unit 6 • modeling geometry lesson 2: using coordinates to...

UNIT 6 • MODELING GEOMETRYLesson 2: Using Coordinates to Prove Geometric Theorems About Circles and Parabolas

Instruction

CCGPS Analytic Geometry Teacher Resource © Walch EducationU6-69

IntroductionA 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.

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.

Prerequisite Skills

This lesson requires the use of the following skills:

• applying the distance, midpoint, and slope formulas

• using slope to determine whether lines are parallel, perpendicular, or neither

• identifying congruent angle pairs formed by parallel lines and a transversal

• simplifying expressions involving square roots

• applying the standard form of an equation of a circle

• applying the standard forms of equations of parabolas

• converting between different forms of equations of parabolas that represent functions

• using an equation of a parabola that represents a function to identify the vertex

• applying properties involving circles, tangent lines, and inscribed polygons

• understanding function notation


Instruction

CCGPS Analytic Geometry Teacher Resource U6-70

© Walch Education

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 y k r−( ) + −( ) =2 2 2 . This is based on the fact that any point (x, y) is on the circle if and only

if x h y k r−( ) + −( ) =2 2.

• Completing the square is the process of determining the value of b

2

2

and adding it to

x2 + bx to form the perfect square trinomial x bxb

2

2

2+ +

.

• A quadratic function can be represented by an equation of the form f (x) = ax2 + bx + c, where a ≠ 0.

• 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.

• 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.

• The axis of symmetry intersects the parabola at the vertex.

• The x-coordinate of the vertex is −b

a2.

• The y-coordinate of the vertex is fb

a−

2

.

• 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.


Instruction


• 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.

• 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.

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


Instruction


© Walch Education

Guided Practice 6.2.1Example 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, .

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.

G (–2, –4 2)√

y

xP (0, 0)


Instruction


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 x x y y= −( ) + −( )2 1

2

2 1

2 Distance formula

r = − −[ ] + −( ) − ( )

( ) ( )2 0 4 2 0

2 2 Substitute (0, 0) and

− −( )2 4 2, for (x1, y

1) and

(x2, y

2).

r = −( ) + −( )2 4 22 2 Simplify, then solve.

r

r

r

= +

==

4 32

36

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.


Instruction


© Walch Education

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 x y y= −( ) + −( )2 1

2

2 1

2 Distance formula

[ ] [ ]= − − + −AP ( 6) (0) (0) (0)2 2 Substitute (0, 0) and (–6, 0)

for (x1, y

1) and (x

2, y

2).

AP = −( ) + ( )6 02 2 Simplify, then solve.

AP

AP

AP

= −

==

( )6

36

6

2

Point A is 6 units from the center, and since the radius of the circle is 6 units, point A is on the circle.

The original statement has been proved, so it is a theorem.

G (–2, –4 2)√

y

xP (0, 0)

A (–6, 0)


Instruction


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).

1. Sketch the graph using the given information.

(0, 8)

(–4, 0)x

y


Instruction


© Walch Education

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.

The vertex is (–4, 0), so h = –4 and k = 0.

(x – h)2 = 4p(y – k) Standard form of an equation for a parabola that opens up or down

[x – (–4)]2 = 4p(y – 0) Substitute the vertex (–4, 0) into the equation.

(x + 4)2 = 4py Simplify, but do not expand the binomial.

The equation of the parabola is (x + 4)2 = 4py.

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)2 = 4py Simplified equation from step 2

(0 + 4)2 = 4p(8) Substitute the point (0, 8) into the equation.

16 = 32p Simplify and solve for p.

p =1

2


Instruction


4. Use the value of p to determine the focus.

p is positive, so the focus is directly above the vertex.

p =1

2, so the focus is

1

2 unit above the vertex.

The vertex is (–4, 0), so the focus is −

41

2, .

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 −

41

2, .

(0, 8)

(–4, 0)x

y

F

V

(–4, )12


Instruction


© Walch Education

Example 3

The following information is given about a parabola:

• The vertex V is at (0, 0).

• The focus F is at (p, 0), with p > 0.

• The line segment through F is perpendicular to the axis of symmetry and connects two points of the parabola.

Prove that the line segment through F has length 4p.

1. Make a sketch using the given information.

B (p, y2)

A (p, y1)

F (p, 0)V (0, 0)x

y

The vertex is (0, 0) and the focus is F (p, 0) with p > 0, so F is on the positive x-axis. The axis of symmetry is the x-axis.

The line segment through F, perpendicular to the axis of symmetry, and connecting two points of the parabola is the vertical line segment, AB . Because the segment is vertical, both A and B have p as their x-coordinate, matching the x-coordinate of F. The y-coordinates of A and B are unknown; they are named y

1 and y

2 in the diagram.


Instruction


2. Because the parabola opens to the right, its equation has the form (y – k)2 = 4p(x – h). Use this equation to solve for y

1 and y

2 in terms of p.

(y – k)2 = 4p(x – h) Standard form of the equation for a parabola that opens right or left

(y – 0)2 = 4p(x – 0) Substitute (0, 0) for (h, k).

y2 = 4px Simplify.

(y1)2 = 4p • p Substitute the coordinates of point A.

(y1)2 = 4p2 Simplify.

y p124= y

1 is positive, so take the positive square root

of 4p2.

y1 = 2p Solve for y

1.

(y2)2 = 4p • p Substitute the coordinates of point B into the

equation y2 = 4px.

(y2)2 = 4p2 Simplify.

y p224= − y

2 is negative, so take the negative square root

of 4p2.

y2 = –2p Solve for y

2.

3. Use the results from step 2 to find the length of AB .

AB = y1 – y

2 = 2p – (–2p) = 2p + 2p = 4p

For the parabola with vertex V (0, 0) and focus F (p, 0) with p > 0, the line segment through F has length 4p. The original statement has been proved; therefore, it is a theorem.


Instruction


© Walch Education

Example 4

Prove or disprove that the points A (4, 2), B (–2, 5), C (6, 5), and D (–4, 10) are all on the quadratic function graph with vertex V (2, 1) that passes through E (0, 2).

1. Make a sketch with the given information.

A quadratic function graph is a parabola. You do not yet know if any of the points A, B, C, or D lie on the parabola, so do not show them in your sketch.

E (0, 2)

V (2, 1)

y

x


Instruction


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, so we’ll use the form (x – h)2 = 4p(y – k) and the vertex to begin deriving the equation.

The vertex is (2, 1), so h = 2 and k = 1.

(x – h)2 = 4p(y – k) Standard form of the equation for a parabola that opens up or down

(x – 2)2 = 4p(y – 1) Substitute (2, 1) for (h, k).

3. Continue to derive the equation of the parabola by finding p.

Use the given point (0, 2) and the equation derived from step 2.

(x – 2)2 = 4p(y – 1) Derived equation

[(0) – 2]2 = 4p[(2) – 1] Substitute point E (0, 2) for (x, y).

(–2)2 = 4p(1) Simplify, then solve for p.

4 = 4p

p = 1


Instruction


© Walch Education

4. Convert the standard form of the equation into the general form of the equation.

Use the value of p = 1 and the standard form of the equation derived in step 2.

(x – 2)2 = 4p(y – 1) Standard form of the equation with the vertex substituted

(x – 2)2 = 4(1)(y – 1) Standard form of the equation with the value of p substituted

x2 – 4x + 4 = 4(y – 1) Expand the binomial.

1

41 12x x y− + = − Divide both sides of the equation by 4.

1

422x x y− + = Add 1 to both sides.

The equation of the parabola is y x x= − +1

422 .

5. Determine whether the points A, B, C, and D are on the parabola by substituting their coordinates into the equation.

A (4, 2): ( ) ( )= − +21

44 4 22 Yes, the equation is true, so A is

on the parabola.

B (–2, 5): ( ) ( )= − − − +51

42 2 22 Yes, the equation is true, so B is

on the parabola.

C (6, 5): ( ) ( )= − +51

46 6 22 Yes, the equation is true, so C is

on the parabola.

D (–4, 10): ( ) ( )= − − − +101

44 4 22 Yes, the equation is true, so D is

on the parabola.

A, B, C, and D are all on the parabola.

The statement has been proved, so it is a theorem.


Instruction


Example 5

Prove or disprove that P (–2, 1), Q (6, 5), R (8, 1), and S (0, –3) are vertices of a rectangle that is inscribed in the circle centered at C (3, 1) and passing through A 1 1 21, +( ) .

1. Make sketches using the given information.

You do not yet know if any of the points P, Q, R, or S lie on the circle, so show the polygon on a separate coordinate system. To help in plotting point A, you can use a calculator to find a decimal approximation.

A (1, 1 + 21)√

y

x

C (3, 1)

y

x

R (8, 1)P (–2, 1)

S (0, –3)

Q (6, 5)

2. Find the radius of the circle.

radius= = −( ) + + −( ) = −( ) + ( ) = + = =AC 1 3 1 21 1 2 21 4 21 252 2 2 255

radius 4 21 25 5= = + = =AC

The radius of the circle is 5 units.


Instruction


© Walch Education

3. Find the distance of points P, Q, R, and S from the center C to determine whether they are on the circle.

PC = − −( ) + −( ) = ( ) + ( ) = =3 2 1 1 5 0 25 52 2 2 2

QC = −( ) + −( ) = −( ) + −( ) = + = =3 6 1 5 3 4 9 16 25 52 2 2 2

RC = −( ) + −( ) = −( ) + ( ) = =3 8 1 1 5 0 25 52 2 2 2

SC = −( ) + − −( ) = ( ) + ( ) = + = =3 0 1 3 3 4 9 16 25 52 2 2 2

P, Q, R, and S are each 5 units from the center, and since the radius is also 5 units, the points are all on the circle.

4. Determine whether PQRS is a rectangle. Use slopes to identify parallel and perpendicular segments.

slope of PQ =−

− −( ) = =5 1

6 2

4

8

1

2 slope of QR =

−−

=−

= −1 5

8 6

4

22

slope of RS =− −

−=

−−

=3 1

0 8

4

8

1

2 slope of SP =

− −( )− −

=−

= −1 3

2 0

4

22

PQ is parallel to RS and QR is parallel to SP because they have equal slopes.

Therefore, PQRS is a parallelogram because it has both pairs of opposite sides parallel.

PQ is perpendicular to QR because the product of their slopes is –1: 1

22 1

−( ) = − .

Thus, PQRS is a rectangle because it is a parallelogram with a right angle.


Instruction


5. Steps 2 and 3 allowed us to determine that P, Q, R, and S are all on the circle. Step 4 shows that PQRS is a rectangle. Therefore, P (–2, 1), Q (6, 5), R (8, 1), and S (0, –3) are vertices of a rectangle that is inscribed in the circle centered at C (3, 1) and passing through A 1 1 21, +( ).

The statement has been proved, so it is a theorem.

A (1, 1 + 21)√

y

x

C (3, 1) R (8, 1)P (–2, 1)

S (0, –3)

Q (6, 5)

unit 6 • modeling geometry lesson 2: using coordinates to...

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