1. Analyzing Conic Sections
Jennifer Huss

2. 7-1 The Distance and Midpoint Formulas
To find the distance between any two points (a, b) and (c, d), use the distance formula:
Distance = (c – a)2 + (d – b)2
The midpoint of a line is halfway between the two endpoints of a line
To find the midpoint between (a, b) and (c, d), use the midpoint formula:
Midpoint = (a + c) , (b + d)

3. Distance = [(-8) – (-4)]2 + (4 – 2)2
7-1 Example
Find the distance between (-4, 2) and (-8, 4). Then find the midpoint between the points.
Distance = [(-8) – (-4)]2 + (4 – 2)2
Distance = (-8 + 4)2 + (2)2
Distance = (-4)2 + (2)2
Distance =
Distance= 20
Midpoint = (-4) + (-8) , 2 + 4
Midpoint = , 6
Midpoint = ( -6, 3)

4. 7-1 Problems Find the distance between (0, 1) and (1, 5).
Find the midpoint between (6, -5) and (-2, -7).
Find the value for x if the Distance = 53 and the endpoints are (-3, 2) and (-10, x).
If you are given an endpoint (3, 2) and midpoint (-1, 5), what are the coordinates of the other endpoint?
1) ) (2, -6) ) x = 0 or x = ) (-5, 8)

5. 7-2 Parabolas
A parabola is a set of points on a plane that are the same distance from a given point called the focus and a given line called the directrix
The axis of symmetry is perpendicular to the directrix and passes through the parabola at a point called the vertex
The latus rectum goes through the focus and is perpendicular to the axis of symmetry
If the equation of the parabola begins with x= then the parabola is not a function (fails the vertical line test)
Axis of Symmetry
Parabola
Focus
Latus Rectum
Vertex
Directrix

6. 7-2 Parabolas (cont.) Important Information About the Parabolas
Form of the equation
y = a (x – h)2 + k
x = a (y – k)2 + h
Axis of Symmetry
x = h
y = k
Vertex
(h, k)
Focus
(h, k + 1/4a)
(h + 1/4a, k)
Directrix
y = k – (1/4)a
x = h – (1/4)a
Direction of Opening
Opens upward when a > 0 and downward when a < 0
Opens to the right when a > 0 and to the left when a < 0
Length of Latus Rectum
1/a units

7. 7-2 Example
Write y = x2 + 4x + 1 in the form y = a (x – h)2 + k and name the vertex, axis of symmetry, and the direction the parabola opens.
You can always check your answers by graphing.
y = x2 + 4x + 1
y = (x2 + 4x + o ) + 1 – o
y = (x2 + 4x + 4) + 1 – 4
y = (x + 2)2 – 3
Vertex: (-2, -3)
Axis of Symmetry: x = -2
The parabola opens up because a = 1 so a > 0.

8. 7-2 Problems Graph the equation x2 = 8y.
For the parabola y2 = -16x name the vertex, focus, length of latus rectum, and direction of opening. Also, give the equations of the directrix and axis of symmetry.
Given the vertex (4, 1) and a point on the parabola (8, 3), find the equation of the parabola.
Graph for #2
Graph for #1
2) Vertex: (0,0) Focus: (-4,0) Latus rectum: 16 Direction: left Directrix: x = 4 Axis of symmetry: y = ) y = (1/8)(x – 4)2 + 1

9. 7-3 Circles
A circle is a set of points equidistant from a center point
The radius is a line between the center and any point on the circle
The equation of a circle is (x – h)2 + (y – k)2 = r2 where the radius is r and the vertex is (h, k)
Sometimes you need to complete the square twice to get the equation in this form (once for x and once for y)
Radius (r)
Vertex (h, k)

10. Radius = 7 and Center is (-2, 6)
7-3 Examples
Find the center and radius of x2 + y2 + 4x – 12y – 9 = 0 and then graph the circle.
x2 + 4x + o + y2 – 12y + o = 9 + o + o
x2 + 4x y2 – 12y + 36 =
(x + 2)2 + (y – 6)2 = 49
Radius = 7 and Center is (-2, 6)

11. 7-3 Examples (cont.)
If a circle has a center (3, -2) and a point on the circle (7, 1), write the equation of the circle.
Find the radius by the distance formula.
Radius = (7 – 3)2 + (1 – (-2))2
r = (4)2 + (3)2
r =
r = 25
r = 5
The equation of the circle will be (x – 3)2 + (y + 2)2 = 25

12. 1) x2 + (y + 2)2 = 4 Center (0, -2) and Radius 2 2) x2 + y2 = 20
7-3 Problems
Find the center and radius of x2 + y2 + 4y = 0. Then graph the circle.
If a circle has a center (0, 0) and a point on the circle (-2, -4) write the equation of the circle.
1) x2 + (y + 2)2 = 4 Center (0, -2) and Radius ) x2 + y2 = 20 #1

13. 7-4 Ellipses
An ellipse is the set of all points in a plane such that the sum of the distances from the foci is constant
An ellipse has two axes of symmetry
The axis of the longer side of the ellipse is called the major axis and the axis of the shorter side is the minor axis
The focus points always lie on the major axis
The intersection of the two axes is the center of the ellipse
Major Axis
Focus
Center
Minor Axis
Focus

14. 7-4 Ellipses (cont.) Important Notes: In the above chart, c = a2 – b2
Equation of the Ellipse
Foci Points
Is the major axis horizontal or vertical?
Center of the Ellipse
(x – h)2 + (y – k)2
a b2
( h + c, k) and (h – c, k)
Horizontal
(h, k)
b a2
(h, k + c) and (h, k – c)
Vertical
= 1
= 1
Important Notes:
In the above chart, c = a2 – b2
a2 > b2 always so a2 is always the larger number
If the a2 is under the x term, the ellipse is horizontal, if the a2 is under the y term the ellipse is vertical
You can tell that you are looking at an ellipse because: x2 is added to y2 and the x2 and y2 are divided by different numbers (if numbers were the same, it’s a circle)

15. 7-4 Example
Given an equation of an ellipse 16y2 + 9x2 – 96y – 90x = -225 find the coordinates of the center and foci as well as the lengths of the major and minor axis. Then draw the graph.
16 (y2 – 6y + o) + 9 (x2 – 10x + o) = (o) + 9(o)
16 (y2 – 6y + 9) + 9 (x2 – 10x + 25) = (9) + 9(25)
16 (y – 3)2 + 9 (x – 5)2 = 144
(y – 3)2 + (x – 5)2
= 1
Center: (5, 3)
16 > 9 so the foci are on the vertical axis
c = – 9
c = 7
Foci: ( , 3) and (5 – 7, 3)
Major Axis Length = 4 (2) = 8
Minor Axis Length = 3 (2) = 6

16. 7-4 Problems
For 49x2 + 16y2 = 784 find the center, the foci, and the lengths of the major and minor axes. Then draw the graph.
Write an equation for an ellipse with foci (4, 0) and (-4, 0). The endpoints of the minor axis are (0, 2) and (0, -2).
X2 + y2
Foci: (0, - 33) (0, 33) Center: (0, 0) Length of major= 14 Length of minor= 8
= 1 #1

17. 7-5 Hyperbolas
A hyperbola is a set of all points on a plane such that the absolute value of the difference (subtraction) of the distances from a point to the two foci is constant
The center is the midpoint of the segment connecting the foci
The vertex is the point on the hyperbola closest to the center
The asymptotes are lines the hyperbola can approach but never touch
The transverse axis goes through the foci
The conjugate axis is perpendicular to the transverse axis at the center point
Hyperbola
Conjugate Axis
Transverse Axis
Asymptotes
Focus
Center
Vertex

18. 7-5 Hyperbolas (cont.)
Equation of Hyperbola
Center
Foci Points
Equation of Asymptote
Vertex
Transverse Axis
(x – h)2 _ (y – k)2
a b2
(h, k)
(h – c, k) and (h + c, k)
y = +/- (b/a) x
(h +a, k) and (h – a, k)
Horizontal
(y – k)2 _ (x – h)2
a b2
(h, k – c) and (h, k + c)
(c = a2 + b2 )
y = +/- (a/b) x
(h, k + a) and
(h, k – a)
Vertical
= 1
= 1
You must be looking at a hyperbola because the x2 and y2 terms are subtracted (x2 – y2) or (y2 – x2)

19. 7-5 Example
Write the standard form of the equation of the hyperbola y2 – 25x2 – 576y – 150x = Then find the coordinates of the center, the vertices, the foci, and the equation of the asymptotes. Graph the hyperbola and the asymptotes.
144(y2 – 4y + o) – 25(x2 + 6x + o) = (o) + 25(o)
144(y2 – 4y + 4) – 25(x2 + 6x + 9) = (4) + 25(9)
144(y – 2)2 – 25(x + 3)2 = 3600
(y-2)2 _ (x + 3)2
= 1
Center: (-3, 2) a = 5 so the vertices are (-3, 7) and (-3, -3)
a2 + b2 = c2
= c2
c = 13
The foci are (-3, 15) and (-3, -11).

20. y – 2 = (5/12) x + 15/12 y – 2 = (-5/12) x + -15/12
7-5 Example (cont.)
Asymptotes have the formula y = +/- a/b x and we have center (-3, 2) and slopes +/- 5/12.
y – 2 = 5/12 (x + 3) y – 2 = -5/12 (x + 3)
y – 2 = (5/12) x + 15/12 y – 2 = (-5/12) x + -15/12
y = (5/12) x + 13/4 y = (-5/12) x + 3/4

21. 7-5 Problems
Find the coordinates of the vertices and the foci. Give the asymptote slopes for each hyperbola. Then draw the graph.
x2 _ y2
2) 25x2 – 4y2 = 100
= 1
Vertices: (-2, 0) (2, 0) Foci: ( 29, 0) (- 29, 0) Slope = +/- 5/2
Vertices: (-3, 0) (3, 0) Foci: (- 58, 0) ( 58, 0) Slope = +/- 7/3 2) 1)

22. 7-6 Conic Sections
Circles, ellipses, parabolas, and hyperbolas are all formed when a double cone is sliced by a plane
The general equation of any conic section is : Ax2 + Bxy + Cy2 + Dx + Ey + F = 0
The standard equations for each specific conic section are listed in previous sections
If B = 0 and you look at A and C in the equations:
Conic Section Name
Relationship of A and C
Parabola
A = 0 or C= 0, but never both equal to 0
Circle
A = C
Ellipse
A and C have the same sign, but A = C
Hyperbola
A and C have opposite signs

23. 7-6 Example
Identify 9x2 + 16y2 – 54x + 64y + 1 = 0 as one of the four conic sections. Then graph the conic section.
9x2 + 16y2 – 54x + 64y = -1
9 (x2 – 6x + o) + 16(y2 + 4y + o) = (o) + 16(o)
9 (x2 – 6x + 9) + 16(y2 + 4y + 4) = (9) + 16(4)
9(x – 3)2 + 16(y + 2)2 = 144
(x – 3) (y + 2)2 9
This conic section is an ellipse. +
= 1

24. 1) (x + 3)2 + (y)2 = 16 circle 2) (x – 3)2 _ (y + 1)2 hyperbola
7-6 Problems
Write the equation in standard form and decide if the conic section is a parabola, a circle, an ellipse, or a hyperbola. Then graph the equation.
x2 + y2 + 6x = 7
5x2 – 6y2 – 30x – 12y + 9 = 0
1) (x + 3)2 + (y)2 = 16 circle ) (x – 3)2 _ (y + 1) hyperbola
= 1 2) 1)

25. 7-7 Solving Quadratic Systems
When you solve a system of quadratic equations the method is almost the same as solving a system of linear equations
If the system has one equation of a conic section and one equation of a straight line, you can get zero, one, or two solutions to the system
If both the equations are conic sections, the system should have zero, one, two, three, or four solutions

26. 7-7 Example
Solve this system of equations using algebraic methods and by graphing the equations.
y = (x – 2)2 + 1
y = -4x + 5
The system of equations has one solution, (0, 5). The graphs of these equations confirms this.
Set the equations equal to each other to solve for x.
-4x + 5 = (x – 2)2 + 1
-4x + 5 = x2 – 4x
-4x + 5 = x2 – 4x + 5
5 = x2 + 5
x2 = 0
x = 0
Then put x = 0 back in to solve for y.
y = -4(0) + 5
y = 5

27. 7-7 Problems
Solve these systems of equations by using algebra and graphing the equations.
4x2 + y2 = ) x2 + y2 = 10
2x2 – y2 = y = x2 – 4
1) Four solutions: { (2, 3) (2, -3) (-2, 3) (-2, -3) } ) Four solutions: { (1, -3) (-1, -3) ( 6, 2) (- 6, -2) }