Conic Sections Lecture Notes by Tita Afumbom Samuel

Conic Sections — Lecture Notes
CONIC SECTIONS
Lecture Notes
Author: TITA AFUMBOM SAMUEL
1 Introduction: The Conic Sections
Aconic section (or simply conic) is the curve obtained by intersecting a plane with a
right circular double cone. Depending on the angle of the cutting plane relative to the
axis of the cone, four distinct curves arise: the circle, the ellipse, the parabola and the
hyperbola.
apex
Circle
Ellipse
Parabola
Hyperbola
Trick (angle test for the cutting plane). Let βbe the angle the cutting plane makes
with the axis of the cone, and αthe half-angle of the cone. Then:
•β=90◦(plane perpendicular to axis) ⇒Circle
•α<β<90◦⇒Ellipse
•β=α(plane parallel to a slant edge) ⇒Parabola
•β<α(plane cuts both nappes) ⇒Hyperbola
All four curves are also unified algebraically: every conic is the locus of points satis-
fying a second-degree (quadratic) equation in two variables xand y. This algebraic
viewpoint is what we develop next, since it is the one most useful for computation.
2 The General Second-Degree Equation
Every conic in the plane can be written in the general form
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Conic Sections — Lecture Notes
Ax2+Bxy +Cy2+Dx +Ey +F=0, (A,B,Cnot all zero).
When the axes of the conic are parallel to the coordinate axes, B=0, giving the simpler
form Ax2+Cy2+Dx +Ey +F=0, which is the form we concentrate on in this lecture
(the rotated case B̸=0 is treated by axis rotation and lies beyond today’s scope).
Trick 1 (the discriminant test). Compute ∆=B2−4AC.
Condition Conic
B2−4AC <0, A=C,B=0 Circle
B2−4AC <0, (A̸=Cor B̸=0) Ellipse
B2−4AC =0 Parabola
B2−4AC >0 Hyperbola
Trick 2 (fast eyeball test when B=0), i.e. comparing coefficients Aof x2and Cof
y2:
•A=C̸=0 (same sign, equal magnitude) ⇒Circle
•A̸=C, same sign ⇒Ellipse
•Aand Copposite signs ⇒Hyperbola
• Exactly one of A,Cis zero ⇒Parabola
Example 2.1. Classify 3x2+3y2−6x+12y−4=0.
Here A=C=3, B=0⇒Circle.
Example 2.2. Classify 4x2−9y2+16x+18y−29 =0.
A=4, C=−9: opposite signs ⇒Hyperbola.
Example 2.3. Classify y2−8x−6y+25 =0.
Here A=0 (no x2term), C=1: exactly one of A,Cvanishes ⇒Parabola.
3 The Circle
Definition. A circle is the locus of all points P(x,y)equidistant from a fixed point
C(h,k), called the centre. The fixed distance ris the radius.
Derivation. If P(x,y)lies on the circle, then |PC|=r, so by the distance formula
q(x−h)2+ (y−k)2=r=⇒(x−h)2+ (y−k)2=r2.
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Conic Sections — Lecture Notes
x
y
C(h,k)
rP(x,y)
Trick (completing the square). Given x2+y2+2gx +2f y +c=0, group and
complete the square:
(x+g)2+ (y+f)2=g2+f2−c.
So centre = (−g,−f)and radius =pg2+f2−c (provided the right side is positive).
Simply read off half the coefficient of xand half the coefficient of y, negate both, to
get the centre instantly.
Example 3.1. Find the centre and radius of x2+y2−6x+4y−3=0.
Half of −6 is −3; half of 4 is 2. Centre = (3, −2). Radius =p(−3)2+ (2)2−(−3) =
√9+4+3=√16 =4.
Example 3.2. Find the equation of the circle with centre (−1, 2)passing through (3, 5).
r2= (3−(−1))2+ (5−2)2=16 +9=25. Equation: (x+1)2+ (y−2)2=25.
Example 3.3. Determine whether 2x2+2y2+8x−4y+1=0 represents a real circle.
Divide by 2: x2+y2+4x−2y+1
2=0. Centre (−2, 1);r2=4+1−1 2=4.5 >0. Yes,
a real circle with r=√4.5 ≈2.12.
4 The Parabola
Definition. A parabola is the locus of points Pequidistant from a fixed point F(the
focus) and a fixed line ℓ(the directrix) not containing F:
|PF|=|PD|,D=foot of perpendicular from Pto ℓ.
Derivation (vertex at origin, axis along x-axis). Let F(a, 0)and directrix x=−a. Then
q(x−a)2+y2=x+a=⇒(x−a)2+y2= (x+a)2=⇒y2=4ax.
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Conic Sections — Lecture Notes
directrix: x=−a
F(a, 0)
O
latus rectum
x
y
Equation Axis Opens Focus Directrix
y2=4ax x-axis right (a>0) (a, 0)x=−a
y2=−4ax x-axis left (a>0) (−a, 0)x=a
x2=4ay y-axis up (a>0) (0, a)y=−a
x2=−4ay y-axis down (a>0) (0, −a)y=a
Key characteristics: vertex at (0, 0)(or (h,k)after translation); eccentricity e=1 al-
ways; length of latus rectum =4a.
Trick (translated vertex). For (y−k)2=4a(x−h), the vertex is simply (h,k)—
read off the numbers that make each bracket zero, with signs flipped. The same 4a,
focus and directrix rules apply relative to the new vertex.
Example 4.1. Find the vertex, focus, and directrix of y2=12x.
4a=12 ⇒a=3. Vertex (0, 0), focus (3, 0), directrix x=−3, latus rectum length =12.
Example 4.2. Find the vertex and focus of (x−2)2=−8(y+1).
Vertex (2, −1); 4a=8⇒a=2; opens downward ⇒focus = (2, −1−2) = (2, −3);
directrix: y=−1+2=1.
Example 4.3. A parabolic reflector has focus 4 cm from the vertex and axis along the
x-axis, opening right. Find its equation and the width of the opening 10 cm from the
vertex.
a=4⇒y2=16x. At x=10: y2=160 ⇒y=±√160 =±4√10. Width =8√10 ≈
25.3 cm.
5 The Ellipse
Definition. An ellipse is the locus of points Psuch that the sum of the distances
from Pto two fixed points F1,F2(the foci) is constant:
|PF1|+|PF2|=2a.
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Conic Sections — Lecture Notes
Derivation (centre at origin, foci on x-axis). With F1(−c, 0),F2(c, 0), expanding |PF1|+
|PF2|=2aand simplifying (two applications of squaring) gives, with b2=a2−c2,
x2
a2+y2 b2=1, a>b>0.
F1(−c, 0)F2(c, 0)
(a, 0)(−a, 0)
(0, b)
(0, −b)
x
y
Key characteristics (major axis along x,a>b):
• Relation: c2=a2−b2
• Vertices: (±a, 0); Co-vertices: (0, ±b)
• Foci: (±c, 0)
• Eccentricity: e=c
a∈(0, 1)
• Directrices: x=±a
e
• Length of latus rectum: 2b2
a
If the y2-term has the larger denominator, the major axis is vertical and the roles of
a,b(and x,y) swap.
Trick (which axis is major?). In x2
p+y2 q=1, the larger denominator always marks
the major axis direction: if p>q, major axis is horizontal with a2=p,b2=q; if
q>p, it is vertical with a2=q,b2=p. Never assume xalways carries a2.
Trick (fast eccentricity). Since b2=a2−c2⇒e2=1−b2
a2. So e=r1−b2 a2— no
need to compute cfirst.
Example 5.1. Find a,b,c,eand foci of x2
25 +y2 9=1.
25 >9: major axis horizontal, a=5, b=3. c=√25 −9=4. e=4/5 =0.8. Foci
(±4, 0). Latus rectum =2(9)/5 =3.6.
Example 5.2. Find the equation of the ellipse with vertices (0, ±10)and eccentricity
e=3/5.
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