Class 11 Maths · Chapter 10
Short answer: Class 11 Maths Chapter 10, "Conic Sections," coordinate geometry ka wo chapter hai jisme ek hi cone ko alag-alag angles pe cut karke circle, parabola, ellipse aur hyperbola nikalte hain. Iska pura structure predictable hai: har conic ke liye ek definition, ek standard equation, aur usse focus/directrix/eccentricity/latus rectum nikaalne ka fixed tareeka — isliye ye scoring chapter maana jaata hai agar 4 formulas clean yaad ho. In NCERT solutions mein Exercise 10.1 se 10.4 tak aur Miscellaneous Exercise ke saare question-types step-by-step CBSE marking style mein cover kiye gaye hain, jaisa baaki class 11 maths ncert solutions mein hota hai.
Class 9-10 Coordinate Geometry mein humne sirf seedhi lines aur points padhe the. Class 11 Chapter 9 "Straight Lines" (dekho class 11 maths straight lines ncert solutions) ne is line-based geometry ko continue kiya. Chapter 10 "Conic Sections" ab curve-based geometry introduce karta hai — aur ye Class 11 Maths ke coordinate geometry unit ka sabse visually rich chapter hai.
Idea simple hai: ek right circular double cone lo, aur usse ek plane se alag-alag angles pe kaato. Depending on cutting angle, cross-section circle, ellipse, parabola, ya hyperbola ban sakta hai — ye chaaron "conic sections" kehlaate hain kyunki ye sab cone se aate hain. Chapter phir har ek conic ki apni algebraic definition deta hai (ek fixed point se distance ki condition ke through) aur unse ek clean standard equation derive karta hai.
Jo cheez ye chapter ko exam-friendly banati hai wo hai uska pattern: circle ke liye centre-radius, parabola ke liye focus-directrix-axis-latus rectum, ellipse aur hyperbola ke liye foci-vertices-eccentricity-latus rectum — chaaron conics same "find these 4-5 properties from the equation" ya "in properties se equation banao" ke around ghoomte hain. Trigonometric ratios wale angle-based reasoning (jo class 11 maths chapter 3 trigonometric functions ncert solutions mein aata hai) yahan bhi kabhi-kabhi kaam aata hai, jaise equilateral-triangle-in-parabola type miscellaneous questions mein.
Ye chapter directly agle chapter "Introduction to Three Dimensional Geometry" aur baad mein "Limits and Derivatives" (dekho class 11 maths limits and derivatives ncert solutions) ki neev banata hai, kyunki curve-equations se properties nikaalne ki practice yahin se shuru hoti hai. Official class 11 maths syllabus 2026-27 pdf aur class 11 maths ncert book pdf download ncert.nic.in par verify kiya ja sakta hai — is 14-chapter rationalised syllabus mein Conic Sections Chapter 10 hai.
Chapter 10 Summary — 5 Minute Revision
Chapter 10 Conic Sections ek cone ko cut karke 4 curves banata hai — circle, parabola, ellipse, hyperbola — aur har ek ki apni fixed-distance-based definition, standard equation(s), aur unse nikalne wali properties (focus, directrix, axis, eccentricity, vertices, latus rectum) sikhata hai. Circle sabse simple hai (centre-radius), parabola ke 4 orientations hote hain (e = 1 hamesha), ellipse aur hyperbola dono "sum/difference of distances constant" definition se aate hain aur inke a, b, c ka relationship (a²=b²+c² vs c²=a²+b²) hi inhe differentiate karta hai. Miscellaneous Exercise real-life applications (arch bridge, rod locus) tak extend karti hai. Poori chapter ek hi skill test karti hai baar-baar: equation ↔ geometric properties ke beech aana-jaana, jo ise high-scoring banata hai agar formulas clean rakhe jaayein.
In-Text Questions — Solutions
Exercise 10.1, Circle — Q1. Centre (0, 2) aur radius 2 wale circle ka equation likho.
Circle ka standard equation: (x − h)² + (y − k)² = r²
Yahan h = 0, k = 2, r = 2:
x² + (y − 2)² = 4
x² + y² − 4y + 4 = 4
∴ x² + y² − 4y = 0
Exercise 10.1, Circle — Q2. Centre (−2, 3) aur radius 4 wale circle ka equation likho.
(x + 2)² + (y − 3)² = 16
Expand karke:
x² + 4x + 4 + y² − 6y + 9 = 16
∴ x² + y² + 4x − 6y − 3 = 0
Exercise 10.1, Circle — Q3. Circle x² + y² − 8x + 10y − 12 = 0 ka centre aur radius nikaalo.
General form x² + y² + 2gx + 2fy + c = 0 se compare karo:
2g = −8 ⇒ g = −4; 2f = 10 ⇒ f = 5; c = −12
Centre = (−g, −f) = (4, −5)
Radius = √(g² + f² − c) = √(16 + 25 + 12) = √53
Exercise 10.1, Circle — Q4. Un points (4, 1) aur (6, 5) se guzarne wale circle ka equation nikaalo jiska centre line 4x + y = 16 par hai.
Centre (h, k) dono points se equidistant hoga (dono radius hain):
(h − 4)² + (k − 1)² = (h − 6)² + (k − 5)²
Expand aur simplify karne par:
h + 2k = 11 ...(i)
Centre line par hai: 4h + k = 16 ...(ii)
(ii) se k = 16 − 4h, (i) mein daalo: h + 2(16 − 4h) = 11 ⇒ −7h = −21 ⇒ h = 3, k = 4
Radius² = (3−4)² + (4−1)² = 1 + 9 = 10
∴ (x − 3)² + (y − 4)² = 10 ⇒ x² + y² − 6x − 8y + 15 = 0
Exercise 10.1, Circle — Q5. Radius 5 wala circle, jiska centre x-axis par hai aur jo point (2, 3) se guzarta hai — equation nikaalo.
Centre x-axis par hai, so centre = (h, 0). Point (2, 3) se distance = 5:
(h − 2)² + 9 = 25 ⇒ (h − 2)² = 16 ⇒ h = 6 ya h = −2
Do possible circles:
(x − 6)² + y² = 25 ya (x + 2)² + y² = 25
Exercise 10.2, Parabola — Q6. y² = 12x ke liye focus, axis, directrix, aur latus rectum ki length nikaalo.
Compare with y² = 4ax: 4a = 12 ⇒ a = 3
Focus = (3, 0); Axis: y = 0 (x-axis); Directrix: x = −3
Latus rectum = 4a = 12
Exercise 10.2, Parabola — Q7. x² = −16y ke liye focus, directrix aur latus rectum nikaalo.
Compare with x² = −4ay: 4a = 16 ⇒ a = 4
Focus = (0, −4); Axis: x = 0; Directrix: y = 4
Latus rectum = 4a = 16
Exercise 10.2, Parabola — Q8. Focus (2, 0) aur directrix x = −2 wale parabola ka equation likho.
Focus positive x-axis par hai aur directrix x = −a form mein hai, so parabola right-opening hai: a = 2
y² = 4ax = 4(2)x = 8x
Exercise 10.2, Parabola — Q9. Vertex (0,0), axis x-axis ke saath, aur point (2, 3) se guzarne wale parabola ka equation nikaalo.
Form: y² = 4ax. Point (2,3) satisfy karega:
9 = 4a(2) = 8a ⇒ a = 9/8
∴ y² = 4(9/8)x ⇒ 2y² = 9x
Exercise 10.2, Parabola — Q10. Focus (0, −3) aur directrix y = 3 wale parabola ka equation likho.
Parabola neeche khulta hai (downward): a = 3
x² = −4ay = −12y
Exercise 10.3, Ellipse — Q11. x²/36 + y²/16 = 1 ke liye foci, vertices, eccentricity, axes ki length aur latus rectum nikaalo.
a² = 36, b² = 16 ⇒ a = 6, b = 4 (a > b, so major axis x-axis par hai)
c² = a² − b² = 36 − 16 = 20 ⇒ c = 2√5
Foci = (±2√5, 0); Vertices = (±6, 0)
e = c/a = 2√5/6 = √5/3
Major axis = 12, Minor axis = 8
Latus rectum = 2b²/a = 2(16)/6 = 16/3
Exercise 10.3, Ellipse — Q12. x²/4 + y²/25 = 1 ke liye foci, vertices aur eccentricity nikaalo.
y² ke neeche bada denominator (25 > 4) hai, so major axis y-axis par hai: a = 5, b = 2
c² = a² − b² = 25 − 4 = 21 ⇒ c = √21
Foci = (0, ±√21); Vertices = (0, ±5)
e = √21/5; Latus rectum = 2b²/a = 8/5
Exercise 10.3, Ellipse — Q13. Vertices (±5, 0) aur foci (±4, 0) wale ellipse ka equation nikaalo.
a = 5, c = 4
b² = a² − c² = 25 − 16 = 9
∴ x²/25 + y²/9 = 1
Exercise 10.3, Ellipse — Q14. Vertices (0, ±13) aur foci (0, ±5) wale ellipse ka equation likho.
Major axis y-axis par hai: a = 13, c = 5
b² = a² − c² = 169 − 25 = 144
∴ x²/144 + y²/169 = 1
Exercise 10.3, Ellipse — Q15. Major axis ki length 20 hai aur foci (0, ±5) hain — ellipse ka equation nikaalo.
2a = 20 ⇒ a = 10; c = 5
b² = a² − c² = 100 − 25 = 75
Foci y-axis par hain, so major axis y-axis par:
x²/75 + y²/100 = 1
Exercise 10.4, Hyperbola — Q16. x²/16 − y²/9 = 1 ke liye eccentricity, foci, vertices, latus rectum nikaalo.
a² = 16, b² = 9 ⇒ a = 4, b = 3
c² = a² + b² = 16 + 9 = 25 ⇒ c = 5
Foci = (±5, 0); Vertices = (±4, 0)
e = c/a = 5/4
Latus rectum = 2b²/a = 18/4 = 9/2
Exercise 10.4, Hyperbola — Q17. y²/9 − x²/27 = 1 ke liye eccentricity aur foci nikaalo.
a² = 9, b² = 27 ⇒ a = 3, b = 3√3
c² = a² + b² = 9 + 27 = 36 ⇒ c = 6
Foci = (0, ±6); Vertices = (0, ±3)
e = c/a = 6/3 = 2
Latus rectum = 2b²/a = 2(27)/3 = 18
Exercise 10.4, Hyperbola — Q18. Vertices (±2, 0) aur foci (±3, 0) wale hyperbola ka equation nikaalo.
a = 2, c = 3
b² = c² − a² = 9 − 4 = 5
∴ x²/4 − y²/5 = 1
Exercise 10.4, Hyperbola — Q19. Foci (0, ±13) aur conjugate axis ki length 24 wale hyperbola ka equation nikaalo.
c = 13; conjugate axis 2b = 24 ⇒ b = 12
a² = c² − b² = 169 − 144 = 25 ⇒ a = 5
Foci y-axis par hain:
y²/25 − x²/144 = 1
Exercise 10.4, Hyperbola — Q20. Vertices (0, ±5) aur foci (0, ±8) wale hyperbola ka equation nikaalo.
a = 5, c = 8
b² = c² − a² = 64 − 25 = 39
∴ y²/25 − x²/39 = 1
Poore Class 11 Maths ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q10)
Miscellaneous Exercise — Q1 (Application). Ek arch semi-ellipse ke shape mein hai. Wo 8 m wide hai aur centre par 2 m high hai. Ek end se 1.5 m door arch ki height nikaalo.
Semi-major axis a = 4 (half of width), semi-minor axis b = 2 (centre height). Upper half ellipse:
x²/16 + y²/4 = 1, y ≥ 0
Ek end ko x = −4 lo. 1.5 m door point: x = −4 + 1.5 = −2.5
(−2.5)²/16 + y²/4 = 1 ⇒ 6.25/16 + y²/4 = 1 ⇒ y²/4 = 0.609375
y² = 2.4375 ⇒ y ≈ 1.56 m
Miscellaneous Exercise — Q2 (Application). 12 cm lambi ek rod ke dono ends coordinate axes par slide karte hain. Rod par ek point P, jo x-axis wale end se 3 cm door hai, uska locus nikaalo.
A = (a, 0) x-axis par, B = (0, b) y-axis par, AB = 12 ⇒ a² + b² = 144
P, A se 3 cm aur B se 9 cm door hai, so P divides AB in ratio 3:9 = 1:3 from A:
P = A + (3/12)(B − A) = (3a/4, b/4)
So x = 3a/4 ⇒ a = 4x/3; y = b/4 ⇒ b = 4y
(4x/3)² + (4y)² = 144 ⇒ 16x²/9 + 16y² = 144
144 se divide karo: x²/9 + y² = 9, phir 9 se divide:
∴ x²/81 + y²/9 = 1
Miscellaneous Exercise — Q3. Parabola y² = 4ax mein ek equilateral triangle inscribe kiya gaya hai jiska ek vertex parabola ke vertex par hai. Triangle ke side ki length nikaalo.
Vertex O(0,0) par, doosre do vertices (x₁,y₁) aur (x₁,−y₁), symmetric about x-axis. Apex angle 60° hai, so har side x-axis se 30° banata hai:
tan 30° = y₁/x₁ ⇒ y₁ = x₁/√3
Parabola par point hone se y₁² = 4a x₁:
x₁²/3 = 4a x₁ ⇒ x₁ = 12a, y₁ = 12a/√3 = 4a√3
Side length = 2y₁:
∴ Side = 8a√3
Miscellaneous Exercise — Q4. Foci (±2, 0) aur eccentricity 1/2 wale ellipse ka equation nikaalo.
c = 2, e = c/a = 1/2 ⇒ a = 4
b² = a² − c² = 16 − 4 = 12
∴ x²/16 + y²/12 = 1
Miscellaneous Exercise — Q5. Points (2, 3) aur (−1, 1) se guzarne wale circle ka equation nikaalo, jiska centre line x − 3y − 11 = 0 par hai.
Centre (h, k), dono points se equal distance:
(h−2)² + (k−3)² = (h+1)² + (k−1)²
Simplify karne par: 6h + 4k = 11 ...(i)
Line se: h = 3k + 11 ...(ii)
(ii) ko (i) mein daalo: 6(3k+11) + 4k = 11 ⇒ 22k = −55 ⇒ k = −5/2, h = 7/2
Radius² = (7/2 − 2)² + (−5/2 − 3)² = 9/4 + 121/4 = 65/2
∴ (x − 7/2)² + (y + 5/2)² = 65/2
Miscellaneous Exercise — Q6. Vertices (±7, 0) aur e = 4/3 wale hyperbola ka equation nikaalo.
a = 7, e = c/a = 4/3 ⇒ c = 28/3
b² = c² − a² = (784/9) − 49 = 343/9
∴ x²/49 − 9y²/343 = 1
Miscellaneous Exercise — Q7 (Application). Ek race-course par daudte hue admi note karta hai ki do flag-posts se uski distances ka sum hamesha 10 m rehta hai, aur posts ke beech distance 8 m hai. Uske path ka equation nikaalo.
Ye ellipse ki classic definition hai (sum of distances from two foci = constant):
2a = 10 ⇒ a = 5; 2c = 8 ⇒ c = 4
b² = a² − c² = 25 − 16 = 9
∴ x²/25 + y²/9 = 1
Miscellaneous Exercise — Q8. Origin se guzarne wale aur x-axis, y-axis par respectively a, b intercepts banane wale circle ka equation nikaalo.
General form: x² + y² + 2gx + 2fy + c = 0. Origin (0,0) se guzarta hai ⇒ c = 0
Point (a, 0) se: a² + 2ga = 0 ⇒ g = −a/2
Point (0, b) se: b² + 2fb = 0 ⇒ f = −b/2
∴ x² + y² − ax − by = 0
Miscellaneous Exercise — Q9. Parabola x² = 6y ke liye latus rectum aur directrix nikaalo.
Compare with x² = 4ay: 4a = 6 ⇒ a = 3/2
Focus = (0, 3/2); Directrix: y = −3/2
Latus rectum = 4a = 6
Miscellaneous Exercise — Q10. Foci (0, ±3) aur semi-minor axis 4 wale ellipse ka equation nikaalo.
Foci y-axis par, so semi-major axis bhi y-axis direction mein hai. Semi-minor b = 4, c = 3:
a² = b² + c² = 16 + 9 = 25
∴ x²/16 + y²/25 = 1
Important Equations — Ek Nazar Me
| Conic | Standard Equation(s) | Key Relations |
|---|---|---|
| Circle | (x−h)² + (y−k)² = r² General: x²+y²+2gx+2fy+c=0 | Centre = (h,k) ya (−g,−f) Radius = r ya √(g²+f²−c) |
| Parabola (4 forms) | y²=4ax (right) | y²=−4ax (left) x²=4ay (up) | x²=−4ay (down) | Focus at distance a from vertex Directrix at distance a on opposite side Latus rectum = 4a | e = 1 always |
| Ellipse | x²/a² + y²/b² = 1 | a > b ⇒ major axis x-axis, c²=a²−b² b > a ⇒ major axis y-axis, c²=a²−b² (a here = larger) e = c/a (0 < e < 1) Latus rectum = 2b²/a |
| Hyperbola | x²/a² − y²/b² = 1 (foci on x-axis) y²/a² − x²/b² = 1 (foci on y-axis) | c² = a² + b² e = c/a (e > 1) Latus rectum = 2b²/a |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Parabola ke 4 standard forms (y²=4ax, y²=−4ax, x²=4ay, x²=−4ay) mix kar dena — isse axis aur opening-direction dono galat ho jaate hain.
- Ellipse mein bina check kiye maan lena ki major axis x-axis par hai — real rule ye hai ki jis variable ke neeche bada denominator hai, major axis usi axis ke saath hoga.
- Ellipse ka c²=a²−b² aur hyperbola ka c²=a²+b² formula aapas mein swap kar dena.
- Circle ke general equation x²+y²+2gx+2fy+c=0 se centre nikaalte waqt sign bhool jaana — centre (−g,−f) hota hai, (g,f) nahi.
- Agar circle equation mein x², y² ka coefficient 1 nahi hai (jaise 2x²+2y²−x=0), pehle poori equation ko us coefficient se divide na karna, jisse g, f, c galat nikalte hain.
- Latus rectum ka formula confuse kar dena — parabola ke liye 4a hota hai, ellipse/hyperbola ke liye 2b²/a, aur a, b interchange karne se answer double ya half ho jaata hai.
Board-Style Important Questions
- 1 mark: Parabola y² = 8x ki eccentricity nikaalo.
Solution: Har parabola ki eccentricity hamesha e = 1 hoti hai (parabola ki hi definition e = 1 wale conic ke roop mein hoti hai), chahe y²=4ax mein a ki value kuch bhi ho. - 2 marks: Circle x² + y² + 6x − 4y − 12 = 0 ka centre aur radius nikaalo.
Solution: x²+y²+2gx+2fy+c=0 se compare: 2g=6⇒g=3; 2f=−4⇒f=−2; c=−12. Centre=(−3,2), radius=√(9+4+12)=√25=5. - 3 marks: Vertex origin par, axis y-axis ke saath, aur point (2,−3) se guzarne wale parabola ka equation nikaalo.
Solution: Negative y wale point se guzarta hai, so parabola neeche khulta hai: x²=−4ay. (2,−3) daalo: 4=−4a(−3)=12a ⇒ a=1/3. Equation: 3x²=−4y. - 4 marks: Ellipse 9x² + 4y² = 36 ke foci, vertices, eccentricity aur latus rectum nikaalo.
Solution: 36 se divide: x²/4+y²/9=1. a²=9,b²=4 (major axis y-axis, kyunki 9>4), a=3,b=2. c²=9−4=5, c=√5. Foci(0,±√5), Vertices(0,±3), e=√5/3, LR=2(4)/3=8/3. - 4 marks: Foci (0, ±√10) wale hyperbola ka equation nikaalo jo point (2,3) se guzarta hai.
Solution: y²/a² − x²/b²=1, c²=a²+b²=10. Point daalo: 9/a² − 4/b²=1, b²=10−a². Solve karne par u=a² ke liye u²−23u+90=0 ⇒ u=5 ya 18; c²=10 se sirf u=5 valid hai (b²=5). Equation: y²/5 − x²/5 = 1. - 6 marks: Ek semi-elliptical arch 8 m wide aur centre par 2 m high hai. Ek end se 1.5 m door height nikaalo.
Solution: a=4, b=2, equation x²/16+y²/4=1 (y≥0). End x=−4 se 1.5m door: x=−2.5. (6.25/16)+y²/4=1 ⇒ y²=2.4375 ⇒ y≈1.56 m.
Aksar Poochhe Jaane Wale Sawaal
Class 11 Maths Chapter 10 Conic Sections mein kitne exercises hain?
Chapter mein 4 numbered exercises hain — 10.1 (Circle), 10.2 (Parabola), 10.3 (Ellipse), 10.4 (Hyperbola) — aur chapter ke end mein ek Miscellaneous Exercise jo application-based aur mixed questions cover karti hai. Puri chapter-wise breakdown ke liye baaki class 11 maths ncert solutions bhi isi pattern follow karte hain.
Conic Sections ka sabse important formula kaun sa hai jo exam mein baar-baar aata hai?
Sabse zyada use hone wale formulas hain: parabola ka latus rectum = 4a, ellipse/hyperbola ka latus rectum = 2b²/a, aur eccentricity e = c/a (parabola ke liye e hamesha 1 hota hai). In sab ka ek consolidated table upar formulas section mein diya gaya hai — quick revision ke liye ise class 11 maths formulas pdf download resource ke saath cross-check karo.
Conic Sections chapter 3 Trigonometric Functions se kaise related hai?
Kuch miscellaneous-type questions (jaise parabola mein inscribed equilateral triangle) angle-based reasoning use karte hain, jahan tan/sin ratios se coordinates nikalte hain — wahi foundational skills jo class 11 maths chapter 3 trigonometric functions ncert solutions mein build hoti hain.
Class 11 Maths ke 2026-27 syllabus mein Conic Sections chapter number kya hai?
Current rationalised NCERT textbook (2026-27 session) mein Conic Sections Chapter 10 hai. Ye ncert.nic.in par available official class 11 maths syllabus 2026-27 pdf aur class 11 maths ncert book pdf download se directly verify kiya ja sakta hai — purane 16-chapter edition ki numbering (jahan Induction/Reasoning chapters bhi the) ab apply nahi hoti.
Exam preparation ke liye Conic Sections ke extra important questions kahan milenge?
Iss chapter ke saath practice ke liye class 11 maths important questions with solutions aur class 11 maths exemplar solutions dono useful hain, kyunki exemplar mein thode tougher application-type conic problems hote hain. Quick revision ke liye class 11 maths conic sections notes bhi saath rakho.
Conic Sections ka board exam mein weightage kitna hota hai?
Exact marks-weightage yahan claim nahi kar rahe — different sources alag-alag numbers dikhate hain aur bina official CBSE curriculum document verify kiye figure fabricate karna galat hoga. Coordinate Geometry cluster (Straight Lines + Conic Sections) generally ek solid-weightage unit maana jaata hai — precise number ke liye current class 11 maths chapter wise weightage 2026-27 CBSE ki official curriculum PDF se check karo.
Class 11 Maths — Saare Chapters
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