📐 Class 11 Mathematics • Chapter 10

Conic
Sections

Master circles, parabolas, ellipses, and hyperbolas — the four elegant curves born from slicing a cone, with real-world applications in astronomy, optics, and engineering.

4
Conic Curves
12
MCQs
NCERT
Source
11th
Standard
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Conic Sections — Full Chapter Video Lecture
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What are Conic Sections?

📌 Origin of Conic Sections

Conic sections (or conics) are curves obtained by intersecting a right circular double-napped cone with a plane. The type of curve depends on the angle β the plane makes with the vertical axis of the cone.

β = 90°

⭕ Circle

Plane is perpendicular to the cone's axis. All points equidistant from centre. e = 0.

α < β < 90°

🥚 Ellipse

Plane cuts one nappe at an angle. Sum of distances from two foci is constant. 0 < e < 1.

β = α

🪃 Parabola

Plane is parallel to one generator of the cone. Equidistant from focus and directrix. e = 1.

0 ≤ β < α

〰️ Hyperbola

Plane cuts both nappes. Difference of distances from two foci is constant. e > 1.

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IMAGE SPACE — Sections of a Cone

Add diagram showing double-napped cone with planes cutting it to form circle, ellipse, parabola, and hyperbola
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1
Circle

📌 Definition

A circle is the set of all points in a plane that are equidistant from a fixed point called the centre. The fixed distance is called the radius (r).

🎯 Circle Formulas
(x – h)² + (y – k)² = r²
Centre: (h, k) | Radius: r
For x² + y² + Dx + Ey + F = 0:
Centre = (–D/2, –E/2), Radius = √(D²/4 + E²/4 – F)
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IMAGE SPACE — Circle diagram

Add diagram showing circle with centre C(h,k), point P(x,y) on circumference, and radius r = CP
📝 MCQ Practice

Circle — Test Yourself

1 Find the centre and radius of the circle x² + y² + 8x + 10y – 8 = 0.
  • A Centre (4, 5), r = 7
  • B Centre (–4, –5), r = 7
  • C Centre (4, –5), r = 7
  • D Centre (–4, 5), r = 49
Solution: (x² + 8x) + (y² + 10y) = 8
(x + 4)² – 16 + (y + 5)² – 25 = 8
(x + 4)² + (y + 5)² = 49 = 7²
Centre = (–4, –5), Radius = 7 ✓
2 The equation of circle with centre (–3, 2) and radius 4 is:
  • A (x+3)² + (y–2)² = 16
  • B (x–3)² + (y+2)² = 16
  • C (x+3)² + (y–2)² = 4
  • D (x–3)² + (y–2)² = 16
Solution: Standard form: (x – h)² + (y – k)² = r²
h = –3, k = 2, r = 4 → r² = 16
Equation: (x + 3)² + (y – 2)² = 16 ✓
3 Does the point (–2.5, 3.5) lie inside, outside or on the circle x² + y² = 25?
  • A Outside
  • B On the circle
  • C Inside
  • D Cannot determine
Solution: Distance² from origin = (–2.5)² + (3.5)² = 6.25 + 12.25 = 18.5
Radius² = 25
Since 18.5 < 25, the point lies inside the circle ✓
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2
Parabola

📌 Definition

A parabola is the set of all points equidistant from a fixed line (directrix) and a fixed point (focus). The line through the focus perpendicular to the directrix is the axis. The intersection with the axis is the vertex.

EquationOpensFocusDirectrixLatus Rectum
y² = 4ax→ Right(a, 0)x = –a4a
y² = –4ax← Left(–a, 0)x = a4a
x² = 4ay↑ Up(0, a)y = –a4a
x² = –4ay↓ Down(0, –a)y = a4a
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IMAGE SPACE — Four standard parabolas

Add diagram showing all 4 orientations: y²=4ax (right), y²=–4ax (left), x²=4ay (up), x²=–4ay (down)
💡 Quick Rule: If the equation has y², the axis is along the x-axis. If it has x², the axis is along the y-axis. Positive coefficient → opens right/up. Negative → opens left/down.
📝 MCQ Practice

Parabola — Test Yourself

4 Find the focus, directrix, and latus rectum length of the parabola y² = 8x.
  • A Focus (4,0), directrix x=–4, LR=16
  • B Focus (2,0), directrix x=–2, LR=8
  • C Focus (8,0), directrix x=–8, LR=32
  • D Focus (2,0), directrix x=2, LR=8
Solution: y² = 8x → compare with y² = 4ax: 4a = 8 → a = 2
Focus = (2, 0); Directrix: x = –2; Latus rectum = 4a = 8 ✓
5 Find the equation of the parabola with vertex (0,0), focus at (0, 2).
  • A y² = 8x
  • B x² = 4y
  • C x² = 8y
  • D y² = 4x
Solution: Focus (0, 2) is on y-axis → axis is y-axis → form x² = 4ay
a = 2 → x² = 4(2)y = x² = 8y ✓
6 A parabola is symmetric about the y-axis and passes through (2, –3). Its equation is:
  • A x² = –4y
  • B x² = 4y
  • C 3x² = –4y
  • D 3x² = 4y
Solution: Symmetric about y-axis, vertex at origin → x² = ±4ay
Point (2, –3) is in 4th quadrant → parabola opens downward → x² = –4ay
Substituting: 4 = –4a(–3) = 12a → a = 1/3
Equation: x² = –4(1/3)y → 3x² = –4y ✓
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3
Ellipse

📌 Definition

An ellipse is the set of all points P such that the sum of distances from P to two fixed points (foci F₁ and F₂) is constant: PF₁ + PF₂ = 2a.

🎯 Standard Equations of Ellipse
x²/a² + y²/b² = 1   (a > b; major axis along x-axis)

x²/b² + y²/a² = 1   (a > b; major axis along y-axis)
For both: c² = a² – b² | e = c/a | Latus rectum = 2b²/a
Foci: (±c, 0) for x-axis major | (0, ±c) for y-axis major
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IMAGE SPACE — Ellipse with major & minor axes

Add diagram: ellipse with centre O, foci F₁ and F₂, vertices A and B, semi-major axis a, semi-minor axis b
📝 MCQ Practice

Ellipse — Test Yourself

7 For the ellipse x²/25 + y²/9 = 1, find the foci, eccentricity, and latus rectum.
  • A Foci (±3,0), e=3/5, LR=18/5
  • B Foci (±4,0), e=4/5, LR=18/5
  • C Foci (±5,0), e=1, LR=9/5
  • D Foci (±4,0), e=4/5, LR=9/5
Solution: a² = 25 → a = 5; b² = 9 → b = 3
c = √(a²–b²) = √(25–9) = √16 = 4 → Foci (±4, 0)
e = c/a = 4/5 → e = 4/5
Latus rectum = 2b²/a = 2(9)/5 = 18/5 ✓
8 The equation of ellipse with vertices (±13, 0) and foci (±5, 0) is:
  • A x²/25 + y²/169 = 1
  • B x²/144 + y²/169 = 1
  • C x²/169 + y²/144 = 1
  • D x²/25 + y²/144 = 1
Solution: Vertices on x-axis → form x²/a² + y²/b² = 1
a = 13 → a² = 169; c = 5 → c² = 25
b² = a² – c² = 169 – 25 = 144
Equation: x²/169 + y²/144 = 1 ✓
9 For 9x² + 4y² = 36, find the eccentricity.
  • A √5/2
  • B √5/3
  • C 2/3
  • D 1/3
Solution: 9x²+4y²=36 → x²/4 + y²/9 = 1
a² = 9 (larger) → a = 3 (major axis along y-axis); b² = 4 → b = 2
c = √(a²–b²) = √(9–4) = √5
e = c/a = √5/3 ✓
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4
Hyperbola

📌 Definition

A hyperbola is the set of all points P such that the difference of distances from P to two fixed foci is constant: |PF₁ – PF₂| = 2a.

🎯 Standard Equations of Hyperbola
x²/a² – y²/b² = 1   (transverse axis along x-axis)

y²/a² – x²/b² = 1   (transverse axis along y-axis)
For both: c² = a² + b² | e = c/a > 1 | Latus rectum = 2b²/a
Foci: (±c, 0) for x-axis | (0, ±c) for y-axis form
Vertices: (±a, 0) for x-axis | (0, ±a) for y-axis form
⚠️ Key Difference from Ellipse: Ellipse: c² = a² – b² | Hyperbola: c² = a² + b². The positive term in the equation determines which axis the transverse axis is on.
🖼️

IMAGE SPACE — Hyperbola diagram

Add diagram: hyperbola x²/a² – y²/b² = 1 with foci F₁ and F₂, vertices, transverse and conjugate axes labeled
📝 MCQ Practice

Hyperbola — Test Yourself

10 For the hyperbola x²/9 – y²/16 = 1, find the foci, eccentricity, and latus rectum.
  • A Foci (±5,0), e=5/3, LR=32/3
  • B Foci (±4,0), e=4/3, LR=16/3
  • C Foci (±5,0), e=5/4, LR=9/2
  • D Foci (±3,0), e=1, LR=32/3
Solution: a² = 9 → a = 3; b² = 16 → b = 4
c = √(a²+b²) = √(9+16) = √25 = 5 → Foci (±5, 0)
e = c/a = 5/3 → e = 5/3
LR = 2b²/a = 2(16)/3 = 32/3 ✓
11 Find the equation of the hyperbola with foci (0, ±3) and vertices (0, ±√11/2).
  • A y²/4 – x²/25 = 1
  • B 100y² – 44x² = 275
  • C y²/9 – x²/4 = 1
  • D 4y² – 11x² = 44
Solution: Foci on y-axis → form y²/a² – x²/b² = 1
a = √11/2 → a² = 11/4; c = 3 → c² = 9
b² = c² – a² = 9 – 11/4 = 25/4
y²/(11/4) – x²/(25/4) = 1 → 100y² – 44x² = 275 ✓
12 The eccentricity of a hyperbola is always:
  • A Equal to 1
  • B Between 0 and 1
  • C Greater than 1
  • D Equal to 0
Solution: For hyperbola: c² = a² + b², so c > a always
Therefore e = c/a > 1 always.
(Circle: e=0; Ellipse: 0<e<1; Parabola: e=1; Hyperbola: e>1) ✓
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Quick Comparison
PropertyCircleParabolaEllipseHyperbola
Eccentricitye = 0e = 10 < e < 1e > 1
Standard Eqnx²+y²=r²y²=4axx²/a²+y²/b²=1x²/a²–y²/b²=1
Key Relationr = radiusPF = PBc²=a²–b²c²=a²+b²
Latus Rectum2r (diameter)4a2b²/a2b²/a
FociCentre(a, 0)(±c, 0)(±c, 0)
ApplicationWheels, ClocksParabolic mirror, Satellite dishPlanetary orbitsRadio navigation
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