Intuition The big pictureA conic is what you get when a plane slices a double cone. Tilt the plane and you get ellipses, parabolas, hyperbolas. Hold the plane perfectly horizontal (perpendicular to the cone's axis) and the slice is a perfect circle .
In the focus–directrix language, every conic is controlled by one number, the eccentricity e e e e = 0 e = 0 e = 0
Definition Conic (focus–directrix definition)A conic is the set of points P P P focus S S S directrix ℓ \ell ℓ e e e P S P M = e \frac{PS}{PM} = e P M P S = e P M PM P M P P P
e = 0 ⇒ e = 0 \Rightarrow e = 0 ⇒ circle 0 < e < 1 ⇒ 0 < e < 1 \Rightarrow 0 < e < 1 ⇒ e = 1 ⇒ e = 1 \Rightarrow e = 1 ⇒ e > 1 ⇒ e > 1 \Rightarrow e > 1 ⇒
A circle is the locus of all points at a fixed distance r r r from a fixed point C C C . There is no directrix and no privileged focus direction — every diameter is equivalent.
The subtle part: if e = P S P M = 0 e = \dfrac{PS}{PM} = 0 e = P M P S = 0 P S = 0 ⋅ P M = 0 PS = 0 \cdot PM = 0 P S = 0 ⋅ P M = 0 P = S P = S P = S e = 0 e=0 e = 0 careful limit .
Worked example Example 1 — Ellipse becoming a circleAn ellipse has a = 5 a = 5 a = 5 e e e b b b e = 0.6 e = 0.6 e = 0.6 e → 0 e \to 0 e → 0
b 2 = a 2 ( 1 − e 2 ) = 25 ( 1 − 0.36 ) = 16 ⇒ b = 4 b^2 = a^2(1-e^2) = 25(1 - 0.36) = 16 \Rightarrow b = 4 b 2 = a 2 ( 1 − e 2 ) = 25 ( 1 − 0.36 ) = 16 ⇒ b = 4 (a genuine ellipse)
Why this step? We use b 2 = a 2 ( 1 − e 2 ) b^2 = a^2(1-e^2) b 2 = a 2 ( 1 − e 2 ) As e → 0 e \to 0 e → 0 b 2 → 25 b^2 \to 25 b 2 → 25 b → 5 = a b \to 5 = a b → 5 = a x 2 + y 2 = 25 x^2 + y^2 = 25 x 2 + y 2 = 25 Why? Equal axes ⇒ \Rightarrow ⇒
Worked example Example 3 — Circle through 3 points usinge = 0 e=0 e = 0 e = 0 e=0 e = 0 ( 0 , 0 ) , ( 2 , 0 ) , ( 0 , 2 ) (0,0), (2,0), (0,2) ( 0 , 0 ) , ( 2 , 0 ) , ( 0 , 2 )
Centre ( h , k ) (h,k) ( h , k ) h 2 + k 2 = ( h − 2 ) 2 + k 2 ⇒ 0 = − 4 h + 4 ⇒ h = 1 h^2+k^2 = (h-2)^2+k^2 \Rightarrow 0 = -4h+4 \Rightarrow h=1 h 2 + k 2 = ( h − 2 ) 2 + k 2 ⇒ 0 = − 4 h + 4 ⇒ h = 1 Why? Equidistant from ( 0 , 0 ) (0,0) ( 0 , 0 ) ( 2 , 0 ) (2,0) ( 2 , 0 )
Similarly k = 1 k = 1 k = 1 ( 1 , 1 ) (1,1) ( 1 , 1 ) r = 1 + 1 = 2 r = \sqrt{1+1} = \sqrt2 r = 1 + 1 = 2
Equation: ( x − 1 ) 2 + ( y − 1 ) 2 = 2 (x-1)^2 + (y-1)^2 = 2 ( x − 1 ) 2 + ( y − 1 ) 2 = 2
Recall Predict before you computeBefore reading on: for the ellipse x 2 9 + y 2 9 = 1 \dfrac{x^2}{9} + \dfrac{y^2}{9} = 1 9 x 2 + 9 y 2 = 1 e e e Forecast: equal denominators ⇒ a = b ⇒ \Rightarrow a=b \Rightarrow ⇒ a = b ⇒ ⇒ e = 0 \Rightarrow e=0 ⇒ e = 0 Verify: b 2 = a 2 ( 1 − e 2 ) ⇒ 9 = 9 ( 1 − e 2 ) ⇒ e 2 = 0 ⇒ e = 0. b^2 = a^2(1-e^2) \Rightarrow 9 = 9(1-e^2) \Rightarrow e^2 = 0 \Rightarrow e=0. b 2 = a 2 ( 1 − e 2 ) ⇒ 9 = 9 ( 1 − e 2 ) ⇒ e 2 = 0 ⇒ e = 0.
e = 0 e=0 e = 0 P S / P M = 0 PS/PM = 0 P S / P M = 0 Why it feels right: the formula literally says e = P S / P M e = PS/PM e = P S / P M e = 0 e=0 e = 0 Why it's wrong: P S / P M = 0 PS/PM = 0 P S / P M = 0 P S = 0 PS = 0 P S = 0 is the focus — a single point, not a circle.
Fix: Treat e = 0 e=0 e = 0 limit of the ellipse: foci merge, directrix flees to infinity, a = b a=b a = b degenerate endpoint, not a literal substitution.
Common mistake "The radius isg 2 + f 2 + c \sqrt{g^2+f^2+c} g 2 + f 2 + c Why it feels right: the general form ends with + c +c + c Fix: It's r = g 2 + f 2 − c r=\sqrt{g^2+f^2-\mathbf{c}} r = g 2 + f 2 − c g 2 + f 2 − c < 0 g^2+f^2-c < 0 g 2 + f 2 − c < 0 no real circle (an "imaginary" circle) and if it = 0 =0 = 0 point circle (radius 0) — another degenerate case.
Common mistake "A circle has one focus, like a parabola."Fix: As e → 0 e\to0 e → 0 centre . So the "focus" of a circle is just its centre, and every direction is equivalent — that total symmetry is e = 0 e=0 e = 0
Recall Explain to a 12-year-oldImagine tying a ball to a string pinned at one spot and drawing it around — you get a perfectly round circle, and it doesn't matter which way you turn, it always looks the same. Other shapes like ovals (ellipses) have a "long way" and a "short way." The number e e e squashed the shape is. A circle isn't squashed at all, so its e e e 0 0 0
"Zero squash, zero e e e
e e e E qual axes when it's E mpty-of-stretch (Ellipse → circle).
What is the eccentricity of a circle? Why can't we set P S / P M = 0 PS/PM = 0 P S / P M = 0 It would force
P S = 0 PS=0 P S = 0 , giving a single point; instead take the limit
e → 0 e\to0 e → 0 of an ellipse.
As e → 0 e\to 0 e → 0 They merge into a single point — the centre.
As e → 0 e\to 0 e → 0 To infinity (
x = a / e → ∞ x = a/e \to \infty x = a / e → ∞ ), so a circle has no directrix.
Relation between a a a b b b e e e b 2 = a 2 ( 1 − e 2 ) b^2 = a^2(1-e^2) b 2 = a 2 ( 1 − e 2 ) What does b 2 = a 2 ( 1 − e 2 ) b^2=a^2(1-e^2) b 2 = a 2 ( 1 − e 2 ) e = 0 e=0 e = 0 b = a b=a b = a (equal axes) → a circle.
Standard equation of a circle, centre ( h , k ) (h,k) ( h , k ) r r r ( x − h ) 2 + ( y − k ) 2 = r 2 (x-h)^2+(y-k)^2=r^2 ( x − h ) 2 + ( y − k ) 2 = r 2 General form of a circle? x 2 + y 2 + 2 g x + 2 f y + c = 0 x^2+y^2+2gx+2fy+c=0 x 2 + y 2 + 2 g x + 2 f y + c = 0 Centre and radius from general form? Centre
( − g , − f ) (-g,-f) ( − g , − f ) , radius
g 2 + f 2 − c \sqrt{g^2+f^2-c} g 2 + f 2 − c When is x 2 + y 2 + 2 g x + 2 f y + c = 0 x^2+y^2+2gx+2fy+c=0 x 2 + y 2 + 2 g x + 2 f y + c = 0 When
g 2 + f 2 − c = 0 g^2+f^2-c=0 g 2 + f 2 − c = 0 (radius 0).
Centre of circle through ( 0 , 0 ) , ( 2 , 0 ) , ( 0 , 2 ) (0,0),(2,0),(0,2) ( 0 , 0 ) , ( 2 , 0 ) , ( 0 , 2 ) ( 1 , 1 ) (1,1) ( 1 , 1 ) , radius
2 \sqrt2 2 .
Conic Sections — the parent family, cut by a plane through a cone.Eccentricity — the single parameter e e e Ellipse — the circle is its e → 0 e\to0 e → 0 a = b a=b a = b Parabola (e = 1 e=1 e = 1 Hyperbola (e > 1 e>1 e > 1 Distance Formula — the tool that builds the circle equation.Completing the Square — converts general form ↔ standard form.Degenerate Conics — point circle, single point, pair of lines: limiting cases.
Focus-directrix ratio PS/PM
x-h squared plus y-k squared equals r squared
General form x2+y2+2gx+2fy+c=0
Intuition Hinglish mein samjhoDekho, saare conics — circle, ellipse, parabola, hyperbola — ek hi cheez se aate hain: ek cone ko plane se kaato. Agar plane thoda tilt karo to ellipse, aur zyada tilt karo to parabola/hyperbola. Lekin agar plane bilkul horizontal rakho (cone ke axis ke perpendicular), to jo slice milta hai wahi ek perfect circle hai. Yahi reason hai ki circle ko "degenerate conic" bolte hain.
Ab focus–directrix wali definition mein har conic ka ek number hota hai, eccentricity e e e e = 0 e = 0 e = 0 P S / P M = 0 PS/PM = 0 P S / P M = 0 ellipse ki limit ke roop mein samjho: jaise-jaise e → 0 e \to 0 e → 0 b 2 = a 2 ( 1 − e 2 ) b^2 = a^2(1-e^2) b 2 = a 2 ( 1 − e 2 ) b = a b = a b = a
Circle ka equation banana easy hai kyunki definition simple hai: centre se har point ki distance fixed (= r =r = r ( x − h ) 2 + ( y − k ) 2 = r 2 (x-h)^2+(y-k)^2 = r^2 ( x − h ) 2 + ( y − k ) 2 = r 2 x 2 + y 2 + 2 g x + 2 f y + c = 0 x^2+y^2+2gx+2fy+c=0 x 2 + y 2 + 2 g x + 2 f y + c = 0 ( − g , − f ) (-g,-f) ( − g , − f ) g 2 + f 2 − c \sqrt{g^2+f^2-c} g 2 + f 2 − c minus c , plus nahi! Agar g 2 + f 2 − c = 0 g^2+f^2-c=0 g 2 + f 2 − c = 0
Exam ke liye 80/20: bas do cheezein pakki karo — (1) e = 0 ⇒ e=0 \Rightarrow e = 0 ⇒
