1. No category

Introduction
A theorem is statement that is shown to be true. Some
important theorems have names, such as the
Pythagorean Theorem, but many theorems do not have
names. In this lesson, we will apply various geometric
and algebraic concepts to prove and disprove
statements involving circles and parabolas in a
coordinate plane. If a statement is proven, it is a
theorem. If a statement is disproved, it is not a theorem.
1
6.2.1: Using Coordinates to Prove Geometric Theorems About Circles and Parabolas
Introduction, continued
The directions for most problems will have the form
“Prove or disprove…,” meaning we will work through
those problems to discover whether each statement is
true or false. Then, at the end of the work, we will state
whether we have proved or disproved the statement.
2
6.2.1: Using Coordinates to Prove Geometric Theorems About Circles and Parabolas
Key Concepts
• A theorem is any statement that is proven or can be
proved to be true.
• The standard form of an equation of a circle with
center (h, k) and radius r is (x – h)2 + (y – k)2 = r 2. This
is based on the fact that any point (x, y) is on the
circle if and only if
( x - h) + ( y - k )
2
2
= r.
3
6.2.1: Using Coordinates to Prove Geometric Theorems About Circles and Parabolas
Key Concepts, continued
• Completing the square
is the process of determining
2
æ bö
the value of ç ÷ and adding it to x2 + bx to form the
2
è 2ø
æ bö
2
perfect square trinomial x + bx + ç ÷ .
è 2ø
• A quadratic function can be represented by an
equation of the form f(x) = ax2 + bx + c, where a ≠ 0.
4
6.2.1: Using Coordinates to Prove Geometric Theorems About Circles and Parabolas
Key Concepts, continued
• The graph of any quadratic function is a parabola that
opens up or down.
• A parabola is the set of all points that are equidistant
from a fixed line, called the directrix, and a fixed point
not on that line, called the focus.
• The parabola, directrix, and focus are all in the same
plane.
5
6.2.1: Using Coordinates to Prove Geometric Theorems About Circles and Parabolas
Key Concepts, continued
• The distance between the focus and a point on the
parabola is the same as the distance from that point
to the directrix.
• The vertex of the parabola is the point on the parabola
that is closest to the directrix.
• Every parabola is symmetric about a line called the
axis of symmetry.
6
6.2.1: Using Coordinates to Prove Geometric Theorems About Circles and Parabolas
Key Concepts, continued
• The axis of symmetry intersects the parabola at the
vertex.
b
• The x-coordinate of the vertex is - .
2a
æ bö
• The y-coordinate of the vertex is f ç - ÷ .
è 2a ø
• The standard form of an equation of a parabola that
opens up or down and has vertex (h, k) is
(x – h)2 = 4p(y – k), where p ≠ 0 and p is the distance
between the vertex and the focus and between the
vertex and the directrix.
6.2.1: Using Coordinates to Prove Geometric Theorems About Circles and Parabolas
7
Key Concepts, continued
• Parabolas that open up or down represent functions,
and their equations can be written in either of the
following forms: y = ax2 + bx + c or
(x – h)2 = 4p(y – k). If one form is known, the other
can be found.
• The standard form of an equation of a parabola that
opens right or left and has vertex (h, k) is
(y – k)2 = 4p(x – h), where p ≠ 0 and p is the distance
between the vertex and the focus and between the
vertex and the directrix.
8
6.2.1: Using Coordinates to Prove Geometric Theorems About Circles and Parabolas
Key Concepts, continued
• In any parabola:
• The focus and directrix are each |p| units from the
vertex.
• The focus and directrix are 2|p| units from each
other.
9
6.2.1: Using Coordinates to Prove Geometric Theorems About Circles and Parabolas
Common Errors/Misconceptions
• neglecting to square the radius for the equation of the
circle
• using the equation of a parabola that opens up or down
for a parabola that opens left or right, and vice versa
10
6.2.1: Using Coordinates to Prove Geometric Theorems About Circles and Parabolas
Guided Practice
Example 1
Given the point A (–6, 0), prove or disprove that point A
is on the circle centered at the origin and passing
through -2, - 4 2 .
(
)
11
6.2.1: Using Coordinates to Prove Geometric Theorems About Circles and Parabolas
Guided Practice: Example 1, continued
1. Draw a circle on a coordinate plane using
the given information.
You do not yet know if point A lies on the circle, so
don’t include it in your diagram.
In the diagram that follows, the name P is assigned
to the origin and G is assigned to the known point on
the circle.
To help in plotting points, you can use a calculator to
find decimal approximations.
12
6.2.1: Using Coordinates to Prove Geometric Theorems About Circles and Parabolas
Guided Practice: Example 1, continued
13
6.2.1: Using Coordinates to Prove Geometric Theorems About Circles and Parabolas
Guided Practice: Example 1, continued
2. Find the radius of the circle using the
distance formula.
Use the known points, P and G, to determine the
radius of the circle.
r=
( x2 - x1 ) + ( y 2 - y1 )
2
(
2
Distance formula
)
r = éë( -2) - ( 0 ) ùû + é -4 2 - ( 0 ) ù
ë
û
2
2
Substitute (0, 0) and
-2,-4 2 for
(x1, y1) and (x2, y2).
(
)
14
6.2.1: Using Coordinates to Prove Geometric Theorems About Circles and Parabolas
Guided Practice: Example 1, continued
r=
( -2)
2
(
+ -4 2
)
2
Simplify, then solve.
r = 4 + 32
r = 36
r =6
The radius of the circle is 6 units.
For point A to be on the circle, it must be precisely
6 units away from the center of the circle.
15
6.2.1: Using Coordinates to Prove Geometric Theorems About Circles and Parabolas
Guided Practice: Example 1, continued
3. Find the distance of point A from the
center P to determine whether it is on the
circle.
The coordinates of point P are (0, 0).
The coordinates of point A are (–6, 0).
AP =
(x
2
- x1 ) + ( y 2 - y1 )
2
2
AP = éë( -6 ) - ( 0 ) ùû + éë( 0 ) - ( 0 ) ùû
2
Distance formula
2
Substitute (0, 0) and
(–6, 0) for (x1, y1)
and (x2, y2).
16
6.2.1: Using Coordinates to Prove Geometric Theorems About Circles and Parabolas
Guided Practice: Example 1, continued
AP =
( -6) + ( 0)
2
2
Simplify, then solve.
AP = (-6)2
AP = 36
AP = 6
Point A is 6 units from the center, and since the
radius of the circle is 6 units, point A is on the circle.
17
6.2.1: Using Coordinates to Prove Geometric Theorems About Circles and Parabolas
Guided Practice: Example 1, continued
The original statement has been proved, so it is a
theorem.
✔
18
6.2.1: Using Coordinates to Prove Geometric Theorems About Circles and Parabolas
Guided Practice: Example 1, continued
19
6.2.1: Using Coordinates to Prove Geometric Theorems About Circles and Parabolas
Guided Practice
Example 2
Prove or disprove that the quadratic function graph with
vertex (–4, 0) and passing through (0, 8) has its focus at
(–4, 1).
20
6.2.1: Using Coordinates to Prove Geometric Theorems About Circles and Parabolas
Guided Practice: Example 2, continued
1. Sketch the graph
using the given
information.
21
6.2.1: Using Coordinates to Prove Geometric Theorems About Circles and Parabolas
Guided Practice: Example 2, continued
2. Derive an equation of the parabola from
its graph.
The parabola opens up, so it represents a function.
Therefore, its equation can be written in either of
these forms: y = ax2 + bx + c or (x – h)2 = 4p(y – k).
We were given a vertex and a point on the parabola;
therefore, we’ll use the form (x – h)2 = 4p(y – k) and
the vertex to begin deriving the equation.
22
6.2.1: Using Coordinates to Prove Geometric Theorems About Circles and Parabolas
Guided Practice: Example 2, continued
The vertex is (–4, 0), so h = –4 and k = 0.
( x - h)
2
= 4p ( y - k )
éë x - ( -4 ) ùû = 4p ( y - 0 )
2
( x + 4)
2
= 4py
Standard form of an equation
for a parabola that opens up
or down
Substitute the vertex (–4, 0)
into the equation.
Simplify, but do not expand
the binomial.
The equation of the parabola is (x + 4)2 = 4py.
23
6.2.1: Using Coordinates to Prove Geometric Theorems About Circles and Parabolas
Guided Practice: Example 2, continued
3. Substitute the given point on the parabola
into the standard form of the equation to
solve for p.
The point given is (0, 8).
( x + 4)
( 0 + 4)
2
2
= 4py
Simplified equation from step 2
= 4p ( 8 )
Substitute the point (0, 8) into
the equation.
16 = 32p
p=
Simplify and solve for p.
1
2
6.2.1: Using Coordinates to Prove Geometric Theorems About Circles and Parabolas
24
Guided Practice: Example 2, continued
4. Use the value of p to determine the focus.
p is positive, so the focus is directly above the vertex.
1
1
2
2
p = , so the focus is
unit above the vertex.
æ
1ö
The vertex is (–4, 0), so the focus is ç -4, ÷ .
2ø
è
25
6.2.1: Using Coordinates to Prove Geometric Theorems About Circles and Parabolas
Guided Practice: Example 2, continued
This result disproves the statement that the quadratic
function graph with vertex (–4, 0) and passing
through (0, 8) has its focus at (–4, 1).
The statement has been disproved, so it is not a
theorem.
Instead, the following statement has been proved:
The quadratic function graph with vertex (–4, 0) and
passing through (0, 8) has its focus at æ -4, 1ö .
ç
è
2 ÷ø
26
6.2.1: Using Coordinates to Prove Geometric Theorems About Circles and Parabolas
Guided Practice: Example 2, continued
✔
27
6.2.1: Using Coordinates to Prove Geometric Theorems About Circles and Parabolas
Guided Practice: Example 2, continued
28
6.2.1: Using Coordinates to Prove Geometric Theorems About Circles and Parabolas