P.o.D. – Find the inclination, in
radians, of the described line.
1.) (12,8), (-4,-3)
2.) (-2,20), (10,0)
3.) 6x-2y+8=0
4.) 4x+5y=9
5.) x-y=10
6.) Find the angle of inclination
between the lines 2x-y=2 and
4x+3y=24.
1.)
2.)
3.)
4.)
0.6023 radians
2.1112 radians
1.2490 radians
2.4669 radians
5.)
𝜋
4
radians
6.) 1.071 radians
10.2 – Introduction to Conics:
Parabolas
Learning Target: I can recognize
a conic as the intersection of a
plane and a cone; write
equations of parabolas in
standard form
*Read the first two paragraphs
on page 735 and study the
pictures.
Standard Equation of a
Parabola:
{Vertical axis = opens up/down}
(𝑥 − ℎ)2 = 4𝑝(𝑦 − 𝑘)
Directrix: y=k-p
Focus: (h,k+p)
{Horizontal axis = opens
left/right}
(𝑦 − 𝑘)2 = 4𝑝(𝑥 − ℎ)
Directrix: x=h-p
Focus: (h+p,k)
Where (h,k) is the vertex and p
is the distance from the vertex
to the focus or the directrix.
(draw a detailed parabola on the
whiteboard)
Avoiding Common Errors:
Do NOT confuse
(𝑥 − ℎ)2 = 4𝑝(𝑦 − 𝑘) with 𝑦 =
2
1
𝑎(𝑥 − ℎ) + 𝑘. Note that 𝑎 = 𝑝
4
EX: Find the standard equation
of the parabola with vertex
(3,2) and focus (1,2)
Begin by plotting these points to
determine if the axis is
horizontal or vertical.
(plot the points on the board)
This parabola is horizontal.
(𝑦 − 𝑘)2 = 4𝑝(𝑥 − ℎ)
Substitute what you know.
(𝑦 − 2)2 = 4𝑝(𝑥 − 3)
We need to find p. Remember, p
is the distance between the
vertex and the focus. Since this
graph opens to the left, p will be
negative.
𝑝 = √(3 − 1)2 + (2 − 2)2 = √4 = −2
Now write the final equation.
(𝑦 − 2)2 = 4(−2)(𝑥 − 3)
(𝑦 − 2)2 = −8(𝑥 − 3)
*I would not multiply 4p.
EX: Find the focus of the
parabola given by
𝑦=
−1
6
(𝑥 2 + 4𝑥 − 2)
We need to complete the square.
Begin by distributing.
−1 2 2
1
𝑦=
𝑥 − 𝑥+
6
3
3
Now we can begin the
completing the square process.
1 −1 2 2
𝑦− =
𝑥 − 𝑥
3
6
3
1 −1 2
𝑦− =
(𝑥 + 4𝑥)
3
6
1
−1
−1 2
2
𝑦 − + (2 ) ( ) =
(𝑥 + 4𝑥 + 22 )
3
6
6
−1
𝑦−1=
(𝑥 + 2)2
6
Put the equation in standard
form.
−1
(𝑥 + 2)2 = 𝑦 − 1
6
(𝑥 + 2)2 = −6(𝑦 − 1)
We need to find p.
4𝑝 = −6
−6 −3
𝑝=
=
4
2
Find the focus.
−3
𝐹 (ℎ, 𝑘 + 𝑝) = 𝐹 (−2,1 +
)
2
1
= 𝐹 (−2, − )
2
EX: Find the standard form of
the equation of the parabola
with vertex (1,3) and focus
(1,5).
Plot the points to determine if
the parabola is horizontal or
vertical.
This parabola is vertical.
(𝑥 − ℎ)2 = 4𝑝(𝑦 − 𝑘)
Find p.
p=2
Substitute what you know.
(𝑥 − 1)2 = 4(2)(𝑦 − 3)
(𝑥 − 1)2 = 8(𝑦 − 3)
Write this parabola as a
standard form quadratic
equation.
(𝑥 − 1)(𝑥 − 1) = 8𝑦 − 24
𝑥 2 − 2𝑥 + 1 = 8𝑦 − 24
𝑥 2 − 2𝑥 + 25 = 8𝑦
1 2 1
25
𝑦= 𝑥 − 𝑥+
8
4
8
EX: Find the equation of the
tangent line to the parabola
given by 𝑥 2 = 2𝑦 at the point
9
(3, ).
2
*We will do this mostly on the
calculator.
Solve the equation for y and
graph.
𝑥2
𝑦=
2
Use the “TANGENT” feature on
the calculator to find the
equation of the tangent line.
𝑦 = 3𝑥 −
9
2
EX: Consider the equation 𝑦 2 =
8𝑥 + 48. Find the coordinates of
the vertex and the focus and the
equations of the directrix and
the axis of symmetry.
First, write the equation in
standard form.
𝑦 2 = 8(𝑥 + 6) → 𝑦 2 = 4(2)(𝑥 + 6)
→ (𝑦 − 0)2 = 4(2)(𝑥 + 6)
Vertex: V(-6,0)
Focus: F(-6+2,0)=F(-4,0)
Directrix: x=-6-2=-8
Axis of Symmetry: y=0
EX: Consider the equation 2𝑥 2 −
8𝑥 + 𝑦 + 6 = 0. Write the equation
in standard form.
2𝑥 2 − 8𝑥 = −𝑦 − 6
−1
𝑥 − 4𝑥 =
𝑦−3
2
−1
2
2
𝑥 − 4𝑥 + 2 =
𝑦−3+4
2
−1
2
(𝑥 − 2) =
𝑦+1
2
−1
2
(𝑥 − 2) =
(𝑦 − 2)
2
−1
2
(𝑥 − 2) = 4 ( ) (𝑦 − 2)
8
Now find the coordinates of the
vertex and the focus.
2
V(2,2)
F(2, 2-1/8)=F(2, 15/8)
Now find the equations for the
directrix and the axis of
symmetry.
Directrix: y=2+1/8=17/8
Axis of symmetry: x=2
EX: Given the equation
𝑥 2 = 12𝑦 + 60. Find the vertex
and focus and the equations for
the directrix and axis of
symmetry.
𝑥 2 = 12(𝑦 + 5)
(𝑥 + 0)2 = 4(3)(𝑦 + 5)
Vertex: V(0,-5)
Focus: F(0,-5+3)=F(0,-2)
Directrix: y=-5-3=-8
Axis of Symmetry: x=0
EX: Write the equation 𝑦 2 + 6𝑦 +
8𝑥 + 25 = 0 in standard form.
Then find the coordinates of the
vertex and focus and the
equations of the directrix and
axis of symmetry.
𝑦 2 + 6𝑦 = −8𝑥 − 25
𝑦 2 + 6𝑦 + 32 = −8𝑥 − 25 + 9
(𝑦 + 3)2 = −8𝑥 − 16
(𝑦 + 3)2 = −8(𝑥 + 2)
(𝑦 + 3)2 = 4(−2)(𝑥 + 2)
Vertex: V(-2,-3)
Focus: F(-2-2,-3)=F(-4,-3)
Directrix: x=-2+2=0
Axis of Symmetry: y=-3
EX: Write the equation of a
parabola that meets each set of
conditions.
a.) Vertex(-2,4); Focus(-2,3)
b.) Focus(2,1); directrix x=-2
a.) Opens down; vertical
(𝑥 − ℎ)2 = 4𝑝(𝑦 − 𝑘 )
→ (𝑥 + 2)2 = 4(−1)(𝑦 − 4)
→ (𝑥 + 2)2 = (−4)(𝑦 − 4)
b.) Opens right; horizontal
V(0,1)
(𝑦 − 𝑘)2 = 4𝑝(𝑥 − ℎ) → (𝑦 − 1)2
= 4(2)(𝑥 − 0) → (𝑦 − 1)2 = 8𝑥
EX: Suppose that the receiver in
a parabolic satellite dish is 2-ft
from the vertex and located at
the focus. Assume that the
vertex is at the origin and the
dish is pointed upward. Find an
equation that models the cross
section of this dish.
V(0,0)
F(0,2)
p=2
(𝑥 − ℎ)2 = 4𝑝(𝑦 − 𝑘 ) → 𝑥 2 = 4(2)(𝑦)
→ 𝑥 2 = 8𝑦
Upon completion of this lesson,
you should be able to:
1. Find the vertex, focus,
directrix, and direction of
opening for a parabola.
2. Write the equation of a
parabola given the vertex
and a point.
For more information, visit
https://www.youtube.com/watch?v=Fz27rCt
B2Bo
HW Pg.740 6-57 3rds, 61, 66-67,
79-81