Essential Question: How are the distance and midpoint formulas used to solve problems?
Distance formula: just like Pythagorean theorem
Ex) find the distance between (4, 3) and (6, 2)
Draw a triangle and show Pythagorean theorem
Show formula:
CLASSIFYING TRIANGLES
Ex) determine if the triangle with the vertices (4, 2), (0, 5), (7, 2) is scalene, isosceles, or equilateral
Find the distances between each pair of points
so the triangle is isosceles
Midpoint formula: (average of the x's, average of the y's)
Ex) find the midpoint of the segment joining (4, 3) and (6, 2)
Midpoint =
Ex) find the midpoint of the segment joining (1, 7) and (5, 9) (3, 8)
Ex) find the equation for the perpendicular bisector of the line segment joining (0, 3) and (6, 7).
a) find the midpoint (3, 5)
b) find the slope 4/6 = 2/3
c) find the perpendicular slope = 3/2
d) write the equation of a line with slope 3/2 through (3, 5): y – 5 = 3/2(x + 3)
y – 5 = 3/2x + 9/2 y = 3/2x + 19/2
ex) solve for x: say the distance between the points (6, x) and (2, 3) is 4
square both sides: 32 = 16 + (x – 3)^{2}
16 = (x – 3)^{2}
x – 3 = ±4 x = 3 ± 4 = 7, 1
Essential Question: What are the features of parabolas and how are they expressed in algebraic and graphical form?
Draw a parabola, show the focus, directrix, and axis of symmetry
4 types:
Direction it opens 
Equation 
Focus 
directrix 
Up 
x^{2} = 4py, p > 0 
(0, p) 
y = p 
Down 
x^{2} = 4py, p < 0 
(0, p) 
y = p 
Right 
y^{2} = 4px, p > 0 
(p, 0) 
x = p 
Left 
y^{2} = 4px, p < 0 
(p, 0) 
x = p 
For each of the following, determine if the parabola opens up, down, left, or right. Then identify the focus and directrix. Then graph it.
Ex) 2y^{2} = 24x
Y2 = 12x, so the parabola opens to the left
12x = 4px, 12 = 4p, p = 3
so the focus is at (3, 0) and the directrix is at x = 3
Ex) x^{2} = 20y
The parabola opens up
20y = 4py, 20 = 4p, p = 5
so the focus is at (0, 5) and the directrix is at y = 5
Ex) Write the standard form of the equation of the parabola with a focus at (3, 0) and the vertex at (0, 0).
The parabola opens left. (just draw the points if you are unsure)
P = 3
Y^{2} = 4px, so the equation is y^{2} = 12x
Ex) Write the standard form of the equation of the parabola with a directrix of y = 4 and the vertex at (0, 0).
The parabola opens down.
P = 4
X^{2} = 4py, so the equation is x^{2} = 16y
Circles
Essential Question: What are the feature of a circle and how equations of circles written?
Standard form of the equation of a circle: x^{2} + y^{2} = r^{2}
Ex) name the radius of the circle with equation x^{2} + y^{2} = 16. then graph the circle
Ex) write the standard form of the equation of a circle with a radius of 5 whose center is at the origin.
Ex) ditto but the radius is 3Ö2.
Ex) write the standard form of the equation of a circle whose center is at the origin that passes through (2, 5)
Essential Question: What are the features of an ellipse and how are equations of ellipses written?
Draw an ellipse both of the following ways:
Equation 
Major axis 
Vertices 
Covertices 
Foci 
where a > b 
Horizontal 
(a, 0), (a, 0) 
(0, b), (0, b) 
C2 = a^{2} – b^{2} (c, 0), (c, 0) 
where a < b 
Vertical 
(0, b), (0, b) 
(a, 0), (a, 0) 
C2 = b^{2} – a^{2} (0, c), (0, c) 
Ex) Write the equation 9x^{2} + 100y^{2} = 900 in standard form. Then identify the vertices, covertices, and foci of the ellipse. Then graph it.
Equation is . Vertices are (10, 0) and (10, 0).
Covertices are (0, 3) and (0, 3). Since c^{2} = 100 – 9, c^{2} = 91, so c = ±Ö91.
So the foci are at (Ö91, 0) and (Ö91, 0)
Ex) Write the equation in standard form. Then identify the vertices, covertices, and foci of the ellipse. Then graph it.
Equation is . Vertices are (0, 6) and (0, 6).
Covertices are (Ö12, 0) and (Ö12, 0). Since c^{2} = 36 – 12, c^{2} = 24, so c = ±Ö24 = ±2Ö6.
So the foci are at (0, 2Ö6) and (0, 2Ö6)
Ex) write the equation of the ellipse with the center at (0, 0), a vertex at (5, 0), and a focus at (3, 0).
A^{2} = 25, c^{2} = a^{2} – b^{2}, 9 = 25 – b^{2}, b^{2} = 16 equation is
Ex) write the equation of the ellipse with the center at (0, 0), a vertex at (0, 30), and a focus at (0, 20).
B^{2} = 900, 900 – a^{2} = 400, a^{2} = 500
Essential Question: What are the features of a hyperbola and how are equations of hyperbolas written?
Equation 
Transverse Axis 
Vertices 
Asymptotes 
Foci 

Horizontal 
(a, 0), (a, 0) 

C^{2} = a^{2} + b^{2} (c, 0), (c, 0) 

Vertical 
(0, b), (0, b) 

C^{2} = a^{2} + b^{2} (0, c), (0, c) 
Ex) Write the equation 9x^{2} – 16y^{2} = 144 in standard form. Identify the vertices, foci, and asymptotes. Then graph it.
Divide through by 144 to get x2 / 16 – y2 / 9 = 1
Vertices: (4, 0), (4, 0)
Asymptotes: + 3/4x
Foci: (5, 0), (5, 0)
Ex) write the equation of a hyperbola with foci (0, 5) and (0, 5) and vertices (0, 3) and (0, 3). Then graph it
B = 3, so b^{2} = 9
C = 5 so c^{2} = 25
C^{2} = a^{2} + b^{2} so 25 = a^{2} + 9, a^{2} = 16
Equation y^{2} / 9 – x^{2} / 16 = 1
Asymptotes: + 9/16x = + 3/4x
Ex) write the equation of a hyperbola with foci (4, 0) and (4, 0) and vertices (1, 0) and (1, 0). Then graph it.
C = 4 so c^{2} = 16
A = 1 so a^{2} = 1
C^{2} = a^{2} + b^{2} so 16 = 1 + b^{2}, b^{2} = 15
Equation x^{2} / 1 – y^{2} / 15 = 1
Asymptotes: + 15x
LESSON PLAN Day 6 106a
Essential Question: How can conic sections be classified based on a given equation?
Classifying conics
Parabola: only 1 of the variables is squared
Circle: x2 and y2 are added and coefficients are both the same
Ellipse: x2 and y2 are being added and coefficients are different
Hyperbola: either x2 or y2 is being subtracted
classify the following:
ex) 4x^{2} – 9y^{2} + 32x – 144y – 548 = 0
ex) 2x^{2} + 2y^{2} – 12x + 4y + 2 = 0
ex) 12x^{2} – 85y + 285 = 0
ex) 2x^{2} + y^{2} – 4x – 4 = 0
Equations with the center at (h, k) (or vertex (h, k) for parabolas)
Circle: (x – h)^{2} + (y – k)^{2} = r^{2}
Center = (h, k), radius is r
Parabola: (y – k)^{2} = 4p(x – h) (x – h)^{2} = 4p(y – k)
Vertex = (h, k), up or right if p is positive, down or left if p is negative
Ellipse: , a > b , a < b
Center = (h, k)
Vertices: (h + a, k), (h – a, k) vertices: (h, k + b), (h, k – b)
Covertices: (h, k + b), (h, k – b) covertices: (h + a, k), (h – a, k)
Hyperbola:
Center = (h, k)
Vertices: (h + a, k), (h – a, k) vertices: (h, k + b), (h, k – b)
Asymptotes: still y = + a/b, but use the center as your center and go from
there with the slope
HW: 106a pg 628 – 629 #21 – 49 odds
LESSON PLAN Day 7 106b
Writing conic equations in standard form (completing the square)
Essential Question: how do you write the equation of a conic section in standard form if it is not centered at the origin?
Classify the conic section and write its equation in standard form
Parabolas only: Get all the y's on one side and all the x's on the other side
Ex) y^{2} – 2x – 20y + 94 = 0 parabola form (y – k)^{2} = 4p(x – h)
Y^{2} – 20y = 2x – 94
Complete the square
Y^{2} – 20y + 100 = 2x – 94 + 100
(y – 10)^{2} = 2x + 4
factor out whatever number is in front of the x
(y – 10)^{2} = 2(x + 2)
circles, ellipses, hyperbolas: put all the x's together and all the y's together
ex) x^{2} + y^{2} – 6x – 8y + 24 = 0
x^{2} – 6x + y^{2} – 8y = 24
complete the square on both of them
x^{2} – 6x + 9 + y^{2} – 8y + 16 = 24 + 9 + 16
(x – 3)^{2} + (y – 4)^{2} = 1
ex) 4x^{2} + y^{2} – 48x – 4y + 48 = 0
4x^{2} – 48x + y^{2} – 4y = 48
4(x^{2} – 12x + 26) + y^{2} – 4y + 4 = 48 + 104 + 4
4(x – 6)^{2} + (y – 2)^{2} = 60
divide through by the constant
ex) 9x^{2} + 4y^{2} – 36x – 16y – 164 = 0
put the negative variable second 4y^{2} – 16y – 9x^{2} – 36x = 164
4(y^{2} – 4y + 4) – 9(x^{2} + 4x + 4) = 164 + 16 – 36
4(y – 2)^{2} – 9(x + 2)^{2} = 144
HW: 106 pg 629 #51 – 61 odds