1. Chapter 10 Conic Sections
10.1 – The Parabola and the Circle
10.2 – The Ellipse
10.3 – The Hyperbola
10.4 – Nonlinear Systems of
Equations and Their Applications

2. 10.1 The Parabola and the Circle
Graph parabolas of the form
x = a(y – k)2 + h.
Find the distance and midpoint
between two points.
Graph circles of the form
(x – h)2 + (y – k)2 = r2.
Find the equation of a circle with a given center and radius.
x2 + y2 + dx + ey + f = 0.
Chapter 1 Outline

3. Identify and Describe the Conic Sections
Conic Section: A curve in a plane that is the result of intersecting the plane with a cone.
Types of conic sections include parabolas, circles, ellipses, and hyperbolas.

4. Equations of Parabolas Opening Left or Right
The graph of an equation in the form x = a(y – k)2 + h is a parabola with vertex at (h, k). The parabola opens to the right if a > 0 and to the left if a < 0. The equation of the axis of symmetry is y = k.
Vertex (h, k)
Vertex (h, k)
y = k
y = k
x = a(y – k)2 + h, where a > 0
x = a(y – k)2 + h, where a < 0

5. Recall from Section 8.5 Example 6:
f(x) = ax2 +bx + C = a(x – h)2 + k
Recall from Section 8.5 Example 6:
Given the function f(x) = x2 – 6x + 8 = (x – 3)2 – 1, determine which way the graph opens, find the vertex and axis of symmetry, and draw the graph. Find the domain and range.
a = 1 > 0 up
V (3, 1)
Axis: x = 3 x y
(x, y) 2 3 4 6 8 –1
(0, 8)
(2, 0)
(3, –1)
(4, 0)
(6, 8)
y-intercept
x-intercept
vertex
Domain: (−∞,∞ ),
Range: [−1,∞ )
x-intercept

6. f(x) = ax2 +bx + C = a(x – h)2 + k
Recall Sec 8.5 Example 7:
f(x) = ax2 +bx + C = a(x – h)2 + k
Given f(x) = – x2 +8x – 12 = – (x – 4)2 + 4, determine which way the graph opens, find the vertex and axis of symmetry, and draw the graph. Find the domain and range.
Solution
a = – 1 < 0, Down
V (4, 4) & Axis: x = 4 x y
(x, y) 2 4 6 8
–12
(0, – 12)
(2, 0)
(4, 4)
(6, 0)
(8, – 12)
y-intercept
x-intercept
vertex
x-intercept
Domain: (−∞,∞ ) & Range: (−∞,4]

7. Example 3:
Determine which way the graph opens, find the vertex and axis of symmetry, and draw the graph. x = 2y2 + 4y – 1
Solution a = 2 > 0, Parabola opens to the right
Let us rewrite the equation: x = a(y – k)2 + h.
x = 2y2 + 4y – 1
x + 1 = 2y2 + 4y
x + 1 = 2(y2 + 2y)
x = 2(y2 + 2y + 1)
x + 3 = 2(y + 1)2
x = 2(y + 1)2 – 3
V (–3, –1)
Axis: y = –1.
X-intercept: (– 1,0)
Y-intercepts:
X=0: 0 = 2y2 + 4y – 1: (0, 0.22), (0, –2.22) x y –1 –2 5 1 –3

8. Example 4: Sketch the graph of x = – 2(y + 4)2 – 1. V (– 1, – 4).
a = – 2 → Parabola opens to the left. x y
–33 –8 –1 –4 –
V (– 1, – 4).
Axis: y = –4.
x-int: (– 33,0)
y-int: (0,?)
x=0: -2(y + 4)2 – 1 =0

9. Learn the Distance and Midpoint Formulas
The horizontal distance between the two points (x1, y1) and (x2, y2), indicated by the blue dashed line is |x2 – x1| and the vertical distance indicated by the red dashed line is |y2 – y1|.
We can use the Pythagorean theorem along with the square root principle to derive:
The Distance Formula
The distance, d, between two points with coordinates (x1, y1) and (x2, y2), can be found using the formula:

10. Example 5:
Find the distance between (–3, 7) and (–11, 9). If the distance is an irrational number, also give a decimal approximation rounded to three places.
Solution
Note: It does not matter which pair is (x1, y1) and
which is (x2, y2).

11. Learn the Distance and Midpoint Formulas
Given any two points (x1, y1) and (x2, y2), the point halfway between the given points can be found by the midpoint formula:
Example 6:
Find the midpoint of the line segment whose endpoints are
(–4, 3) and (6, 7).

12. Graph Circles with Centers at the Origin
A circle is the set of points in a plane that are the same distance, called the radius, from a fixed point, called the center.
Circle with Its Center at the Origin and Radius r
Circle with Its Center at (h, k) and Radius r

13. Graph Circles with Centers at the Origin
Example 7:
x2 + y2 = 16 is a circle
C(0, 0) & r=4
Example 8:
x2 + y2 = 10 is a circle
C(0, 0) & r= 10 ≈3.16

14. Graph Circles with Centers at (h, k)
Example 9:
Determine the equation of the circle shown below.
C(-3, 2) & r=3

15. Graph Circles with Centers at (h, k)
Example 10:
Find the center and radius of the circle and draw the graph.
(x + 2)2 + (y  1)2 = 9
Solution
C(2, 1).
9 = r2, r = 3
4 points:
(  2, 4), (  2,  2)
(  5, 1), ( 1, 1 )
r = 3
(-2, 1)

16. Example 11: Write the equation of the circle in standard form.
center: (–5, 3); radius 4
Solution: (x – h)2 + (y – k)2 = r2
(x  (5))2 + (y – 3)2 = 42
(x + 5)2 + (y – 3)2 = 16
Example 12: C(0, 0); r = 5
Solution: x2 + y2 = r2
x2 + y2 = 52
x2 + y2 = 25

17. Example 13:
Find the center and radius of the circle and draw the graph.
x2 + 8x + y2  6y + 16 = 0
Solution
C(4, 3) & r = 3
4 points:
(  4,0), (  4,6)
(  7, 3), ( 1, 3 )

18. Summary: Distance & Midpoint Formulas Circles Parabolas

19. Example 1:
Given determine which way the graph opens, find the vertex and axis of symmetry, and draw the graph.
Solution
a = 2, h = 2, and k = 3.
a > 0, the parabola opens up.
V (2, 3)
Axis: x = 2
y-int (0, 5)

20. Note that for h = 0 and k = 0, the circle is centered at the origin
Note that for h = 0 and k = 0, the circle is centered at the origin. Otherwise, the circle is translated |h| units horizontally and |k| units vertically.

21. The center is at (–1, 3) and the radius is 2.
Find the center and the radius and then graph the circle x2 + y2 + 2x – 6y + 6 = 0.
x2 + 2x + y2 – 6y = –6
x2 + 2x y2 – 6y + 9 = –
(–1, 3) x y
(x + 1)2 + (y – 3)2 = 4
(x – (–1))2 + (y – 3)2 = 22
The center is at (–1, 3) and the radius is 2.