Analytic Geometry
Unit 6: Modeling Geometry
Excerpts from Georgia Department of
Education Webinar August 8, 2013
melissa.stewart@hallco.org
August 2013
Warm-Up
Use the Pythagorean theorem to find an equation in x and y
whose solutions are the points on the circle of radius 2 with
center (1,1) and explain why it works.
Solution
For any point (x, y) on the circle. The horizontal length is |x – 1|
and the vertical length is |y – 1|. The hypotenuse is 2.
melissa.stewart@hallco.org
August 2013
Concepts & Skills to Maintain from Previous Grades
 number sense
• computation with whole numbers and decimals, including
application of order of operations
• addition and subtraction of common fractions with like
denominators
• applications of the Pythagorean Theorem
• usage of the distance formula, including distance between a
point and a line.
• finding a midpoint
• graphing on a coordinate plane
• completing the square
• operations with radicals
• methods of proof
Websites to help with the above:
http://www.crctlessons.com/
www.aplusmath.com
www.aaamath.com
melissa.stewart@hallco.org
August 2013
What’s the main idea of Unit 6?
• Translate between the geometric description and the
equation for a conic section
• Use coordinates to prove simple geometric theorems
algebraically
• Solve system of equations
Enduring Understandings from this Unit
• Write and interpret the equation of a circle
• Derive the formula for a circle using the Pythagorean
Theorem
• Recognize, write, and interpret equations of parabolas
• Prove properties involving parabolas
• Prove properties involving circles
• Apply algebraic formulas and ideas to geometric figures
and definitions
• The intersection of a line and a quadratic figure is the
point where the two equations are equal.
melissa.stewart@hallco.org
August 2013
Examples & Explanations
1.
A flashlight mirror has the shape of a paraboloid of
diameter 4 inches and depth 2 inches. Where should the
bulb be placed so that the emitted light rays are parallel to
the axis of the paraboloid?
Solution:
The bulb would need to be placed ½ inch from the rear of the
flashlight mirror.
melissa.stewart@hallco.org
August 2013
2.
Annika wonders why we are suddenly thinking about
parabolas in a completely different way than when we did
quadratic functions. She wonders how these different ways
of thinking match up. For instance, when we talked about
quadratic functions earlier we started with
.
“Hmmmm. …. I wonder where the focus and directrix would
be on this function,” she thought. Help Annika find the focus
and directrix for
.
Solution:
The vertex is at the origin, so the focus will be located at (0, p)
and the directrix will be located at
The vertex is at the origin, so the focus will be located at (0, ¼ )
and the directrix will be located at
melissa.stewart@hallco.org
August 2013
3.
You may know that the smaller |a| in the standard form
equation of a parabola , the wider the parabola . In other
words y = .1x² is a wider parabola than y = .2x². How does
this relate to the directrix and focus?
Solution:
Some Generalizations:
1st:
2nd:
y
3rd:
4th:
Notice as |a| decreases p increases, ie. the distance between the
vertex and directrix and vertex and focus increases.
melissa.stewart@hallco.org
August 2013
 Websites to assist and enrich:
http://brightstorm.com/
http://www.khanacademy.org/
 The student edition for Unit 6 can be found at
https://www.georgiastandards.org/CommonCore/Pages/Math-9-12.aspx On the left side, please
look under mathematics, Accelerated Geometry
B/Advanced Algebra. Then, the right side has a pulldown menu to access the units.
 Additional parent guides will be posted to the parent
resource page on
http://www.hallco.org/boe/index.php (right hand
menu) as they become available.
melissa.stewart@hallco.org
August 2013