperbola)
Reads as: e equals c over a
Variables: e = eccentricity (e > 1); c = distance from centre to focus; a = semi-transverse axis
Use when: Calculating eccentricity of a hyperbola.
⚠ Forgetting that e > 1 for hyperbola; if e < 1, it's an ellipse.
Length of latus rectum of hyperbola = 2b²/a
Reads as: Length of latus rectum of hyperbola equals 2 b square over a
Variables: a = semi-transverse axis; b = semi-conjugate axis
Use when: Finding latus rectum length for any hyperbola.
⚠ Same formula as ellipse but with different meaning of a and b — ensure correct identification of axes.
How It Works
Formation of conic sections by plane-cone intersection
  1. Step 1: Consider a double-napped right circular cone with vertex V, axis l, and generator m at angle α to axis.
  2. Step 2: Let a plane intersect the cone at angle β to the vertical axis.
  3. Step 3: If β = 90°, the section is a circle (plane horizontal).
  4. Step 4: If α < β < 90°, the section is an ellipse (plane cuts one nappe at an angle).
  5. Step 5: If β = α, the section is a parabola (plane parallel to generator).
  6. Step 6: If 0 ≤ β < α, the section is a hyperbola (plane cuts both nappes).
  7. Step 7: If plane passes through vertex, degenerated conics occur: point (α < β ≤ 90°), line (β = α), or pair of lines (0 ≤ β < α).
💡 The angle β is measured from the vertical axis, not from the generator. Circle requires β=90°, parabola requires β=α.
Deriving equation of a circle with centre (h,k) and radius r
  1. Step 1: Let C(h,k) be centre and r radius.
  2. Step 2: Let P(x,y) be any point on circle.
  3. Step 3: By definition, distance CP = r.
  4. Step 4: Apply distance formula: √[(x-h)² + (y-k)²] = r.
  5. Step 5: Square both sides: (x-h)² + (y-k)² = r².
💡 The equation is derived directly from the distance formula — no other assumptions needed.
Deriving standard equation of parabola y² = 4ax
  1. Step 1: Place vertex at origin O, focus F at (a,0) with a>0, directrix x = -a.
  2. Step 2: Let P(x,y) be any point on parabola.
  3. Step 3: By definition, PF = distance from P to directrix (PB).
  4. Step 4: PF = √[(x-a)² + y²]; PB = |x + a| (since directrix is x = -a).
  5. Step 5: Equate: √[(x-a)² + y²] = |x + a|.
  6. Step 6: Square both sides: (x-a)² + y² = (x+a)².
  7. Step 7: Expand: x² - 2ax + a² + y² = x² + 2ax + a².
  8. Step 8: Simplify: y² = 4ax.
💡 The derivation uses the locus definition — distance to focus equals distance to directrix. The absolute value is handled by squaring.
Deriving standard equation of ellipse x²/a² + y²/b² = 1
  1. Step 1: Place foci at F1(-c,0) and F2(c,0), centre at origin.
  2. Step 2: Let P(x,y) be any point on ellipse.
  3. Step 3: By definition, PF1 + PF2 = 2a (constant).
  4. Step 4: Write: √[(x+c)² + y²] + √[(x-c)² + y²] = 2a.
  5. Step 5: Isolate one radical, square both sides, simplify.
  6. Step 6: Isolate remaining radical, square again.
  7. Step 7: Simplify using c² = a² - b² to get x²/a² + y²/b² = 1.
💡 The derivation involves two squaring steps. The constant 2a is chosen because at vertex (a,0), sum of distances = (a+c)+(a-c)=2a.
Deriving standard equation of hyperbola x²/a² - y²/b² = 1
  1. Step 1: Place foci at F1(-c,0) and F2(c,0), centre at origin.
  2. Step 2: Let P(x,y) be any point on hyperbola.
  3. Step 3: By definition, |PF1 - PF2| = 2a (constant).
  4. Step 4: Write: √[(x+c)² + y²] - √[(x-c)² + y²] = ±2a.
  5. Step 5: Isolate one radical, square both sides, simplify.
  6. Step 6: Isolate remaining radical, square again.
  7. Step 7: Simplify using c² = a² + b² to get x²/a² - y²/b² = 1.
💡 The difference (not sum) is constant. The constant 2a is the distance between vertices. For hyperbola, c > a, so e > 1.
Finding centre and radius from general circle equation
  1. Step 1: Write equation in form x² + y² + Dx + Ey + F = 0.
  2. Step 2: Group x-terms and y-terms: (x² + Dx) + (y² + Ey) = -F.
  3. Step 3: Complete square for x: add (D/2)² to both sides.
  4. Step 4: Complete square for y: add (E/2)² to both sides.
  5. Step 5: Write as (x + D/2)² + (y + E/2)² = (D²/4 + E²/4 - F).
  6. Step 6: Centre is (-D/2, -E/2); radius = √(D²/4 + E²/4 - F).
💡 The centre coordinates have opposite signs of the terms in brackets. Radius must be positive — if RHS is negative, it's not a real circle.
Exam Traps — Don't Make These Mistakes
⚠ Misidentifying major axis in ellipse
✗ In ellipse 9x² + 4y² = 36, students write a² = 9, b² = 4, so a = 3, b = 2, and major axis is along x-axis.
→ The equation must be written in standard form x²/4 + y²/9 = 1. Since denominator under y² (9) is larger, the major axis is along y-axis, not x-axis.
✓ Rewrite as x²/4 + y²/9 = 1. Here a² = 9 (under y²), so a = 3, major axis is y-axis. b² = 4, b = 2, minor axis is x-axis.
⚠ Sign error in circle centre from general equation
✗ For circle x² + y² + 8x + 10y - 8 = 0, centre is (8, 10).
→ After completing squares, the equation becomes (x+4)² + (y+5)² = 49. The centre is (-4, -5), not (8,10). Students forget the sign change.
✓ Complete the square: (x²+8x+16) + (y²+10y+25) = 8+16+25 → (x+4)² + (y+5)² = 49. Centre = (-4, -5), radius = 7.
⚠ Confusing parabola opening direction with sign of coefficient
✗ For parabola y² = -8x, students say 'opens to the right because coefficient is negative'.
→ For y² = -4ax, the parabola opens to the left (negative x-direction). The sign of coefficient determines direction: positive → right/up, negative → left/down.
✓ y² = -8x → 4a = 8 → a = 2. Since equation is y² = -4ax, parabola opens to the left. Focus is (-2,0), directrix x = 2.
⚠ Using ellipse formula c² = a² + b² instead of c² = a² - b²
✗ For ellipse with a=5, b=3, students calculate c = √(25+9) = √34.
→ For ellipse, c² = a² - b² (since a > b). Using addition gives c > a, which would make eccentricity > 1, impossible for ellipse.
✓ c² = a² - b² = 25 - 9 = 16, so c = 4. Eccentricity e = c/a = 4/5 = 0.8 (which is < 1, correct for ellipse).
⚠ Misidentifying transverse axis in hyperbola
✗ For hyperbola y²/16 - x²/9 = 1, students say transverse axis is along x-axis because x term comes first.
→ The positive term determines the transverse axis. Here y² term is positive, so transverse axis is along y-axis, not x-axis.
✓ In y²/16 - x²/9 = 1, the positive term is y², so transverse axis is y-axis. a² = 16, a = 4 (semi-transverse axis along y). b² = 9, b = 3.
⚠ Forgetting to convert to standard form before identifying parameters
✗ For 9x² + 4y² = 36, students directly take a² = 9, b² = 4.
→ The equation must be divided by 36 to get standard form x²/4 + y²/9 = 1. Only then can a and b be correctly identified.
✓ Divide both sides by 36: x²/4 + y²/9 = 1. Now a² = 9 (larger denominator), b² = 4. So a = 3, b = 2, major axis along y-axis.
Compare & Contrast
Definition (sum vs difference of distances)
Ellipse
Hyperbola
Ellipse: sum of distances from two foci is constant (2a). Hyperbola: absolute difference of distances from two foci is constant (2a).
Relationship between a, b, c
Ellipse
Hyperbola
Ellipse: c² = a² - b² (so c < a, eccentricity e < 1). Hyperbola: c² = a² + b² (so c > a, eccentricity e > 1).
Axis of symmetry
Parabola y² = 4ax
Parabola x² = 4ay
y² = 4ax has axis along x-axis (opens right/left). x² = 4ay has axis along y-axis (opens up/down).
Eccentricity
Circle
Ellipse
Circle has eccentricity e = 0 (special case of ellipse where a = b). Ellipse has 0 < e < 1.
Exam Focus
Marks pattern: 2 MCQs (1 mark each) + 1 SA-I (2 marks) + 1 SA-II (3 marks) + 1 LA (5 marks) = 11-12 marks typically
Most asked: Finding equation of circle given centre and radius or general form · Identifying focus, directrix, vertex, latus rectum of parabola from equation · Finding foci, vertices, axes lengths, eccentricity, latus rectum of ellipse · Finding foci, vertices, eccentricity, latus rectum of hyperbola · Word problems involving parabolic reflectors, elliptical arches, suspension bridges
Don't skip: Long answer (5 marks) on finding equation of ellipse/hyperbola given conditions (e.g., passing through points, foci, vertices); Word problem (3-5 marks) on real-life application of parabola (e.g., reflector, arch, cable); Derivation-based question on standard equations (e.g., derive y²=4ax or ellipse equation)
One-Line Revision (read aloud before the exam)
  1. Conic sections are curves obtained by intersecting a plane with a double-napped right circular cone.
  2. Circle: set of points equidistant from a fixed centre; equation (x-h)²+(y-k)²=r².
  3. Parabola: set of points equidistant from a focus and a directrix; standard forms y²=±4ax and x²=±4ay.
  4. Ellipse: set of points where sum of distances from two foci is constant; equation x²/a²+y²/b²=1 (a>b for x-axis major).
  5. Hyperbola: set of points where absolute difference of distances from two foci is constant; equation x²/a²-y²/b²=1.
  6. For ellipse: c² = a² - b², eccentricity e = c/a (0 < e < 1).
  7. For hyperbola: c² = a² + b², eccentricity e = c/a (e > 1).
  8. Length of latus rectum for parabola y²=4ax is 4a.
  9. Length of latus rectum for ellipse and hyperbola is 2b²/a.
  10. Circle is a special ellipse with e=0 and a=b=r.
  11. When β=90°, section is circle; when β=α, section is parabola; when α<β<90°, ellipse; when 0≤β<α, hyperbola.
  12. Degenerated conics occur when plane passes through vertex: point, line, or pair of lines.
  13. To find centre from general circle equation, complete the square for x and y terms.
  14. In ellipse, the larger denominator gives the major axis direction.
  15. In hyperbola, the positive term gives the transverse axis direction.
  16. Equilateral hyperbola has a=b, so equation becomes x²-y²=a² or y²-x²=a².

MCQs — 10 Questions with Answers

Q1. What conic section is formed when a plane intersects a right circular cone perpendicular to its vertical axis?

  • A. Circle ✓
  • B. Ellipse
  • C. Parabola
  • D. Hyperbola

Answer: A — When β = 90° (plane perpendicular to vertical axis), the intersection produces a circle.

Q2. Find the radius of the circle x² + y² – 6x + 8y – 11 = 0.

  • A. 3
  • B. 4
  • C. 5
  • D. 6 ✓

Answer: D — Completing the square: (x – 3)² + (y + 4)² = 36, so radius = √36 = 6.

Q3. Which of the following is the equation of a circle with centre (–2, 3) and radius 5?

  • A. (x – 2)² + (y – 3)² = 25
  • B. (x + 2)² + (y – 3)² = 25 ✓
  • C. (x + 2)² + (y + 3)² = 25
  • D. (x – 2)² + (y + 3)² = 25

Answer: B — Standard form with centre (h, k) = (–2, 3) and radius r = 5 gives (x – (–2))² + (y – 3)² = 25.

Q4. If a plane cuts a right circular cone such that α < β < 90°, what conic section is obtained?

  • A. Circle
  • B. Ellipse ✓
  • C. Parabola
  • D. Hyperbola

Answer: B — When the plane angle β lies between the semi-vertical angle α and 90°, an ellipse is formed.

Q5. Find the centre of the circle (x – 1)² + (y + 2)² = 49.

  • A. (1, –2) ✓
  • B. (–1, 2)
  • C. (1, 2)
  • D. (–1, –2)

Answer: A — The standard form (x – h)² + (y – k)² = r² has centre (h, k) = (1, –2).

Q6. What is NOT a condition for conic sections formed by a plane cutting one nappe of a cone?

  • A. β = 90° produces a circle
  • B. α < β < 90° produces an ellipse
  • C. 0 ≤ β < α produces a parabola ✓
  • D. β = α produces a parabola

Answer: C — When 0 ≤ β < α, the plane cuts both nappes producing a hyperbola, not a parabola.

Q7. The equation of the circle passing through (4, 1) with centre (2, –3) is which of the following?

  • A. (x – 2)² + (y + 3)² = 18
  • B. (x – 2)² + (y + 3)² = 20 ✓
  • C. (x – 2)² + (y – 3)² = 20
  • D. (x + 2)² + (y + 3)² = 18

Answer: B — Radius r = √[(4–2)² + (1–(–3))²] = √[4 + 16] = √20, so equation is (x – 2)² + (y + 3)² = 20.

Q8. A circle has centre at (0, 0) and passes through (3, 4). Its equation is:

  • A. x² + y² = 9
  • B. x² + y² = 16
  • C. x² + y² = 25 ✓
  • D. x² + y² = 49

Answer: C — Radius r = √(3² + 4²) = √25 = 5, so the equation is x² + y² = 25.

Q9. Which definition correctly describes a parabola?

  • A. Set of points equidistant from two fixed points
  • B. Set of points equidistant from a fixed point and a fixed line ✓
  • C. Set of points at constant distance from a fixed line
  • D. Set of points forming a closed curve

Answer: B — A parabola is defined as the locus of points equidistant from a focus (fixed point) and a directrix (fixed line).

Q10. For a circle x² + y² + 2gx + 2fy + c = 0, if g² + f² – c = 0, then the conic is:

  • A. A circle with positive radius
  • B. A degenerate circle (a point) ✓
  • C. An imaginary circle
  • D. A hyperbola

Answer: B — When g² + f² – c = 0, the radius √(g² + f² – c) = 0, resulting in a degenerate conic—a single point (the centre).

Flashcards

What is the definition of a circle in analytic geometry?

A circle is the set of all points in a plane equidistant from a fixed point called the centre.

Write the standard equation of a circle with centre (h, k) and radius r.

The equation is (x – h)² + (y – k)² = r².

What is the equation of a circle centred at the origin with radius r?

The equation is x² + y² = r².

Define a parabola using focus and directrix.

A parabola is the set of all points equidistant from a fixed point (focus) and a fixed line (directrix).

What is the relationship between the angle β of the intersecting plane and the cone axis angle α for an ellipse?

For an ellipse, the plane angle β satisfies α < β < 90° where the plane cuts one nappe of the cone.

When does the intersection of a plane with a cone produce a parabola?

A parabola is formed when the plane angle β equals the cone semi-vertical angle α.

How do you find the centre and radius of a circle from the general equation x² + y² + 2gx + 2fy + c = 0?

Complete the square: centre is (–g, –f) and radius is √(g² + f² – c).

What is the vertex of a parabola?

The vertex is the point where the parabola intersects its axis of symmetry.

When a plane cuts through both nappes of a cone, what conic section is formed?

A hyperbola is formed when 0 ≤ β < α, where the plane intersects both nappes.

What are degenerate conic sections and when do they occur?

Degenerate conics occur when the plane passes through the vertex: a point (α < β ≤ 90°), a line (β = α), or two intersecting lines (0 ≤ β < α).

Important Board Questions

Define a parabola and write its key geometric properties (focus, directrix, vertex, and axis of symmetry). [2 marks]

State the definition using equidistant property from focus and directrix. Briefly identify each of the four components and their relationship to the axis of symmetry.

Find the equation of the circle passing through the points (2, 3) and (–1, 1) whose centre lies on the line x – 3y – 11 = 0. Show all working steps. [5 marks]

Let centre be (h, k). Use two conditions: (1) equal distance from both given points, (2) centre on given line. Set up and solve the system of three equations to find h, k, and r².

Explain how different conic sections (circle, ellipse, parabola, and hyperbola) are formed by varying the angle β of an intersecting plane with respect to the semi-vertical angle α of a double-napped cone. Include the condition for degenerate cases. [6 marks]

Compare each condition: β = 90° (circle), α < β < 90° (ellipse), β = α (parabola), 0 ≤ β < α (hyperbola). Then explain three degenerate cases when plane passes through vertex. Relate plane angle to nappe intersection.

Assertion-Reason Questions for Conic Sections

Q1. Assertion (A):

The equation (x – 4)² + (y + 3)² = 9 represents a circle with centre at (4, –3).

Reason (R):

The standard form of a circle is (x – h)² + (y – k)² = r², where the centre is (h, k).

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true. The given equation matches the standard form with h = 4, k = –3, and r = 3. The reason correctly explains why the centre is (4, –3), not (–4, 3).

Q2. Assertion (A):

A circle passing through (0, 0), (2, 0), and (0, 2) has its centre on the line y = x.

Reason (R):

All circles have their centres on the perpendicular bisector of any chord.

Answer: B — Both A and R are true, but R is NOT the correct explanation of A — Both are true, but the reason does not directly explain the assertion. The assertion is true because for the three given points, the perpendicular bisectors of two chords happen to intersect on y = x. However, the reason is a general property not specific to why this particular centre lies on y = x.

Q3. Assertion (A):

When a plane cuts a cone such that β = α (where α is the semi-vertical angle), a parabola is formed.

Reason (R):

A parabola results when the plane is parallel to exactly one generator of the cone.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertions are true, and the reason is the correct explanation. When β = α, the cutting plane is indeed parallel to a generator of the cone, which is the precise geometric condition for forming a parabola.

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Case-Based Questions (CBQ) for Conic Sections

Case Study:

A civil engineer is designing a circular water tank in a park. The tank's boundary in the coordinate plane (with unit in metres) is given by the equation x² + y² – 8x + 6y = 0. The park also has a straight walking path that passes through points (2, 0) and (6, 0). The engineer needs to determine if the path will intersect the tank, be tangent to it, or pass entirely outside it.

Q1. What is the centre of the circular water tank?

  • A. (4, –3) ✓
  • B. (–4, 3)
  • C. (4, 3)
  • D. (–4, –3)

Answer: A

Q2. What is the radius of the water tank?

  • A. 5 metres ✓
  • B. 6 metres
  • C. 7 metres
  • D. 8 metres

Answer: A

Q3. What is the equation of the walking path?

  • A. x = 4
  • B. y = 0 ✓
  • C. y = 1
  • D. x + y = 4

Answer: B

Q4. What is the relationship between the walking path and the water tank?

  • A. The path lies entirely inside the tank
  • B. The path is tangent to the tank
  • C. The path intersects the tank at two points ✓
  • D. The path lies entirely outside the tank

Answer: C

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Class 11 Maths Formula Sheet

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