Use the Pythagorean theorem to find the length of the missing side.
1) a = 12, b = 9 c = 15
2) a = 5, c = 13 b = 12
Find the mean of the two numbers.
26
3) 18 and 34
4) 18 and -34 -8

Objective – To be able to find the distance between two points and find the midpoint of the line segment joining two points.
The Distance Formula
The distance “d” between the points (x
1
,y
1
) and (x
2
,y
2
) is as follows:
d = ( x
2
– x
1
) 2 + ( y
2
– y
1
) 2

Example 1
Find the distance between (2,-4) and (-5,-1).
d = ( x
2
– x
1
) 2 + ( y
2
– y
1
) 2
d = ( -5 – 2 ) 2 + ( -1 – -4 ) 2
d = (-7) 2 + (3) 2 d = 49 + 9 d = 58 d

7.62

The Midpoint Formula
The midpoint of the line segment joining A(x
1
,y
1
) and
B(x
2
,y
2
) is as follows:
A(x
1
,y
1
)
Midpoint M
B(x
2
,y
2
)
M = x
1
+ x
2
, y
1
+ y
2 ( )
2 2

Example 3
Find the midpoint of the line segment joining (6,-2) and (2,-9).
M = x
1
+ x
2
, y
1
+ y
2
( )
2 2
M =
6 + 2
,
-2 + -9 ( )
2 2
M =
8
,
-11 ( )
2 2
M =
-11 ( )
2

Warm-up 10.2
1) y² = 2x² + 3 2) y + x² = 7 no
3) y=3x + 9 no
5) y² = 5 no yes
4) y= 6x² - x + 3 yes

10.2 Parabolas

California State Standard
9.0
Students demonstrate and explain the effect that changing a coefficient has on the graph a quadratic functions; that is students can determine how the graph of a parabola changes as a, b, and c vary in the equation y = a ( x - b )² + c.
16.D
Students demonstrate and explain how the geometry of the graph of a conic section (e.g., asymptotes, foci, eccentricity) depends on the coefficient of the quadratic equation representing it.
17.0
Given a quadratic equation of the form ax ² + by ² + cx + dy + e = 0, students can use the method for completing the square to put the equation into standard form and can recognize whether the graph of the equation is a circle, ellipse, parabola, or hyperbola. Students can then graph the equation.

Standard Equation of a parabola (vertex at origin)
The standard form of the equation of a parabola with vertex at (0,0) is as follows.
EQUATION x²=4py y²=4px
FOCUS
(0,p)
(p,0)
DIRECTRIX y= -p x= -p
AXIS OF SYMMETRY
Vertical (x=0)
Horizontal (y=0)
Focus- the point that is equidistant from any point
Directrix-is perpendicular to the axis of symmetry

Extra Example 1
Draw the parabola.
Directrix: x=-2
Focus = (p,0)
Focus = (2,0)

Extra Example 2
Write an equation of the parabola shown below.
Focus = (1/3, 0)

Solve for x when y = 2
1. x
2
+ y
2
= 13 2. Y 2 = 12 – x 2 x 2 + (2) 2 = 13
(2) 2 = 12 – x 2
4 = 12 – x 2 x 2 + (4) = 13
8 = x 2 x 2 = 9
8 = x x = 3, - 3
X = +/- 2 2
Find the slope of the line
4. Y = 4 – 2x 5. 3x + 4y = 5
Y = -2x + 4 4y = -3x + 5 m = -2 Y = -3/4x + 5/4 m = -3/4
3. x 2 = 25 – y 2 x 2 = 25 – (2) 2 x 2 = 25 – (4) x = 21
X = +/- 21

CA. Standard 17.0
Given a quadratic equation of the form ax 2 + bx 2 + cx + dy + e = 0, students can use the method for completing the square to put the equations into standard form and can recognize whether the graph of the equation is a circle, ellipse, parabola, or hyperbola. Students can then graph the equation. (e.g., asymptotes, foci, eccentricity) depends on the coefficients of the quadratic equation representing it.

Circle is the set of all points (x,y) that are equidistant from the center.
Radius = distance between any (x,y,) and the center of a circle.

Standard Equation of a Circle when it’s center is the origin.
x 2 + y 2 = r 2
Only if the center is at the origin (0,0)

• Draw the circle given by y 2 =25-x 2
Solution
Write the equation in standard form.
x 2 + y 2 = 25 r = 25 = 5
Plug in points such as (0,5) (5,0), (0,-
5)…

Draw the circle given by x 2 + y 2 = 16 x 2 + y 2 = r 2 r = 16 r = 4
X 0 0 4 -4
Y 4 -4 0 0

Writing an equation of a circle.
Find the equation of a circle whose center is the origin and contains the point (1,4)
Since we have two points, (0,0), and (1,4), we could use the distance formula to find the radius.
Distance Formula = d = (x
2
– x
1
) 2 + (y
2
-y
1
) 2 r = (1-0) 2 + (4-0) 2 r = 17 x 2 + y 2 = r 2 x 2 + y 2 = 17 2 x 2 + y 2 = 17

Pg. 604-605, 23-37 odd, 47-53 odd, 59-69 odd.

Write the equation of a circle who has (2,3) as one of it’s points and it’s center is (0,0)
Use distance formula to find the radius.
r = (2-0) 2 + (3-0) 2 r = 11
Write standard form x 2 + y 2 = r 2 x 2 + y 2 = 11 2 x 2 + y 2 = 11

with the given characteristics and centered at (0,0)
A. Vertex(0,5)
Co-vertex (4,0) x 2 + y 2 = 1
4 2 5 2 x 2 + y 2 = 1
16 25 .
B. Vertex(-6,0)
Focus (3,0) c 2 =a 2 -b 2
9 = 36-b 2
27 = b 2 x 2 + y 2 = 1
36 27 .

Pgs 612-613
Problems
#18-20, 30-32, 42-44, 51 & 52

An Ellipse
sum of the distance between P and two distinct points, called the foci , is a constant.
The line through the foci intersects the ellipse at two points, the vertices . The line segment joining the vertices is the major axis , and its midpoint is the center of the ellipse. The line perpendicular to the major axis at the center intersects the ellipse at two points called the co-vertices . The line segment that joins these points is the minor axis of the ellipse.

Extra Example 2
Graph the equation: y 2
25
 x
49
2

1
Use this equation: y
2 a
2
 x b
2
2

1 a 2 = 25 b 2 = 49
–10 –8 –6 –4 –2
–2
–4
–6
–8
–10
6
4
10
8
2
2 4 6 8 10

Extra Example 3
Write the equation in standard form:
Foci: (0,-5); (0,5) c = 5
Vertices: (0,-3); (0,3) c
2  a
2  b
2 a = 3 y
2 a
2 x
2
 b
2
5
2 
3
2
16
 b
2
 b
25

9
 b
2
2 y
2
9

1 x
2

16

1

Step one: Write the equation in standard form.
Step two: Then determine the direction of the transverse axis.
Step three: Find the a and b value; plot the points and draw the rectangle
Step four: Draw the two diagonals of the rectangle.
Step five: Finally, draw the hyperbolas by using the diagonals as asymptotes while going through the vertices.

Vo
• Hyperbola -
The set of all points P such that the difference of the distances from P to two fixed points, called the foci , is constant.
• Vertices -
The line through the foci intersects the hyperbola at two points
• Transverse axis -
The line segment joining the vertices, its midpoint is the center of the hyperbola
.

State Standard
Students demonstrate and explain the effect that changing a coefficient has on the graph of quadratic functions; that is, a student can determine how the graph of a parabola changes as a, b and c vary in the equation y = a(x-b) 2 +c
-Jonathan Guerrero-
-Kyle Kathol-
-Sebastian Morales-
-Fausto Rendon-
-Kevin Wong-

Parabolas, circles, ellipses, and hyperbolas are all curves formed by the intersection of a plane and a double-napped cone.
Therefore, these shapes are called conics .
STANDARD FORM OF EQUATIONS OF TRANSLATED CONICS
In the following equations the point (h,k) is the vertex of the parabola and the center of the other conics.
CIRCLE (x-h) 2 + (y-k) 2 = r 2
Horizontal axis Vertical axis
PARABOLA (y-k) 2 = 4p(x-h) (x-h) 2 = 4p(y-k)
ELLIPSE (x-h) a 2
2 + (y-k) b 2
2 = 1 (x-h) 2 + (y-k) 2 = 1 b 2 a 2
HYPERBOLA (x-h) 2 - (y-k) 2 = 1 (y-k) 2 - (x-h) 2 = 1 a 2 b 2 a 2 b 2

Extra Example 1:
Write an equation of the parabola whose vertex is at
(3,2) and whose focus is at (4,2).
(y-k) 2 = 4p(x-h)
1) Plug in the vertex (h,k)
(y2 ) 2 = 4p(x3 )
(y - 2) 2 = 4(x - 3) 2) To find p, you need to find the distance between the vertex and focus.
1
|p| = (4-3) 2 + (2-2) 2 =
3) Plug p into the equation
(y-2) 2 = 4( 1 )(x-3)

Extra Example 2:
Graph (x-1) 2 + (y-4) 2 = 9
Standard form of circle
(x-h) 2 + (y-k) 2 = r 2
1)Center is (h,k) = ( 1
2)Radius is 3. [9=3 2 ]
, 4 )
Plot several points that are each 3 units from the center
(-2,4)
10
9
8
7
6
5
4
3
2
1
–5 –4 –3 –2 –1
–1
(1,7)
(1,4)
(1,1)
(4,4)
1 2 3 4 5

If the graph of Ax 2 + Bxy + Cy 2 + Dx +Ey + F=0 is a conic, then the type can be determined as the following:
Discriminant Type of Conic
B 2 -4AC<0 B=0, A=C circle
B 2 -4AC<0 and either B=0 or A=C Ellipse
B 2 -4AC=0 Parabola
B 2 -4AC>0 Hyperbola

Pg. 628 – 630
#13-18, 21, 27,
29-39 odd, 68, 71-79

Extra Example 3:
Write an equation of the ellipse with foci at (4,2) and
(4,-6) and vertices at (4,4) and (4,-8).
(x-h) 2 + (y-k) 2 = 1 b 2 a 2
1) Find the center, which is
3) Find c, which is the distance halfway between the two between the focus and the vertices. center
Use the distance formula:
Use the midpoint formula:
(h,k) = 4+4 , 4+(-8) =
2 2
(4,-2) c = (4-4) 2 + (2-(-2)) 2 = 4
4) Find ‘b’ by substituting the values of a and c into the equation b 2 = a 2 - c 2 2) Find a, which is the distance between the vertex and the center.
b 2 = 6 2 – 4 2 b 2 = 20  b = 20
Use distance formula: a = (4-4) 2 + (4-(-2)) 2 = 6
(x-4) 2 + (y+2) 2 = 1
20 36