File: Algebra2B 2011 Answers 8.12.11
Updated: August 12, 2011
Section 5.00
1. What are the four conic sections shapes generated from the intersection of a plane and a cone?
Unless specific instructions are given below, perform the following conic operations:
First state type of conic function, orientation and provide a sketch of the conic function and put equation in standard form.
i) If conic is a Parabola find:
a) Vertex (h,k) =
b) p value and direction of opening
c) Focus point (F = )
d) Equation of Directrix
ii) If conic is a Circle find:
a) Center (h,k) =
b) Radius (r = )
iii) If conic is a Ellipse find:
a) Center (h,k) =
b) Vertices (V = ); as well as (U = )
c) Length of Major
d) Length of Minor axis
e) Foci ordered pairs (F = )
iv) If conic is a Hyperbola find:
a) Center (h,k) =
b) Vertices (V = ); as well as (U = )
c) Length of Transverse axes
d) Length of Conjugate axis
e) Foci ordered pairs (F = )
f) Asmptotes:
2. Explain how the formula of a circle is derived. Use an example to help illustrate your explanation.
Explanation of formula using Pythagorean Theorem:
When (h,k) = (0,0) then: x2 + y2 = r2
If however, the center of the circle (h,k) is moved (as in the example below ) then in order for the Pythagorean Theorem Formula to work, the center of the circle must be moved back to the center of the x and y axis. Therefore in this example below:
Instead of --: x2 + y2 = r2
Circle Equation: (x + 6)2 + (y – 4)2 = 52
(x + 6)2 + (y – 4)2 = 25
Formula for a Circle with center h,k and radius r:
Circle Equation: (x – h)2 + (y – k)2 = r2
3. (x + 5)2 + (y – 3 )2 = 49
a2 + b2 = c2
x2 + y2 = r2
(x – h)2 + (y – k) 2 = r 2
Circle: (h,k) = (-5, 3); radius = 7
4. Complete the following table for circles with the given center and radius
Center(h,k) / Radius
(r) / Circle Equation
a) / (-5, 2) / 6 / (x + 5)2 + (y – 2)2 = 36
b) / (6, -3) / 5 / (x – 6)2 + (y + 3)2 = 25
c) / (4, -1) / 8 / (x – 4)2 + (y + 1)2 = 64
d) / (-5, 6) / 4 / (x + 5)2 + (y – 6)2 = 16
5. Complete the following table for circles with the given equations.
Equation / Center(h, k) / Radius
(r)
a) / (x + 5)2 + (y + 3)2 = 49 / (- 5, - 3) / 7
b) / (x – 7)2 + (y – 2)2 = 81 / (7, 2) / 9
c) / (x – 2)2 + (y + 1)2 = 100 / (2, - 1) / 10
d) / (x + 8)2 + (y – 4)2 = 4 / (- 8, 4) / 2
6. Write the equation for the conic function below.
§ The center of the circle (h, k) is at (-2, 0).
§ The radius = 2
§ (x – h)2 + (y – k) 2 = r2
(x – -2)2 + (y – 0) 2 = 22
(x + 2)2 + y 2 = 4
7.
* This is an ellipse equation and can be recognized by the fact that the denominators are different and it is a sum.
** (x – h)2 + (y – k)2 = 1
a 2 b 2
or (x – h)2 + (y – k)2 = 1
b 2 a 2
- Since the x is not in the form: (x – 0)2 nor is the y in the form of: (y – 0)2 , the center (h, k) is at (0, 0).
- Since the bigger denominator (a2 = 25) is under the y, then the ellipse is stretched along the y-axis. This means the distance (a) from the center is 5. And the major axis (2a) length (the distance from one vertex to the other) is 10.
- Similarly the smaller denominator (b2 = 9) is under the x, then b = 3 and 2b (the minor axis length) = 6
8.
* This circle equation and can be recognized by the fact that the denominators are the same and it is a sum.
(x – 2)2 + (y – 1)2 = 36 Multiply both sides by 36
to put in Standard Form.
Circle: (h,k) = (2, 1); radius = 6
9.
* Visually one can see that this is an ellipse with the major axis vertical.
§ The center of the ellipse (h, k) is at (3, 0).
§ v = (3, + 12); u1 = (-5,0) and u2 = (11,0)
§ 2a = 24; \a = 12 and a2 = 144
§ (11 – -5 = 16) \2b = 16; \b = 8 and b2 = 64
§ (x – h)2 + (y – k) 2 = 1 (“a” is on the y-axis)
b2 a2
§ (x – 3)2 + (y – 0) 2 = 1
64 144
§ (x – 3)2 + y2 = 1
64 144
10.
* This is an ellipse equation and can be recognized by the fact that the denominators are different and it is a sum.
- Since the x is not in the form: (x – 0)2 nor is the y in the form of: (y – 0)2 , the center (h, k) is at (0, 0).
- Since the bigger denominator (a2 = 25) is under the x, then the ellipse is stretched along the x-axis. This means the distance (a) from the center is 5. And the major axis (2a) length (the distance from one vertex to the other) is 10.
- Similarly the smaller denominator (b2 = 9) is under the y, then b = 3 and 2b (the minor axis length) = 6
11.
* This is parabola equation and can be recognized by the fact that it is in (or can be put into) the form of
(x – h)2 = 4p(y – k) x 2 =
** (y – k)2 = 4p(x – h) y 2 =
- Since the equation in the form of (- y 2 ) the parabola will curve to the left.
- 1/2 y 2 = x - Put the squared variable on left side
y 2 = -2x - The vertex h,k is (0,0)
- Multiply by -2 to get y2 by itself
4p = -2 - Divide by 4 to both sides
p = - 1/2 “neg p” also indicates it curves left
Start at vertex (0, 0) and move distance “+ p” to find focus and directrix.
12.
* This is parabola equation and can be recognized by the fact that it is in (or can be put into) the form of
(y – k)2 = 4p(x – h) y 2 =
- Since the equation in the form of (- y 2 ) the parabola will curve to the left.
- 1/2 (y – 2)2 – 4 = x - Put squared variable on left
- 1/2 (y – 2)2 = (x + 4) - Add 4 to both sides
(y – 2)2 = -2(x + 4) - Multiply by -2 to get the
equation in standard form.
Now it can be seen that the vertex (h,k) = (-4, 2)
4p = -2 - Divide by 4 to both sides
p = - 1/2 “neg p” also indicates it curves left
Start at vertex (-4, 2) and move distance “+ p” to find focus and directrix.
13.
* This is parabola equation and can be recognized by the fact that it is in (or can be put into) the form of
(x – h)2 = 4p(y – k) x 2 =
- Since the equation in the form of (x 2 ) the parabola will curve to the upwards.
1/16 x 2 = y - Put squared variable on left
x 2 = 16y - Multiply by 16 to get the
equation in standard form.
Now it can be seen that the vertex (h,k) = (0, 0)
4p = 16 - Divide by 4 to both sides
p = 4 “pos p” also indicates it curves up
Start at vertex (0, 0) and move distance “+ p” to find focus and directrix.
14.
* This is a hyperbola equation and can be recognized by the fact that it is subtraction.
- Unlike ellipse equations the a2 always goes first
** (x – h)2 – (y – k)2 = 1 x2 – y2 =
a 2 b 2
(y – k)2 – (x – h)2 = 1 y2 – x2 =
a 2 b 2
When the equation is x2 – y2 I remember that the graph looks like an “x”
(See question 28 for another simplified approach to hyperbola equations).
15.
* This is a hyperbola equation and can be recognized by the fact that it is subtraction.
- Unlike ellipse equations the a2 always goes first
** (y – k)2 – (x – h)2 = 1 y2 – x2 =
a 2 b 2
When the equation is y2 – x2 I remember is that I see that the top of the graph looks like an “y”
(See question 28 for another simplified approach to hyperbola equations).
16. Below are the equations of four conic sections. Indicate the conic type for each equation and sketch the parameters that are easily obtained.
a) b)
c) d)
17. Write the equation of a circle with center (0, 3) and a radius of 3.
(x – 0)2 + (y – 3)2 = 32
x 2 + (y – 3)2 = 9
18. Write the equation of a circle with center (-1, 2) and a radius of 4.
(x – -1)2 + (y – 2)2 = 42
(x + 1)2 + (y – 2)2 = 16
19. (x – 4)2 + y2 = 16
Circle: Center h,k = (4, 0), radius = 4
20. (x – 1)2 + (y + 2)2 = 25
Circle: Center h,k = (1, -2), radius = 5
21. Determine which of the four conic equation is/are functions.
Determine which conic equations pass the Vertical Line Test (There is only one y for every x value).
Conic function / FunctionParabola x2 / Yes
Parabola y2 / No
Circle / No
Ellipse / No
Hyperbola x2 – y2 / No
Hyperbola y2 – x2 / No
§ Only an x2 = y parabola is a function.
22. Write the equation for the conic function below.
Circle: Center h,k (0, 3), radius 4
(x – h)2 + (y – k) 2 = r2
(x – 0)2 + (y – 3) 2 = 42
x 2 + (y – 3)2 = 16
23. Write the equation for the conic function below.
Circle: Center h,k (-2, -2), radius 3
(x – h)2 + (y – k) 2 = r2
(x – -2)2 + (y – -2) 2 = 32
(x + 2)2 + (y + 2)2 = 9
24. Write an equation of a circle with a diameter of 20 inches and a center at the point (4, -5)
(x – h)2 + (y – k) 2 = r2
(x – 4)2 + (y – -5) 2 = 102
(x – 4)2 + (y + 5)2 = 100
25. Write the equation for the conic function below.
Circle: Center h,k (4, 0), radius 3
(x – h)2 + (y – k) 2 = r2
(x – 4)2 + (y – 0) 2 = 32
(x – 4)2 + y2 = 9
26. Write the equation for the conic function below.
Circle: Center h,k (3, -1), radius 2
(x – h)2 + (y – k) 2 = r2
(x – 3)2 + (y – -1) 2 = 22
(x – 3)2 + (y + 1)2 = 4
Section 5.07
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27. (x – 5)2 + (y + 3)2 = 16
Circle: Center h,k = (5, -3), radius = 4
28. Write the equation for the conic function below. (Each square is 1 x 1)
Circle: Center h,k (0, -5), radius 3
(x – h)2 + (y – k) 2 = r2
(x – 0)2 + (y – -5) 2 = 32
x 2 + (y + 5)2 = 9
29.
* This is circle equation and can be recognized by the fact that the denominators are the same and it is a sum.
(x – 0)2 + (y – 0)2 = 36 Multiply both sides by 36
to put in Standard Form.
Circle: (h,k) = (0, 0); radius = 6
x 2 + y 2 = 36
30. Write the equation of the ellipse with foci (0, +3) and y-intercepts +5.