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
A conic 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.
Circle
Ellipse
apex
Hyperbola
Parabola
Trick (angle test for the cutting plane). Let β be the angle the cutting plane makes
with the axis of the cone, and α the 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 satisfying a second-degree (quadratic) equation in two variables x and 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, C not 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 B ̸= 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
B2 − 4AC < 0, (A ̸= C or B ̸= 0)
B2 − 4AC = 0
B2 − 4AC > 0
Circle
Ellipse
Parabola
Hyperbola
Trick 2 (fast eyeball test when B = 0), i.e. comparing coefficients A of x2 and C of
y2 :
• A = C ̸= 0 (same sign, equal magnitude) ⇒ Circle
• A ̸= C, same sign ⇒ Ellipse
• A and C opposite signs ⇒ Hyperbola
• Exactly one of A, C is 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 x2 term), C = 1: exactly one of A, C vanishes ⇒ 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 r is 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
y
C (rh, k )
P( x, y)
x
Trick (completing the square). Given x2 + y2 + 2gx + 2 f y + c = 0, group and
complete the square:
( x + g)2 + (y + f )2 = g2 + f 2 − c.
p
So centre = (− g, − f ) and radius = g2 + f 2 − c (provided the right side is positive).
Simply read off half the coefficient of x and 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.
p
Half of −6 is −3; half of 4 is 2. Centre = (3, −2). Radius = (−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 + 21 = 0. Centre (−2, 1); r2 = 4 + 1 − 12 = 4.5 > 0. Yes,
√
a real circle with r = 4.5 ≈ 2.12.
4
The Parabola
Definition. A parabola is the locus of points P equidistant from a fixed point F (the
focus) and a fixed line ℓ (the directrix) not containing F:
| PF | = | PD |,
D = foot of perpendicular from P to 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
y
latus rectum
x
O
F ( a, 0)
directrix: x = − a
Equation
Axis
Opens
Focus
Directrix
y2 = 4ax
y2 = −4ax
x2 = 4ay
x2 = −4ay
x-axis right (a > 0)
x-axis left (a > 0)
y-axis up (a > 0)
y-axis down (a > 0)
( a, 0)
(− a, 0)
(0, a)
(0, − a)
x = −a
x=a
y = −a
y=a
Key characteristics: vertex at (0, 0) (or (h, k ) after translation); eccentricity e = 1 always; 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 P such that the sum of the distances
from P to 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 | = 2a and simplifying (two applications of squaring) gives, with b2 = a2 − c2 ,
x 2 y2
+ 2 = 1,
a2
b
a > b > 0.
y
(0, b)
(− a, 0)
( a, x0)
F1 (−c, 0)
F2 (c, 0)
(0, −b)
Key characteristics (major axis along x, a > b):
• Relation: c2 = a2 − b2
• Vertices: (± a, 0); Co-vertices: (0, ±b)
• Foci: (±c, 0)
c
• Eccentricity: e = ∈ (0, 1)
a
a
• Directrices: x = ±
e
2b2
• Length of latus rectum:
a
If the y2 -term has the larger denominator, the major axis is vertical and the roles of
a, b (and x, y) swap.
x 2 y2
+
= 1, the larger denominator always marks
p
q
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 x always carries a2 .
r
2
b
b2
Trick (fast eccentricity). Since b2 = a2 − c2 ⇒ e2 = 1 − 2 . So e = 1 − 2 — no
a
a
need to compute c first.
Trick (which axis is major?). In
x 2 y2
+
= 1.
25
9
√
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 5.1. Find a, b, c, e and foci of
Example 5.2. Find the equation of the ellipse with vertices (0, ±10) and eccentricity
e = 3/5.
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Conic Sections — Lecture Notes
Major axis vertical, a = 10. c = ae = 6. b2 = a2 − c2 = 100 − 36 = 64. Equation:
x2
y2
+
= 1.
64 100
Example 5.3. Identify the conic 9x2 + 4y2 − 36x + 8y + 4 = 0 and find its centre.
Group: 9( x2 − 4x ) + 4(y2 + 2y) = −4. Complete squares: 9( x − 2)2 − 36 + 4(y + 1)2 −
( x − 2)2 ( y + 1)2
4 = −4, so 9( x − 2)2 + 4(y + 1)2 = 36. Divide by 36:
+
= 1. Centre
4
9
(2, −1); since 9 > 4, major axis vertical with a = 3, b = 2.
6
The Hyperbola
Definition. A hyperbola is the locus of points P such that the (absolute) difference
of the distances from P to two fixed foci F1 , F2 is constant:
| PF1 | − | PF2 | = 2a.
Derivation (centre at origin, foci on x-axis). With F1 (−c, 0), F2 (c, 0) and b2 = c2 − a2 ,
the same squaring technique as for the ellipse (with a sign change) gives
x 2 y2
− 2 = 1.
a2
b
y
F1 (−c, 0)
F2 (c, 0) x
(− a, 0)
( a, 0)
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Conic Sections — Lecture Notes
Key characteristics (transverse axis along x):
• Relation: c2 = a2 + b2 (note the +, unlike the ellipse)
• Vertices: (± a, 0); Foci: (±c, 0)
b
• Asymptotes: y = ± x
a
c
• Eccentricity: e = > 1
a
a
• Directrices: x = ±
e
2b2
• Length of latus rectum:
a
2
2
x
y
For 2 − 2 = 1 (transverse axis vertical), swap the roles of x and y; asymptotes
a
b
a
become y = ± x.
b
Trick (asymptotes in one line). To find the asymptotes of
x 2 y2
− 2 = 1, replace the
a2
b
x2
y2
b
−
= 0 ⇒ y = ± x. This "set RHS to zero"
2
2
a
a
b
( x − h )2 ( y − k )2
shortcut also works after translation, e.g. for
−
= 0.
a2
b2
Trick (sign tells the transverse axis). Whichever squared term is positive names the
transverse axis: positive x2 -term ⇒ horizontal transverse axis; positive y2 -term ⇒
vertical.
constant 1 on the right by 0:
x 2 y2
Example 6.1. Find the vertices, foci, eccentricity and asymptotes of
−
= 1.
16
9
√
a = 4, b = 3. c = 16 + 9 = 5. Vertices (±4, 0); Foci (±5, 0); e = 5/4 = 1.25;
asymptotes y = ± 43 x.
Example 6.2. Find the equation of the hyperbola with foci (0, ±13) and vertices (0, ±5).
Transverse axis vertical, a = 5, c = 13. b2 = c2 − a2 = 169 − 25 = 144. Equation:
x2
y2
−
= 1.
25 144
Example 6.3. Identify and analyse x2 − 4y2 − 2x − 16y − 19 = 0.
Group: ( x2 − 2x ) − 4(y2 + 4y) = 19. Complete squares: ( x − 1)2 − 1 − 4[(y + 2)2 − 4] =
( x − 1)2 ( y + 2)2
19 ⇒ ( x − 1)2 − 4(y + 2)2 = 19 + 1 − 16 = 4. Divide by 4:
−
= 1.
4
1
√
Centre (1, −2), a = 2, b = 1, c = 5; transverse axis horizontal; asymptotes: y + 2 =
± 12 ( x − 1).
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Conic Sections — Lecture Notes
7
Tangents and Normals to the Conics
7.1 The unifying “T = 0” trick
Every axes-aligned conic can be written S( x, y) : Ax2 + Cy2 + Dx + Ey + F = 0. The
tangent at a point ( x1 , y1 ) lying on the conic is obtained instantly by the substitution rule
x2 → xx1 ,
y2 → yy1 ,
x→
x + x1
,
2
y→
leaving F unchanged. Writing T ≡ Axx1 + Cyy1 + D x+2x1 + E
( x1 , y1 ) is simply
y + y1
,
2
y + y1
2 + F, the tangent at
T = 0.
This one rule reproduces every tangent formula below — there is no need to memorise
four separate formulas.
Tangent at ( x1 , y1 ) — summary.
Conic
Tangent at ( x1 , y1 )
Circle x2 + y2 = r2
Parabola y2 = 4ax
x 2 y2
Ellipse 2 + 2 = 1
a
b
x 2 y2
Hyperbola 2 − 2 = 1
a
b
xx1 + yy1 = r2
yy1 = 2a( x + x1 )
xx1 yy1
+ 2 =1
a2
b
xx1 yy1
− 2 =1
a2
b
Example 7.1. Verify the T = 0 rule for the parabola.
S = y2 − 4ax, so A = 0, C = 1, D = −4a, E = 0, F = 0. Then T = 1 · yy1 + (−4a) x+2x1 =
yy1 − 2a( x + x1 ). Setting T = 0 gives yy1 = 2a( x + x1 ), exactly the standard formula.
✓
Example 7.2. Find the tangent to x2 + y2 = 25 at (3, 4).
Point lies on the circle since 9 + 16 = 25. Tangent: 3x + 4y = 25.
x 2 y2
Example 7.3. Find the tangent to
+
= 1 at (3 cos 30◦ , 2 sin 30◦ ) =
9
4
√
√
x · 323
y·1
3x y
Tangent:
+
=1⇒
+ = 1.
9
4
6
4
Page 8 of 16
√ !
3 3
,1 .
2
Conic Sections — Lecture Notes
7.2 The differentiation (calculus) method for the gradient of the tangent
The algebraic T = 0 trick gives the tangent line directly, but it is built on the same fact
dy
, found by implicit
used in calculus: the gradient of the tangent at ( x1 , y1 ) is
dx ( x1 ,y1 )
differentiation of the conic’s equation.
Gradient of the tangent by implicit differentiation.
Conic
Differentiate implicitly Gradient at ( x1 , y1 )
x 2 + y2 = r 2
2x + 2y
y2 = 4ax
2y
x 2 y2
+ 2 =1
a2
b
x 2 y2
− 2 =1
a2
b
dy
=0
dx
dy
= 4a
dx
2x 2y dy
=0
+ 2
a2
b dx
2x 2y dy
=0
− 2
a2
b dx
x
m=− 1
y1
2a
m=
y1
b2 x
m=− 2 1
a y1
b2 x1
m= 2
a y1
Once m is known, the tangent at ( x1 , y1 ) is simply y − y1 = m( x − x1 ).
Trick (differentiate term-by-term, treat y as a function of x). For any term in y,
dy
d 2
(y ) = 2y . Every other term differentiates as usual in x.
apply the chain rule:
dx
dx
dy
Then solve algebraically for
— do not attempt to isolate y first.
dx
b2 x
Consistency check. Substituting m = − 2 1 into y − y1 = m( x − x1 ) and using the
a y1
2
y2
x
xx
yy
fact that ( x1 , y1 ) lies on the ellipse (so 21 + 21 = 1) reduces it exactly to 21 + 21 =
a
b
a
b
1 — the same tangent found by the T = 0 trick in Section 7.1. The calculus and
algebraic methods always agree; use whichever is faster for the question in front of
you.
Example 7.4. Use differentiation to find the gradient of the tangent to x2 + y2 = 25 at
(3, 4), and hence write down the tangent.
dy
dy
x
3
Differentiate: 2x + 2y
= 0 ⇒
= − . At (3, 4): m = − . Tangent: y − 4 =
dx
dx
y
4
3
− ( x − 3) ⇒ 3x + 4y = 25 — matches Example 7.2 exactly.
4
Example 7.5. Use differentiation to find the gradient of the tangent to y2 = 12x at
(3, 6).
Page 9 of 16
Conic Sections — Lecture Notes
dy
dy
6
6
= 12 ⇒
= . At (3, 6): m =
= 1. Tangent: y − 6 =
dx
dx
y
6
1( x − 3) ⇒ y = x + 3.
Differentiate: 2y
x2
y2
Example 7.6. Use differentiation to find the gradient of the tangent to
−
= 1 at
16
9
√ (4 sec θ, 3 tan θ ) when θ = 45◦ , i.e. at 4 2, 3 .
√
√
√ 2x 2y dy
dy
9x
9(4 2)
36 2
Differentiate:
−
=0⇒
=
. At 4 2, 3 : m =
=
=
16
9 dx
dx
16y
16(3)
48
√
3 2
.
4
7.3 Condition for tangency of the line y = mx + c
Substituting y = mx + c into each conic and setting the discriminant of the resulting
quadratic to zero gives:
Conic
Tangency condition
x 2 + y2 = r 2
c2 = r 2 (1 + m2 )
a
a
(m ̸= 0), i.e. tangent is y = mx +
c=
m
m
y2 = 4ax
x 2 y2
+ 2 =1
a2
b
x 2 y2
− 2 =1
a2
b
c2 = a2 m2 + b2
c2 = a2 m2 − b2
Trick (how to derive any tangency condition fast). Substitute y = mx + c into the
conic, collect as αx2 + βx + γ = 0, then impose β2 = 4αγ (a tangent line meets the
conic at exactly one repeated point). You never need to solve the quadratic — only
its discriminant.
Example 7.7. For which value of c is y = 2x + c tangent to y2 = 12x?
a
3
Here a = 3, m = 2 ⇒ c =
= .
m
2
x2
Example 7.8. Show that y = x + 3 is tangent to
+ y2 = 1, i.e. a2 = 2, b2 = 1.
2
?
Here m = 1, c = 3: check c2 = a2 m2 + b2 ⇒ 9 = 2(1) + 1 = 3. Since 9 ̸= 3, the line is
not tangent (it does not meet the ellipse at all, since 9 > 3 means the line lies entirely
outside). This shows the condition also acts as a quick real-intersection test.
7.4 Normals (formulas only)
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Conic Sections — Lecture Notes
Conic
Normal at ( x1 , y1 )
Circle
line through ( x1 , y1 ) and the centre
y
y − y1 = − 1 ( x − x1 )
2a
a2 x b2 y
−
= a2 − b2
x1
y1
a2 x b2 y
+
= a2 + b2
x1
y1
Parabola y2 = 4ax
Ellipse
Hyperbola
Example 7.9. Find the normal to y2 = 8x at (2, 4).
4
a = 2: normal is y − 4 = − ( x − 2) = −( x − 2) ⇒ y = − x + 6.
4
7.5 Chord of contact and pair of tangents (beautiful unifying results)
Chord of contact. From an external point ( x1 , y1 ) outside the conic, two tangents
can be drawn, touching the conic at two points. Remarkably, the chord joining
those two points of contact is given by exactly the same expression T = 0 used for
the tangent — even though ( x1 , y1 ) is now external rather than on the curve.
Pair of tangents. The pair of straight lines joining ( x1 , y1 ) to the conic (i.e. the two
tangent lines themselves) is given by
S S1 = T 2 ,
where S = 0 is the conic, S1 is S evaluated at ( x1 , y1 ), and T is the same expression
as above.
Example 7.10. Find the chord of contact of tangents drawn from (5, 5) to the circle
x2 + y2 = 16.
T : 5x + 5y − 16 = 0, i.e. 5x + 5y = 16. (Note (5, 5) is external since 25 + 25 = 50 > 16,
so two real tangents exist.)
Example 7.11. Find the pair of tangents from (4, 0) to the circle x2 + y2 = 9.
S = x2 + y2 − 9, S1 = 16 + 0 − 9 = 7, T = 4x − 9. Pair of tangents: 7( x2 + y2 − 9) =
(4x − 9)2 ⇒ 7x2 + 7y2 − 63 = 16x2 − 72x + 81 ⇒ 9x2 − 7y2 − 72x + 144 = 0.
tangent
O
chord of contact P( x1 , y1 )
Page 11 of 16
Conic Sections — Lecture Notes
8
Unified Focus–Directrix Definition and Polar Form
Unified definition. For any conic (except the circle), there is a focus F, a directrix ℓ,
and a constant e > 0 (the eccentricity) such that for every point P on the curve,
| PF |
= e,
| PD |
D = foot of perpendicular from P onto ℓ.
Value of e
Conic
e=0
0<e<1
e=1
e>1
Circle (limiting case; no finite directrix)
Ellipse
Parabola
Hyperbola
Placing the focus at the pole of a polar coordinate system with the directrix perpendicular to the polar axis, at distance l/e from the focus, gives the single unified polar
equation of a conic:
r=
l
1 + e cos θ
(or with −e cos θ, ±e sin θ depending on directrix orientation),
where l = a(1 − e2 ) for the ellipse/hyperbola is the semi-latus rectum (half the
latus rectum).
This single formula is the modern, unified way conics are treated in orbital mechanics:
planetary and satellite orbits are conics with the Sun/Earth at a focus, and e alone
determines whether the orbit is a closed ellipse (e < 1), a parabolic escape trajectory
(e = 1), or a hyperbolic flyby (e > 1).
4
Example 8.1. A conic has focus at the pole and polar equation r =
. Iden1 + 0.5 cos θ
tify it and find r at θ = 0 and θ = π.
e = 0.5 < 1 ⇒ ellipse. At θ = 0: r = 4/1.5 = 8/3. At θ = π: r = 4/0.5 = 8.
9
Interesting Question Types and Real-World Applications
The tangent theory of Section 7 is not just algebra — it is the engine behind some
of the most elegant results in geometry, and behind real engineering designs. This
section links today’s tools to the kinds of questions you will meet in examinations and
in applied practice.
Page 12 of 16
Conic Sections — Lecture Notes
9.1 The reflective (focusing) properties
Parabola. A ray travelling parallel to the axis strikes the parabola and reflects
through the focus (and vice versa): the tangent at the point of incidence makes equal
angles with the incoming ray and the focal chord.
Ellipse. A ray leaving one focus reflects off the ellipse and passes through the other
focus: the tangent at any point makes equal angles with the two focal radii PF1 , PF2 .
This is why elliptical rooms and bridges produce “whispering-gallery” effects.
Hyperbola. A ray aimed at one focus, on striking the hyperbola, reflects away from
the other focus along the line joining them — used in Cassegrain telescope design.
incoming rays
axis
F ( a, 0)
This is precisely why satellite dishes, headlamp reflectors and radio telescopes are
shaped as paraboloids: every incoming signal parallel to the axis is concentrated exactly at the focus, regardless of where on the dish it strikes.
9.2 Two “beautiful results” worth knowing
Result 1 (foot of perpendicular from the focus). The foot of the perpendicular from
the focus of a parabola to any tangent always lies on the tangent at the vertex. For
y2 = 4ax, that locus is simply the line x = 0.
Result 2 (director circle). The locus of points from which two perpendicular tangents
can be drawn to a conic is its director circle:
Ellipse: x2 + y2 = a2 + b2 ,
Hyperbola: x2 + y2 = a2 − b2 (exists only if a > b).
9.3 Common exam and applied question archetypes
• “Show that the line . . . is a tangent.” → use the tangency condition (Section 7.3).
• “Find the tangents from an external point . . . ” → use SS1 = T 2 (pair of tangents).
Page 13 of 16
Conic Sections — Lecture Notes
• “Find the locus of . . . ” (foot of perpendicular, intersection of perpendicular tangents, midpoints of chords) → often reduces to a circle or another conic (director
circle, auxiliary circle).
• “A satellite dish/headlamp/telescope has . . . ” → parabola + reflective property.
• “An elliptical room/bridge/orbit has . . . ” → ellipse foci, whispering-gallery
property, orbital eccentricity.
• “A ship/aircraft uses two transmitters . . . ” → hyperbola — the constant timedifference locus, basis of LORAN/hyperbolic navigation.
• “Find the shortest distance from a point to a conic.” → the normal at the closest
point passes through the given point (Section 7.4).
Example 9.1. (Applied.) A satellite dish is parabolic with focus 0.6 m from the vertex.
Explain, using the reflective property, why the receiver must sit exactly at the focus.
Every incoming ray parallel to the axis reflects through the focus, regardless of where
it strikes the dish. Placing the receiver at the focus is therefore the only position at
which all reflected rays converge simultaneously; anywhere else on the axis, reflected
rays from different points on the dish cross at different locations and the signal is not
concentrated.
Example 9.2. (Locus/director circle.) Find the locus of the point of intersection of
x 2 y2
+
= 1.
perpendicular tangents to
16
9
a2 = 16, b2 = 9 ⇒ director circle x2 + y2 = 25: a circle of radius 5 centred at the origin.
10
Recognition and Sketching Toolkit
Master comparison table.
Circle
Parabola
Ellipse
Hyperbola
Defn. (loci)
fixed dist.
| PF | = | PD |
sum of dist.= 2a
diff. of dist.= 2a
Std. eqn
Eccentricity
A, C signs
Asymptotes
x 2 + y2 = r 2
y2 = 4ax
2
x2
+ yb2 = 1
a2
2
x2
− yb2 = 1
a2
e=0
A=C
none
e=1
one is 0
none
0<e<1
same, unequal
none
e>1
opposite
y = ± ba x
Page 14 of 16
Conic Sections — Lecture Notes
The 5-step sketching algorithm (works for every conic):
1. Classify: apply the discriminant/coefficient trick (Section 2).
2. Complete the square in x and y to find the centre/vertex (h, k ).
3. Read off a, b (or 4a for a parabola) from the denominators or coefficients.
4. Compute the defining points: for circle — just r; for parabola — focus & directrix; for ellipse/hyperbola — vertices, co-vertices/asymptotes, foci (c via
b2 = a2 − c2 or c2 = a2 + b2 ).
5. Plot the centre/vertex first, then the defining points, then sketch the curve
through them (draw asymptotes as dashed guides first for a hyperbola).
11
Mixed Worked Examples
Example 11.1. Identify and fully analyse 4x2 + y2 − 8x + 4y − 8 = 0.
A = 4, C = 1: same sign, unequal ⇒ ellipse. Complete the square: 4( x − 1)2 − 4 +
( x − 1)2 ( y + 2)2
(y + 2)2 − 4 = 8 ⇒ 4( x − 1)2 + (y + 2)2 = 16. Divide by 16:
+
= 1.
4 √
16
√
Centre
(
1,
−
2
)
;
since
16
>
4,
major
axis
vertical:
a
=
4,
b
=
2,
c
=
16
−
4
=
2
3;
√
√
2 3
3
e=
=
≈ 0.866.
4
2
Example 11.2. Identify and analyse x2 − y2 + 6x − 4y + 4 = 0.
A = 1, C = −1: opposite signs ⇒ hyperbola. Complete the square: ( x + 3)2 − 9 −
√
(y + 2)2 + 4 = −4 ⇒ ( x + 3)2 − (y + 2)2 = 1. Centre (−3, −2), a = 1, b = 1, c = 2,
asymptotes: y + 2 = ±( x + 3).
Example 11.3. Identify and analyse y2 + 6y − 4x + 17 = 0.
Only a y2 term (and no x2 ) ⇒ parabola. (y + 3)2 − 9 − 4x + 17 = 0 ⇒ (y + 3)2 =
4x − 8 = 4( x − 2). Vertex (2, −3), 4a = 4 ⇒ a = 1, axis horizontal, opens right; focus
(3, −3); directrix x = 1.
12
Practice Exercises
Part A — Routine.
1. Classify and sketch x2 + y2 + 2x − 4y − 4 = 0.
2. Find the focus and directrix of x2 = −20y.
3. An ellipse has foci (±6, 0) and passes through (0, 8). Find its equation, e, and
latus rectum.
4. Find the equation of a hyperbola with vertices (±3, 0) and asymptotes y = ± 23 x.
5. Classify, without completing the square, the conic 5x2 − 3y2 + 10x − 6y − 7 = 0
using the coefficient trick, then confirm using the discriminant.
Part B — Tangents and Normals.
Page 15 of 16
Conic Sections — Lecture Notes
6. Find the tangent and normal to x2 + y2 = 20 at (4, 2).
7. Find the tangent to y2 = 16x at the point where x = 4 on the upper branch.
x 2 y2
8. Find the condition on c for the line y = 3x + c to be tangent to
+
= 1, and
4
9
hence find both tangent lines of gradient 3.
x 2 y2
9. Find the chord of contact of tangents drawn from (6, 0) to the ellipse
+
= 1.
9
4
10. Find the pair of tangents from (0, 5) to the circle x2 + y2 = 9, and verify your
answer using the tangency condition on each individual line.
Part C — Locus, Beautiful Results and Challenge.
11. Find the locus of the point of intersection of perpendicular tangents to
x 2 y2
−
=
25 16
1 (state the condition for this locus to exist).
12. Prove, using the reflective property, that a ray leaving the focus of the parabola
y2 = 4ax along the axis reflects straight back along the axis.
13. (Challenge) Show that the tangents at the endpoints of any chord through the
focus of a parabola meet at right angles on the directrix.
Part D — Applications.
14. A satellite orbit has semi-latus rectum l = 7000 km and e = 0.2. Find r at perigee
(θ = 0) and apogee (θ = π).
15. A parabolic reflector (torch/headlamp) is 12 cm wide at its rim and 9 cm deep.
Find the distance of the bulb (focus) from the vertex.
16. An elliptical whispering-gallery room is 30 m long and 18 m wide (major and
minor axes). Find how far each “whispering point” (focus) is from the centre of
the room.
17. Two radio transmitters 200 km apart send synchronised signals. A ship notes a
constant time difference corresponding to a path-difference of 120 km. Identify
the type of curve on which the ship must lie, and find its equation (take the midpoint of the transmitters as origin, transmitters on the x-axis).
End of lecture notes — Conic Sections.
Page 16 of 16