NCERT Plus One (Class 11) Mathematics • Chapter 10

Conic Sections

Q: Find the coordinates of the focus and the equation of the directrix of the parabola x² = 8y.

Compare with x² = 4ay : 4a = 8 ⇒ a = 2 . • Focus: (0, 2) • Directrix: y = −2 .

Q: Find the centre and radius of the circle x² + y² − 8x + 10y − 12 = 0.

Method 1 (Completing Square): (x² − 8x + 16) + (y² + 10y + 25) = 12 + 16 + 25 (x − 4)² + (y + 5)² = 53 . • Centre: (4, −5) , Radius: √53 .

Q: Find the foci, vertices, lengths of major and minor axes, eccentricity, and latus rectum of 9x² + 4y² = 36.

Divide by 36: x² 4 + y² 9 = 1 . Bigger denominator 9 is under y² → Vertical Ellipse: a = 3, b = 2 . c = √(9 − 4) = √5 . • Foci: (0, ±√5) , Vertices: (0, ±3) • Major axis = 6 , Minor axis = 4 , e = √5/3 , LR = 2(4)/3 = 8/3 .

Q: Find the foci, vertices, eccentricity, and length of latus rectum of the hyperbola 16x² − 9y² = 576.

Divide by 576: x² 36 − y² 64 = 1 . Positive term is x² → a² = 36 (a = 6) , b² = 64 (b = 8) . c = √(36 + 64) = √100 = 10 . • Vertices: (±6, 0) , Foci: (±10, 0) • Eccentricity: e = 10/6 = 5/3 , LR = 2(64)/6 = 64/3 .

Q: Find the equation of the circle passing through (4, 1) and (6, 5) whose centre lies on the line 4x + y = 16.

Let centre be (h, k) : 4h + k = 16 ... (1). Equal radii: (h − 4)² + (k − 1)² = (h − 6)² + (k − 5)² −8h + 16 − 2k + 1 = −12h + 36 − 10k + 25 ⇒ 4h + 8k = 44 ⇒ h + 2k = 11 ... (2). Solving (1) & (2): 7k = 28 ⇒ k = 4, h = 3 . r² = (3 − 4)² + (4 − 1)² = 1 + 9 = 10 . • Equation: (x − 3)² + (y − 4)² = 10 .

Q: Find the equation of an ellipse with foci (±3, 0) and semi-major axis 4.

Foci on x-axis → Horizontal Ellipse x²/a² + y²/b² = 1 . Given c = 3 , a = 4 . b² = a² − c² = 16 − 9 = 7 . • Equation: x² 16 + y² 7 = 1 .

Q: A parabolic reflector has a diameter of 20 cm and depth of 5 cm. Find the position of the focus.

Place vertex at (0, 0) opening right along X-axis: y² = 4ax . At depth x = 5 , diameter is 20 ⇒ y = ±10 , so point (5, 10) lies on parabola. 10² = 4a(5) ⇒ 100 = 20a ⇒ a = 5 cm . • Focus is at (5, 0) — right at the midpoint of the circular rim!

Slicing double cones into circles, parabolas, ellipses & hyperbolas — equations, foci, eccentricity & real-world orbits.

📄 Official NCERT Textbook PDF

Topic 10.2: How Slicing a Double Cone Makes Circle, Ellipse, Parabola & Hyperbola

Imagine two ice-cream cones balanced tip-to-tip like an hourglass! If you slice the top cone straight across horizontally, you get a neat round Circle (like slicing a carrot straight!). Tilt your knife slightly and you get an oval Ellipse. Tilt it parallel to the slanted cone wall and it shoots off forever as a Parabola. Slice steeply through BOTH cones at once and you get twin open curves — a Hyperbola!

DOUBLE CONE ANGLESα = cone half-angle with axis;
β = slicing plane angle with axis

4 NON-DEGENERATE CONICSβ = 90° → Circle | α < β < 90° → Ellipse
β = α → Parabola | 0 ≤ β < α → Hyperbola

DEGENERATE (AT VERTEX V)α < β ≤ 90° → Point | β = α → 1 Line
0 ≤ β < α → Pair of Intersecting Lines

🧠 Memory Trick: Watch the knife angle β drop from 90° down to 0°: 90° = Circle → Tilted = Ellipse → Matches Wall (β = α) = Parabola → Cuts Both Cones (β < α) = Hyperbola!

💡 Simple Kid-Check: Picture a rotating slanted line (the generator) spinning around a vertical pole (the axis) at fixed angle α. That makes two hollow cones joined at one middle point called the vertex V (two nappes). Every conic curve is simply the shadow-outline where a flat knife cuts this cone!

1
Circle (β = 90°): Plane is completely perpendicular to the vertical axis — every point on the rim sits equidistant from the center.
2
Ellipse (α < β < 90°): Plane tilts slightly, cutting completely across one nappe in a closed stretched oval.
3
Parabola (β = α): Plane stays parallel to one generator line, so the cut never closes on the far side — it opens infinitely along one nappe!
4
Hyperbola (0 ≤ β < α): Plane is steeper than the cone wall, slicing BOTH upper and lower nappes to create two mirror-image branches!

Students often confuse α and β! Remember α is the fixed semi-vertical angle of the cone, while β is the angle the cutting plane makes with the vertical axis (NOT with the horizontal floor!). Also watch out for vertex cuts: if the plane passes right through vertex V, you get a degenerate conic (point, line, or two crossing lines).

Topic 10.3: Circle Standard & General Equations (Centre & Radius)

Tie a goat to a wooden peg at (h, k) with a tight rope of length r in a Kerala backyard. When the goat walks around keeping the rope stretched, every point (x, y) it steps on stays at exact distance r from (h, k)! Using Pythagoras' distance rule immediately gives the Circle Equation!

STANDARD FORM(x − h)² + (y − k)² = r²
Centre = (h, k), Radius = r

GENERAL FORMx² + y² + 2gx + 2fy + c = 0
Centre = (−g, −f), r = √(g² + f² − c)

FAST CENTRE RULEMake coeffs of x², y² equal to 1, then take −(half coeff of x), −(half coeff of y)!

🧠 Memory Trick: Standard form signs are backwards! Inside (x − 2)² + (y + 3)² = 25, the centre is (+2, −3) and radius is √25 = 5!

💡 Simple Kid-Check: A circle has ONLY TWO secrets: where is its centre (h, k), and how long is its radius r? Everything else in the equation is just Pythagoras squared: horizontal step (x − h) squared plus vertical step (y − k) squared equals r²!

Origin Centre: x² + y² = r² | General Centre/Radius: Divide so x², y² have coeff 1 → Centre (−g, −f), r = √(g² + f² − c)

🎯 Super-Easy 2-Method Example: Find centre & radius of x² + y² − 4x − 8y − 45 = 0.
• General Shortcut: Compare with x² + y² + 2gx + 2fy + c = 0 → 2g = −4 ⇒ g = −2; 2f = −8 ⇒ f = −4; c = −45.
• Centre = (−g, −f) = (2, 4).
• Radius r = √((-2)² + (-4)² − (−45)) = √(4 + 16 + 45) = √65!
• Completing Square Check: (x − 2)² + (y − 4)² = 45 + 4 + 16 = 65 ✓

Two classic Kerala HSE 1-mark traps! (1) Given 2x² + 2y² − x = 0, NEVER pick g = −1/2 directly! First divide the whole equation by 2 to get x² + y² − ½x = 0 → 2g = −½ ⇒ g = −¼, so Centre = (¼, 0) and r = ¼! (2) Always check g² + f² − c > 0: if it is 0 it's a single point, and if < 0 no real circle exists!

Topic 10.4: Parabola: Focus, Directrix, Axis & Latus Rectum (4 Orientations)

Look at a curved DTH dish antenna or a car headlight mirror! Parallel TV signals from the sky bounce off the curved surface and meet at ONE magic receiver point — the Focus F! Why? Because a Parabola is the exact curve where every point P sits equally far from a fixed point Focus F and a straight guide line called the Directrix!

DEFINITIONPF = PM (Focus dist = Directrix dist)
Eccentricity e = 1, Vertex (0, 0)

4 STANDARD FORMS (a > 0)y² = ±4ax (Left/Right)
x² = ±4ay (Down/Up)

LATUS RECTUMFocal chord ⊥ to axis.
Length ALWAYS = 4a!

🧠 Memory Trick: Look at the linear term (power 1)! If x is power 1 (y² = ±4ax), the parabola opens along the X-axis! If y is power 1 (x² = ±4ay), it opens along the Y-axis!

💡 Simple Kid-Check: Imagine the vertex (0, 0) sits right between the Focus F inside the bowl (distance a) and the flat Directrix wall behind the bowl (distance a on the opposite side). The parameter a is always positive (a > 0) and controls how wide the bowl opens!

Equation (a > 0)	Opens Towards	Axis	Focus F	Directrix	Latus Rectum
`y² = 4ax`	Right (+X)	`y = 0`	`(a, 0)`	`x = −a`	`4a`
`y² = −4ax`	Left (−X)	`y = 0`	`(−a, 0)`	`x = a`	`4a`
`x² = 4ay`	Upward (+Y)	`x = 0`	`(0, a)`	`y = −a`	`4a`
`x² = −4ay`	Downward (−Y)	`x = 0`	`(0, −a)`	`y = a`	`4a`

📡 Quick Example: For x² = −16y, compare with x² = −4ay:
4a = 16 ⇒ a = 4. Linear term is −y → opens Downward! Focus = (0, −4), Directrix = y = 4, Latus Rectum = 4a = 16!

Watch two frequent DHSE mistakes on y² = 10x: (1) Students write Focus = (10, 0) — WRONG! Set 4a = 10 ⇒ a = 5/2, so Focus is (5/2, 0) and Directrix is x = −5/2! (2) Latus Rectum length is 4a = 10 (the whole coefficient in front of the linear variable, NOT 4 × 10 = 40!).

Topic 10.5: Ellipse: Major/Minor Axes, Eccentricity e < 1 & Latus Rectum

Hammer two nails F₁ and F₂ into a wooden board, loop a loose string of length 2a around both nails, pull it tight with a pencil, and trace all around! Because the total string length PF₁ + PF₂ = 2a never changes, your pencil draws a perfect Ellipse — just like Earth orbiting the Sun!

GOLDEN ELLIPSE LINKPF₁ + PF₂ = 2a | a² = b² + c²
so c = √(a² − b²) (always a > b!)

ECCENTRICITY (e < 1)e = c / a (0 < e < 1)
Small e → round; e → 1 → flat!

AXES & LATUS RECTUMMajor = 2a, Minor = 2b
Latus Rectum = 2b² / a

🧠 Memory Trick: In an Ellipse, a is ALWAYS the BIGGER number (semi-major axis)! Look under x² and y² — whichever denominator is LARGER is a², and that axis holds the Foci and Vertices!

💡 Simple Kid-Check: Picture a stretched circle! The long stretch from centre to tip is a (Major Axis = 2a), the short squeeze is b (Minor Axis = 2b), and the two focal pins sit inside at distance c = √(a² − b²) from the centre. Because the pins are INSIDE the ellipse (c < a), eccentricity e = c/a is strictly less than 1!

Feature	Horizontal Ellipse (`x²/a² + y²/b² = 1`)	Vertical Ellipse (`x²/b² + y²/a² = 1`)
Bigger denominator	Under `x²` (`a > b`)	Under `y²` (`a > b`)
Vertices & Foci	Vertices `(±a, 0)`, Foci `(±c, 0)`	Vertices `(0, ±a)`, Foci `(0, ±c)`
Axes & Latus Rectum	Major `2a`, Minor `2b`, LR = `2b²/a`	Major `2a`, Minor `2b`, LR = `2b²/a`

🪐 Worked Example: Analyze x²25 + y²169 = 1.
Bigger denominator is 169 under y² → Vertical Ellipse! So a = 13, b = 5.
c = √(169 − 25) = √144 = 12 → Foci (0, ±12), Vertices (0, ±13), e = 12/13, LR = 2(25)/13 = 50/13!

NEVER assume a² always sits under x²! In x²/4 + y²/25 = 1, students often write a = 2, b = 5 and panic when a² − b² turns negative! Remember NCERT's golden convention: a is ALWAYS the larger semi-axis, so here a² = 25 (a = 5) and b² = 4 (b = 2) along a Vertical Ellipse!

Topic 10.6: Hyperbola: Transverse/Conjugate Axes & Eccentricity e > 1

Imagine two coastal radio towers F₁ and F₂ pinging a ship off Kochi. If the ship records a constant TIME DIFFERENCE between the two pings, then the DIFFERENCE of distances |PF₁ − PF₂| = 2a stays constant — placing the ship right on a two-branched Hyperbola!

GOLDEN HYPERBOLA LINK|PF₁ − PF₂| = 2a | c² = a² + b²
(PLUS sign! So c > a always!)

ECCENTRICITY (e > 1)e = c / a > 1
Equilateral (a = b) gives e = √2!

AXES & LATUS RECTUMTransverse = 2a, Conjugate = 2b
Latus Rectum = 2b² / a

🧠 Memory Trick: For a Hyperbola, follow the PLUS sign, NOT the bigger number! Whichever square term is positive (+x²/a² or +y²/a²) owns a², vertices, and foci!

💡 Simple Kid-Check: Swap the + between fractions in an Ellipse into a −, and the closed loop bursts open into two opposite wings! Notice how the focal distance formula flips too: Ellipse used c² = a² − b² (so c < a), while Hyperbola uses c² = a² + b² (so foci sit OUTSIDE the vertices, giving c > a and e = c/a > 1!).

Feature	Horizontal (`x²/a² − y²/b² = 1`)	Vertical (`y²/a² − x²/b² = 1`)
Positive term	`x²/a²` (cuts X-axis!)	`y²/a²` (cuts Y-axis!)
Vertices & Foci	Vertices `(±a, 0)`, Foci `(±c, 0)`	Vertices `(0, ±a)`, Foci `(0, ±c)`
`c`, `e` & LR	`c = √(a² + b²)`, `e = c/a > 1`, LR = `2b²/a`	`c = √(a² + b²)`, `e = c/a > 1`, LR = `2b²/a`

🚀 Classic Exam Example: Find foci & eccentricity of y²9 − x²27 = 1.
Positive term is y²/9 → Vertical Hyperbola! Even though 9 < 27, a² = 9 (a = 3) and b² = 27!
c = √(9 + 27) = √36 = 6 → Foci (0, ±6), Vertices (0, ±3), e = 6/3 = 2, LR = 2(27)/3 = 18!

Two giant Hyperbola traps! (1) Unlike an ellipse, a does NOT have to be larger than b! Look ONLY at which term has the PLUS sign (+y²/a² means vertical, even if a < b!). (2) Don't mix up c²: Ellipse has c² = a² − b² (MINUS), while Hyperbola has c² = a² + b² (PLUS)!

Topic 10.6: Quick Comparison Table & Real-Life Applications (Orbits & Dishes)

Zoom up into space! Every comet or spacecraft flying past our Sun obeys ONE unified conic rule controlled by its Eccentricity e: circular satellite orbit (e = 0), elliptical planet orbit like Earth (0 < e < 1), escape parabolic path (e = 1), or interstellar hyperbolic slingshot like Oumuamua (e > 1)!

ECCENTRICITY LADDERCircle: e = 0 | Ellipse: 0 < e < 1
Parabola: e = 1 | Hyperbola: e > 1

ELLIPSE vs HYPERBOLAEllipse: + in eq, c² = a² − b²
Hyperbola: − in eq, c² = a² + b²

REAL-WORLD TECHParabolic DTH dish & torch;
Elliptical orbits; Hyperbolic towers!

🧠 Memory Trick: Remember the opposite-sign rule! Ellipse equation has + inside, so its c² formula takes − (a² − b²). Hyperbola equation has − inside, so its c² formula takes + (a² + b²)!

💡 Simple Kid-Check: How do all four curves relate to real life? Put a bulb at a Parabola's focus → rays shoot straight out in a car headlight! Put a kidney stone at one Ellipse focus and a shockwave emitter at the other → sound bounces straight onto the stone without cutting skin (lithotripsy)!

Property	Circle	Parabola	Ellipse	Hyperbola
Standard Eq	`x²+y²=r²`	`y²=4ax`	`x²/a²+y²/b²=1`	`x²/a²−y²/b²=1`
Eccentricity `e`	`0`	`1`	`c/a < 1`	`c/a > 1`
Focal Relation	`c = 0`	`PF = PM`	`c² = a² − b²`	`c² = a² + b²`
Latus Rectum	Diameter `2r`	`4a`	`2b²/a`	`2b²/a`
Real Application	Wheels, gears	DTH dish, headlight	Planetary orbits	Cooling towers, LORAN

Fast 1-mark DHSE checklist: Given any equation Ax² + By² = C with positive C: if A = B > 0 → Circle; if A ≠ B and both > 0 → Ellipse; if A and B have opposite signs → Hyperbola; if one squared term is missing (only x² or only y²) → Parabola! Also remember: the Latus Rectum formula 2b²/a is IDENTICAL for both Ellipse and Hyperbola!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March • 1 Marks): Find the coordinates of the focus and the equation of the directrix of the parabola x² = 8y.

Compare with x² = 4ay: 4a = 8 ⇒ a = 2.
• Focus: (0, 2)
• Directrix: y = −2.

Q2 (Kerala HSE Say • 2 Marks): Find the centre and radius of the circle x² + y² − 8x + 10y − 12 = 0.

Method 1 (Completing Square): (x² − 8x + 16) + (y² + 10y + 25) = 12 + 16 + 25
(x − 4)² + (y + 5)² = 53.
• Centre: (4, −5), Radius: √53.

Q3 (Kerala HSE March • 2 Marks): Find the equation of the parabola with vertex at (0, 0), passing through (2, 3) and axis along the x-axis.

Axis along x-axis and (2, 3) is in Quadrant I ⇒ parabola opens right: y² = 4ax.
Substitute (2, 3): 3² = 4a(2) ⇒ 9 = 8a ⇒ 4a = 9/2.
• Equation: y² = (9/2)x (or 2y² = 9x).

Q4 (Kerala HSE March • 3 Marks): Find the foci, vertices, lengths of major and minor axes, eccentricity, and latus rectum of 9x² + 4y² = 36.

Divide by 36: x²4 + y²9 = 1. Bigger denominator 9 is under y² → Vertical Ellipse: a = 3, b = 2.
c = √(9 − 4) = √5.
• Foci: (0, ±√5), Vertices: (0, ±3)
• Major axis = 6, Minor axis = 4, e = √5/3, LR = 2(4)/3 = 8/3.

Q5 (Kerala HSE Say • 3 Marks): Find the foci, vertices, eccentricity, and length of latus rectum of the hyperbola 16x² − 9y² = 576.

Divide by 576: x²36 − y²64 = 1.
Positive term is x² → a² = 36 (a = 6), b² = 64 (b = 8).
c = √(36 + 64) = √100 = 10.
• Vertices: (±6, 0), Foci: (±10, 0)
• Eccentricity: e = 10/6 = 5/3, LR = 2(64)/6 = 64/3.

Q6 (NCERT / DHSE Model • 3 Marks): Find the equation of the circle passing through (4, 1) and (6, 5) whose centre lies on the line 4x + y = 16.

Let centre be (h, k): 4h + k = 16 ... (1).
Equal radii: (h − 4)² + (k − 1)² = (h − 6)² + (k − 5)²
−8h + 16 − 2k + 1 = −12h + 36 − 10k + 25 ⇒ 4h + 8k = 44 ⇒ h + 2k = 11 ... (2).
Solving (1) & (2): 7k = 28 ⇒ k = 4, h = 3.
r² = (3 − 4)² + (4 − 1)² = 1 + 9 = 10.
• Equation: (x − 3)² + (y − 4)² = 10.

Q7 (Kerala HSE March • 3 Marks): Find the equation of an ellipse with foci (±3, 0) and semi-major axis 4.

Foci on x-axis → Horizontal Ellipse x²/a² + y²/b² = 1.
Given c = 3, a = 4.
b² = a² − c² = 16 − 9 = 7.
• Equation: x²16 + y²7 = 1.

Q8 (NCERT / DHSE Model • 4 Marks): A parabolic reflector has a diameter of 20 cm and depth of 5 cm. Find the position of the focus.

Place vertex at (0, 0) opening right along X-axis: y² = 4ax.
At depth x = 5, diameter is 20 ⇒ y = ±10, so point (5, 10) lies on parabola.
10² = 4a(5) ⇒ 100 = 20a ⇒ a = 5 cm.
• Focus is at (5, 0) — right at the midpoint of the circular rim!