Degenerate Conics:

Pre-Calculus
S.Y. 2024 - 2025 : Q1

1.0 CONIC SECTIONS

What is a Conic Section?
➔ A curve formed by the intersection
of a plane and a double right
circular cone. (Apollonius of Perga)

Conic Sections:

GENERAL EQ. OF ANY CONIC SECTION

Ax² + Bxy + Cy² + Dx + Ey + F = 0

Discriminant Formula:
B² - 4AC
★ If the plane is perpendicular to the
axis, the section is a circle. (a
special type of ellipse) ☆ If B² - 4AC < 0, conic is an ellipse
★ An ellipse is formed by slicing a ☆ If B² - 4AC < 0, B=0, A=C, conic is a
single cone with a slanted plane circle
not perpendicular to the axis. ☆ If B² - 4AC = 0, conic is a parabola
★ A parabola is formed by the slicing ☆ If B² - 4AC > 0, conic is a hyperbola
of the plane through the top OR the
bottom of the cone. Examples:
★ A hyperbola is formed by slicing a.) x² + y² + 4x - 2y - 11 = 0
both the top AND bottom of the 0² - 4(1)(1) = -4 < 0 → circle
cone. B=0 , A=C

The conic sections above are called b.) y² - 4x - 12y + 28 = 0

non-degenerate conics. In a 0² - 4(0)(1) = 0 = 0 → parabola
non-degenerate conics, the cutting plane
does not pass through the vertex of the c.) x² + 2y² - 2x + 8y - 11 = 0
cone. When the cutting plane intersects 0² - 4(1)(2) = -8 < 0 → ellipse
the vertex of the cone, the resulting conic
section is called degenerate conic. d.) 9y² - 16x² - 144 = 0
Degenerate conics include; a point, a line, 0² - 4(-16)(9) = 576 > 0 →
and two intersecting lines. hyperbola

PRE-CALCULUS | S.Y. 2024 - 2025 | SEMESTER 1 | QUARTER 1 PAGE 1

 Examples:
1.1 CIRCLE a.) x² + y² + 10y = 0
Is a set of all points on a plane that are D=0, E=10, F=0
equidistant from a given fixed point 0/4² + 10/4² - 0 = 25 > 0 → circle

Equations of Circles: b.) x² + y² - 6x + 2y + 10 = 0

D=-6, E=2, F=10
STANDARD FORM
(-6)/4² + 2/4² - 10 = 0
★ x² + y² = r² 9 + 1 - 10 = 0 = 0 → point circle
★ (x-h)² + (y-k)² = r²
Examples: c.) x² + y² + 4x - 8y - 1 = 0
a.) Center (0,0) ; r = 3 units D=4, E=-8, F=-1
(x-0)² + (y-0)² = 3² 4/4² + (-8)/4² - (-1) = 0
x² + y² = 9 4 + 16 - (-1)
4 + 16 + 1 = 21 > 0 → circle
b.) Center (3,2) ; r = 5 units
(x-3)² + (y-2)² = 5²
TRANSFORMING EQUATIONS (GF TO SF)
(x-3)² + (y-2)² = 25
Step 1: Group the terms with the same
c.) Center (1,-4) ; r = 3 units variable (x,y)
(x-1)² + (y+4)² = 3² Step 2: Move the constant to the right side
(x-1)² + (y+4)² = 9 of the equation
Step 3: Create two perfect square
d.) Center (¼,⅔) ; r = 4 units trinomials by completing the square
(x-¼)² + (y-⅔)² = 4² method. Whatever terms that were added
(x-¼)² + (y-⅔)² = 16 to the left side, add them also to the right
side of the eq.
e.) Center (√5, 3√2) ; r = √11 Step 4: Simplify both sides of the eq.
(x-√5)² + (y-3√2)² = 11 Step 5: Transform the perfect square
trinomials into square of binomials to
make the eq. In the center-radius form
GENERAL FORM
a² ± 2ab + b² → (a ± b)²
★ x² + y² + Dx + Ey + F = 0
★ D, E, and F and constants Examples:
a.) x² + y² + 6x + 8y + 24 = 0
☆ If D/4² + E/4² - F > 0, the equation (x²+6x)+(y²+8y) = -24
has a graph, a circle (x²+6x+9)+(y²+8y+16) = -24 + 9 + 16
☆ If D/4² + E/4² - F = 0, the equation 6/2 = 3 8/2 = 4
has a graph, a point circle 3² = 9 4² = 16
☆ If D/4² + E/4² - F < 0, the equation (x²+6x+9)+(y²+8y+16) = 1
has no graph ↳ √9 = 3 ↳ √16 = 4
(x+3)² + (y+4)² = 1
★ Center (-3,-4) ; r = 1

PRE-CALCULUS | S.Y. 2024 - 2025 | SEMESTER 1 | QUARTER 1 PAGE 2

b.) x² + y² - 14x + 24 = 0 b.) x² + (y-4)² = 49
(x²-14x)+y² = -24 x² + y² - 8y + 16 - 49 = 0
(x²-14x+49)+y² = -24 + 49 ★ x² + y² - 8y - 33 = 0
-14/2 = -7
(-7)² = 49 c.) (x-1/2)² + (y+3)² = 9
(x²-14x+49)+y² = 25 x² - x + 1/4 + y² + 6y + 9 - 9 = 0
(x-7)²+y² = 5² ★ x² + y² - x + 6y + 1/4 = 0
★ Center (7,0) ; r = 5 units
2(x)(1/2) = x

c.) x²+y²-10x+2y = 0
(1/2)² = 1/2 x 1/2 = 1/4
(x²-10x)+(y²+2y) = 0
(x²-10x+25)+(y²+2y+1) = 0 + 25 + 1
-10/2 = -5 2/2 = 1 d.) (x-2/3)² + (y+2)² = 4
(-5)² = 25 1² = 1 x² - 4/3x + 4/9 + y² + 4y + 4 - 4 = 0
(x-5)²+(y+1)² = √26 ★ x² + y² - 4/3x + 4y + 4/9 = 0
★ Center (5,-1) ; r = √26 OR
★ 9x² + 9y² - 12x + 36y + 4 = 0
d.) 4x²+4y²+12x-8y+10 = 0 2(x)/1 (2/3) = 4/3x
(4x²+12x)+(4y²-8y) = -10
4(x²+3x)+4(y²-2y) = -10 2/3 x 2/3 = 4/9
(x²+3x+9/4)+(y²-2y+4) = -10 + 9 + 4
3/2 = (3/2)² = 9/4 -2/2 = -1
(-1)² = 1 1.2 FINDING THE EQUATION OF A
CIRCLE
4(x+3/2)/4² + 4(y-1)/4² = 3/4
(x+3/2)² + (y-1)² = 3/4 Examples:
(x+3/2)² + (y-1)² = (√3/2)² a.) Find the equation of a circle in GF
★ Center (-3/2, 1) ; r = √3/2 units whose center is at (-3,-2) and has a radius
of 3√5 units
C(-3,-2) ; r=3√5
TRANSFORMING EQUATIONS (SF TO GF) (x-h)²+(y-k)² = r²
Step 1: Square the two binomials (x+3)²+(y+2)² = (3√5)²
Step 2: Place all terms to the left side of (x+3)²+(y+2)² = 45
the equation x² + 6x + 9 + y² + 4y + 4 - 45 = 0
Step 3: Simplify the equation by combining ★ x² + y² + 6x + 4y - 32 = 0
all like terms
Step 4: Follow the GF; x²+y²+Dx+Ey+F = 0 b.) Write the equation of a Circle S in SF
whose center is at the origin and has a
Examples: diameter of 14 units (radius is half of
a.) (x-10)² + (y+8)² = 36 diameter)
x² - 20x + 100 + y² + 16y + 64 - 36 = 0 C(0,0) ; r=7
★ x² + y² - 20x + 16y + 128 = 0 x²+y² = 7²
★ x²+y² = 25

PRE-CALCULUS | S.Y. 2024 - 2025 | SEMESTER 1 | QUARTER 1 PAGE 3

c.) Write the equation of a Circle P in SF Axis of Symmetry
whose center is at the origin and passes ★ A line that passes through the
through the point (4,3) focus and is perpendicular to the
directrix

DISTANCE FORMULA (find r)

Latus Rectum
d = √(x₂-x₁)²+(y₂-y₁)²
★ A chord that passes through the
= √(4-0)²+(3-0)²
focus and perpendicular to the axis
= √16+9
of symmetry
= √25
★ r =5
Directrix
C(0,0) ; r=5 ★ A line which is perpendicular to the
x²+y² = 5² axis of symmetry and parallel to
★ x²+y² = 25 latus rectum

1.3 PARABOLA CHARACTERISTICS

Is a set of all coplanar points that are CASE 1: Vertex is at Origin

equidistant from a fixed line called directrix ➔ Standard Form of the equation of a
and a fixed point called focus. It has 4 Parabola
possible openings — upward, downward, y²=4px → left, right
right, left. x²=4py → upward, downward

PROPERTIES OF PARABOLA
❕ (Table on pg. 28)
Examples:
a.) Find the coordinates of focus and the
equation of the directrix
x²=16y
4p/4 = 16/4
★ p=4
★ Focus: (0,4)
★ Directrix: y=-4

Vertex (h,k)
b.) Find the coordinates of focus and the
★ Referred as the lowest or highest
equation of the directrix
point of a parabola
y²=48x
4p/4 = 48/4
★ p=12
★ Focus: (12,0)
★ Directrix: x=-12

PRE-CALCULUS | S.Y. 2024 - 2025 | SEMESTER 1 | QUARTER 1 PAGE 4

c.) Find the coordinates of focus and the (x-8)²=-8(y-2)
equation of the directrix ★ Vertex: (8,2)
y²=-20x 4p/4=-8/4
4p/4 = -20/4 ★ p=-2
★ p=-5 ★ Focus: (8,0)
★ Focus: (-5,0) ★ Directrix: y=4
★ Directrix: x=5

d.) Find the coordinates of focus and the 1.4 SF AND GF OF A PARABOLA
equation of the directrix Standard Form of the eq. of a Parabola
x²-4y=0 ★ y²=4px
x²=4y ★ x²=4py
4p/4 = 4/4 General Form of the eq. of a Parabola
★ p=1 ★ y²+Dx+Ey+F=0
★ Focus: (0,1) ★ x²+Dx+Ey+F=0
★ Directrix: y=-1

TRANSFORMING EQUATIONS (GF TO SF)

CASE 2: Vertex is at (h,k)
➔ Standard Form of the equation of a Examples:
Parabola with vertex at (h,k) a.) y²-8x+12y-20=0
y²+12y=8x+20
(y-k)²=4p(x-h) → left, right
y²+12y+36=8x+20+36
(x-h)²=4p(y-k) → upward,
12/2=6
downward

❕ (Table on pg. 34) 6²=36

y²+12y+36=8x+56
(y+6)²=8(x+7)
Examples: b.) x²-18x+10y+4=0
a.) Find the coordinates of the vertex, the x²-18x=-10y+4
focus, and the equation of the directrix x²-18x+81=-10y-4+81
(y+2)²+4(x-3) -18/2=-9
★ Vertex: (3,-2) (-9)²=81
4p=4 x²-18x+81=-10y+77
★ p=1 (x-9)²=-10(y-77/10)
★ Focus: (4,-2)
★ Directrix: x=2
TRANSFORMING EQUATIONS (GF TO SF)

b. Find the coordinates of the vertex, the Examples:

focus, and the equation of the directrix a.) (y+4)²=-5(x+6)
y=-1/8(x-8)²+2 y²+8y+16=-5x-30
1/8(x-8)²=-y+2 y²+8y+16+5x-30=0
8[1/8(x-8)²=-y+2] y²+5x+8y+46=0
(x-8)²=8(-y+2)

PRE-CALCULUS | S.Y. 2024 - 2025 | SEMESTER 1 | QUARTER 1 PAGE 5

b.) (x-3)²=8(y+2)
x²-6x+9=8y+16
x²-6x+9-8y-16=0
x²-6x-8y-7=0

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