Introduction to Conic Sections and Circles.pptx

Introduction to Conic
Sections and Circles
Quarter 1  Week 1

Learning Outcomes
Upon completion of the lessons, you
should be able to:

Learning Outcomes
should be able to:
1. illustrate the different types of
conic sections: parabola, ellipse,
circle, hyperbola, and degenerate
cases; (STEM_PC11AG-Ia-1)

Learning Outcomes
should be able to:
2. define a circle; (STEM_PC11AG-Ia-2)

Learning Outcomes
should be able to:
2. define a circle; (STEM_PC11AG-Ia-2)
3. determine the standard form of
equation of a circle. (STEM_PC11AG-Ia-
3)

Definition
Conic sections are curves that
result from the intersection of a
plane and a double cone.

Definition
plane and a double cone. The four
conic sections are

Definition
conic sections are a circle,

Definition
conic sections are a circle, an
ellipse,

Definition
ellipse, a parabola,

Definition
ellipse, a parabola, and a
hyperbola.

Definition
ellipse, a parabola, and a
hyperbola. Conics is an abbreviation
for conic sections.

Circle – when the
plane is
horizontal

Ellipse – when the
plane intersects
only one cone to
form a bounded
curve

Parabola – when the
plane intersects
only one cone to
form an unbounded
curve

• Hyperbola – when the
plane (not necessarily
vertical) intercepts
both cones to form two
unbounded curves (each
called a branch of the
hyperbola)

The general form of the second-degree
equation in two variables, 𝑥 and 𝑦, is
given by

given by
𝑨𝒙𝟐
+ 𝑩𝒙𝒚 + 𝑪𝒚𝟐
+ 𝑫𝒙 + 𝑬𝒚 + 𝑭 = 𝟎

given by
𝑨𝒙𝟐
+ 𝑩𝒙𝒚 + 𝑪𝒚𝟐
+ 𝑫𝒙 + 𝑬𝒚 + 𝑭 = 𝟎
If we let 𝐴 = 1, 𝐵 = 0, 𝐶 = 1, 𝐷 = 0, 𝐸 = 0, and 𝐹 =
−𝑟2
,
this general equation reduces to the
equation of a circle centered at the
origin: 𝑥2
+ 𝑦2
= 𝑟2
.

The concept of discriminant is also
applicable to second-degree equations in
two variables. The discriminant 𝐵2
− 4𝐴𝐶
determines the shape of the conic
section.

Example 1 Determining the Type of
Conic
Determine what type of conic corresponds
to each of the following equations:
1. 2𝑥2
+ 4𝑥𝑦 + 3𝑦2
+ 12𝑦 − 1 = 0
2. 3𝑥2
− 2𝑦2
+ 6𝑥 + 10𝑦 − 16 = 0

Solution
1. Write the general form of the
second-degree equation:

Solution
second-degree equation: 𝐴𝑥2
+ 𝐵𝑥𝑦 + 𝐶𝑦2
+
𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0
2𝑥2
+ 4𝑥𝑦 + 3𝑦2
+ 12𝑦 − 1 = 0

Solution
+
𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0
2𝑥2
+ 4𝑥𝑦 + 3𝑦2
+ 12𝑦 − 1 = 0
Identify 𝐴, 𝐵, 𝐶, 𝐷, 𝐸, and 𝐹.

Solution
+
𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0
2𝑥2
+ 4𝑥𝑦 + 3𝑦2
+ 12𝑦 − 1 = 0
𝐴 = 2, 𝐵 = 4, 𝐶 = 3, 𝐷 = 0, 𝐸 = 12, 𝐹 = −1

Solution
+
𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0
2𝑥2
+ 4𝑥𝑦 + 3𝑦2
+ 12𝑦 − 1 = 0
𝐴 = 2, 𝐵 = 4, 𝐶 = 3, 𝐷 = 0, 𝐸 = 12, 𝐹 = −1
Calculate the discriminant

Solution
+
𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0
2𝑥2
+ 4𝑥𝑦 + 3𝑦2
+ 12𝑦 − 1 = 0
𝐴 = 2, 𝐵 = 4, 𝐶 = 3, 𝐷 = 0, 𝐸 = 12, 𝐹 = −1
𝐵2
−4𝐴𝐶 =

Solution
+
𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0
2𝑥2
+ 4𝑥𝑦 + 3𝑦2
+ 12𝑦 − 1 = 0
𝐴 = 2, 𝐵 = 4, 𝐶 = 3, 𝐷 = 0, 𝐸 = 12, 𝐹 = −1
𝐵2
−4𝐴𝐶 = 42
−4 2 3 =

Solution
+
𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0
2𝑥2
+ 4𝑥𝑦 + 3𝑦2
+ 12𝑦 − 1 = 0
𝐴 = 2, 𝐵 = 4, 𝐶 = 3, 𝐷 = 0, 𝐸 = 12, 𝐹 = −1
𝐵2
−4𝐴𝐶 = 42
−4 2 3 = 16 − 24 =

Solution
+
𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0
2𝑥2
+ 4𝑥𝑦 + 3𝑦2
+ 12𝑦 − 1 = 0
𝐴 = 2, 𝐵 = 4, 𝐶 = 3, 𝐷 = 0, 𝐸 = 12, 𝐹 = −1
𝐵2
−4𝐴𝐶 = 42
−4 2 3 = 16 − 24 = −8

Solution
+
𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0
2𝑥2
+ 4𝑥𝑦 + 3𝑦2
+ 12𝑦 − 1 = 0
𝐴 = 2, 𝐵 = 4, 𝐶 = 3, 𝐷 = 0, 𝐸 = 12, 𝐹 = −1
𝐵2
−4𝐴𝐶 = 42
−4 2 3 = 16 − 24 = −8 < 0

Solution
+
𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0
2𝑥2
+ 4𝑥𝑦 + 3𝑦2
+ 12𝑦 − 1 = 0
𝐴 = 2, 𝐵 = 4, 𝐶 = 3, 𝐷 = 0, 𝐸 = 12, 𝐹 = −1
𝐵2
−4𝐴𝐶 = 42
−4 2 3 = 16 − 24 = −8 < 0
Since the discriminant is negative, the
equation is that of an ellipse.

Solution
second-degree equation:

Solution
+
𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0
3𝑥2
− 2𝑦2
+ 6𝑥 + 10𝑦 − 16 = 0

Solution
+
𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0
3𝑥2
− 2𝑦2
+ 6𝑥 + 10𝑦 − 16 = 0
𝐴 = 3, 𝐵 = 0, 𝐶 = −2, 𝐷 = 6, 𝐸 = 10, 𝐹 = −16

Solution
+
𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0
3𝑥2
− 2𝑦2
+ 6𝑥 + 10𝑦 − 16 = 0
𝐴 = 3, 𝐵 = 0, 𝐶 = −2, 𝐷 = 6, 𝐸 = 10, 𝐹 = −16
𝐵2
−4𝐴𝐶 =

Solution
+
𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0
3𝑥2
− 2𝑦2
+ 6𝑥 + 10𝑦 − 16 = 0
𝐴 = 3, 𝐵 = 0, 𝐶 = −2, 𝐷 = 6, 𝐸 = 10, 𝐹 = −16
𝐵2
−4𝐴𝐶 = 02
−4 3 −2 =

Solution
+
𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0
3𝑥2
− 2𝑦2
+ 6𝑥 + 10𝑦 − 16 = 0
𝐴 = 3, 𝐵 = 0, 𝐶 = −2, 𝐷 = 6, 𝐸 = 10, 𝐹 = −16
𝐵2
−4𝐴𝐶 = 02
−4 3 −2 = 0 + 24 =

Solution
+
𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0
3𝑥2
− 2𝑦2
+ 6𝑥 + 10𝑦 − 16 = 0
𝐴 = 3, 𝐵 = 0, 𝐶 = −2, 𝐷 = 6, 𝐸 = 10, 𝐹 = −16
𝐵2
−4𝐴𝐶 = 02
−4 3 −2 = 0 + 24 = 24

Solution
+
𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0
3𝑥2
− 2𝑦2
+ 6𝑥 + 10𝑦 − 16 = 0
𝐴 = 3, 𝐵 = 0, 𝐶 = −2, 𝐷 = 6, 𝐸 = 10, 𝐹 = −16
𝐵2
−4𝐴𝐶 = 02
−4 3 −2 = 0 + 24 = 24 > 0

Solution
+
𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0
3𝑥2
− 2𝑦2
+ 6𝑥 + 10𝑦 − 16 = 0
𝐴 = 3, 𝐵 = 0, 𝐶 = −2, 𝐷 = 6, 𝐸 = 10, 𝐹 = −16
𝐵2
−4𝐴𝐶 = 02
−4 3 −2 = 0 + 24 = 24 > 0
Since the discriminant is negative, the
equation is that of a hyperbola.

Practice Exercises
Determine what type of conic corresponds
to each of the following equations:
1. 2𝑥2
+ 𝑦2
= 4
2. 𝑥2
+ 𝑥𝑦 − 𝑦2
+ 2𝑥 = −3

The Distance Formula can be used to obtain an equation
of a circle whose center is (ℎ, 𝑘) and whose radius is 𝑟.

(𝑥 − ℎ)2+(𝑦 − 𝑘)2= 𝑟

(𝑥 − ℎ)2+(𝑦 − 𝑘)2= 𝑟
(𝑥 − ℎ)2+(𝑦 − 𝑘)2= 𝑟2

Your Turn!
A.Identify the center and radius of the
circle with the given equation in each
item
1. 𝑥2
+ 𝑦2
=
4
9
2. (𝑥 + 4)2
+ 𝑦 −
3
4
2
= 1
3. 𝑥2
− 4𝑥 + 𝑦2
− 4𝑦 − 8 = 0

Your Turn!
B.Find the standard equation of the
circle which satisfies the given
conditions.
1.Center at the origin, radius 4
2.Center at −4, 3 , radius 7
3.The point (1, 4) is on a circle whose
center is at (−2, −3)

Every accomplishment
STARTS with the Decision
to TRY.

Introduction to Conic Sections and Circles.pptx

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