美国高中数学2-6 Analytic Geometry
Section 6-2: Equations of Circles
Definition of a Circle
A circle is the set of all points in a plane equidistant from a fixed point called the center point. We can
derive the equation directly from the distance formula. If we place the center point on the origin point, the equation of a circle with center point (0, 0)
and radius r is:
222
x + y = r
Look at the family of circles drawn:
A simple translation of the circle equation becomes:
222
(x - h) + (y - k) = r
With center at (h, k) and radius r.
Here are some examples:
Sample Problems
1) Find the center, radius and graph the equation: (x
22- 2) + (y + 5) = 17
Solution: Center point : (2, -5), radius = The graph is below:
222) x + y - 8x + 4y - 8 = 0 Find center, radius and graph.
Solution: We need to put the equation into the correct form. We will do this by completing the square!!
22 (x - 8x ) + (y + 4y ) = 8 Complete the square! See 1.6
22 (x - 8x + 16) + (y + 4y + 4) = 8 + 16 + 4
22 (x - 4) + (y + 2) = 28 Now it's in the
correct form!!
Center point (4, -2) with radius = The
graph is:
3) Find the intersection of the line y = x - 1 and the
22circle x + y = 25.
Solution: A line could intersect a circle twice or once (if it is tangent) or not intersect at all. To find the intersection use substitution. Replace y in the circle equation with x - 1.
22 x + (x - 1) = 25
22 x + x - 2x + 1 = 25
2 2x - 2x - 24 = 0
2 x - x - 12 = 0 Now factor
(x - 4)(x + 3) = 0
x = 4 or x = -3
Substituting back in to find y gives the following points:
(4, 3) and (-3, -4) If you check these points in both equations, you will discover they make both true. The graph of the system is:
What would it mean if the solutions were imaginary?
4) Sketch the graph of
Solution: The above graph is part of a
circle. Why? Because it only includes the positive values for y. Thus it is the top half of a circle with radius 4. It is a semi-circle. By only studying this part of the circle, it makes it a function. It now passes the
vertical line test. Circles are not functions! Here is the graph:
That's it for circles. Let's head toward
Ellipses!
Section 6-3: Ellipses
Definition of an Ellipse
If F(c, 0) and F(-c, 0) are two fixed points in the plane 12
and a is a constant, 0< c < a, then the set of all points P
in the plane such that
PF + PF = 2a 12
is an ellipse. F and F are the foci of the ellipse. 12
Reflective Property of Ellipses (Manipula Math)
Notice the two fixed points in the graph, (-4, 0) and (4, 0). These are the foci points for the graph. By the definition, the distance from these points to a point on the ellipse is a constant. In this case the constant is 10. No matter what point you take on the ellipse, the distance to the fixed points will always equal 10. As one distance gets longer the other will get shorter so
that the sum will always be 10.
Equation of the ellipse
This ellipse has the major axis parallel to the x-axis making it open longer across. The length is 2a. The
minor axis is parallel to the y-axis and has length 2b. The foci points are 2c units apart. The center of
the ellipse until we translate it will remain at (0, 0). The vertex points are at the end points of the major axis. Look at the equation. The a value is always the
biggest number!!
Notice the major axis and the minor axis have
reversed. The longer axis is now vertical. What
causes this to happen? Look at the equation closely. The a value is now under the y value rather than the x value in the previous equation. We now
have an easy method to tell which way the ellipse opens!! Look to see whether the larger value is under
the x value or the y value!!
Problems
1) Sketch the graph and find the vertices, end points of
22the minor axis and foci points for: x + 4y = 16
Solution: First put the equation in the correct form by dividing by16.
22 x/16 + y/4 = 1 The larger
22 = 16 and b = 4. Since the larger value value is a
is under x, the ellipse has the major axis horizontal. The values are a = 4, b = 2. To find
c, subtract 16 - 4 and take the square root. Thus
c = 3.5 (rounded to tenths)
Center at (0, 0)
Vertices: (4, 0) and (-4, 0)
End points of minor axis: (0,
2) and (0, -2)
Foci: (3.5, 0) and (-3.5, 0)
2) An ellipse has its center at the origin. Find an equation of the ellipse with Vertex (8, 0) and minor axis 4 units long.
Solution: a = 8 and b = 2 the minor axis is 2b = 4, so b = 2.
22The equation is: x/64 + y/4 = 1 The vertex is on the
x-axis.
3) An ellipse has its center at the origin. Find an equation of the ellipse with vertex (0, -12) and focus ( 0, -4).
Solution: a = 12 and c = 4. Both are on
2the y-axis, so the major axis is vertical. To find b,
2square a and c and subtract. b = 144 - 16 = 128
22Thus the equation is: x/128 + y/144 = 1
Translation of the Ellipse
The center is now at (h, k). All values are now calculated from this point rather than from (0, 0)
1) Sketch the graph and find the vertices, end points of the minor axis and foci for:
Solution: The larger of the values is under
x. The major axis is horizontal. a = 5, b = 4 and c = 3
Center: (2, -1)
Vertices: (7, -1) and (-3, -1) add/subtract 5 from the x-value!
End points of minor: (2, 3) and (2, -5) add/subt 4 from y-value!
Foci points: (5, -1) and (-1, -1) add/subt 3 from x-value!
Notice this ellipse is almost circular. The reason is because a and b are close in value. In truth, a circle is
a special case of an ellipse with a = b!!
2) Find the equation of the ellipse with center (2,5),
one focus (5,5) and one vertex (7, 5).
Solution: The focus and vertex are on
the same horizontal line. (y values are the same!) a is the distance from the center to a vertex, so x = 5
c is the distance from the center to a focus point, so c = 3. This make b = 4.
Thus the equation is:
On to Hyperbolas
Back up to circles
Section 6-4: Hyperbolas
Definition of the hyperbola
A hyperbola is the set of all points P(x, y) in the plane
such that
| PF - PF | = 2a 12
Again F and F are focus points. This time the 12
difference of these distances remain a constant at
2a. The explanation is similar to that of the ellipse. Since the ellipse is the sum of the distances and the hyperbola is the difference of the distances, the equations are very similar. They differ only in the sign and the longest side for a hyperbola is c. (Remember
for the ellipse it was a)
Drawing a Hyperbola (Manipula Math)
The equation of the above hyperbola would have the
form:
The hyperbola opens left and right. Notice it comes in two parts. Different than an ellipse which is a closed figure. Hyperbolas can also open up and down. I am
sure you can guess at the equation!!
This hyperbola has the form:
To get the correct shape of the hyperbola, we need to find the asymptotes of the hyperbola. The asymptotes
are lines that are approached but not touched or
crossed. These asymptotes are boundaries of the hyperbola. This is one difference between a hyperbola and a parabola. For the hyperbolas that open right/left,
the asymptotes are:
and for hyperbolas opening up/down, the asymptotes
are:
To form the asymptotes easily on the graph, all we need
do is form a rectangle using a and b.
Sample Problems
221) Graph the hyperbola x/16 - y/4 = 1 Find the
vertices, foci and equations of the asymptotic lines.
Solution: This hyperbola opens right/left
222because it is in the form x - y. a = 16, b = 4, c = 16 +
4 = 20. Therefore, a = 4, b = 2 and c = 4.5
Vertices: (4, 0) and (-4, 0)
Foci: (4.5, 0) and (-4.5, 0)
Equations of asymptotic lines: y = .5x and y =
- .5x
To graph the hyperbola, go 2 units up/down from center point and 4 units left/right from center point.
222) Graph the hyperbola y/25 - x/9 = 1. Give the
vertices, foci and equations of asymptotic lines.
Solution: This hyperbola opens up and
22down because it is in the form y - x. a = 25, b = 9 and
2c = 25 + 9 = 34. Thus, a = 5, b = 3, c = 5.8
Vertices: (0, 5) and (0, -5)
Foci: (0, 5.8) and (0, -5.8)
Equations of asymptotic lines: y = (5/3)x
and y = (-5/3)x
The box is formed by going 5 units up/down
from center and 3 units left/right from center.
3) Find an equation of a hyperbola with center at the
origin, one vertex at (7, 0) and a focus at (12, 0).
Solution: The vertex and focus are on the x-axis, so the hyperbola opens right/left. a = 7, c =
2212. That makes a = 49, c = 144 and
2b = 144 - 49 = 95. Therefore the equation is:
22x/49 - y/95 = 1
Translations of Hyperbolas
If the hyperbola opens right/left the translation is:
with the equations of the asymptotic lines as:
y - k = + (b/a)(x - h)
If the hyperbola opens up/down the translation is:
with the equations of the asymptotic lines as:
y - k = + (a/b)(x - h)
Sample Problems
1) Graph the equation: Find the
center, vertices, foci and the equations of the asymptotic
lines.
Solution: Since it is y - x it opens
22up/down. a = 36, b = 25 and
2c = 36 + 25 = 61. Thus, a = 6, b = 5 and c = 7.8
Center: (1, 2)
Vertices: (1, 8) and (1, -4) ( six units up
and down from center)
Foci: (1, 9.8) and (1, -5.8) ( 7.8 units
up/down from center)
Equations of asymptotic lines: y - 2 =
6/5)(x - 1) (+
The box is formed by going 6 units up/down
and 5 units right/left from center.
2) Find the equation of a hyperbola with center (1, 1),
vertex (3, 1) and focus at (5, 1).
Solution: The vertex and foci are on the same horizontal line. This makes the hyperbola open right/left. a = 2 (distance from vertex to center), c = 4
22(distance from focus to center). Thus a = 4, c = 16
and
2b = 16 - 4 = 12.
The equation is:
Now for parabolas!!
Let's back up and regroup!
Section 6-5: Parabolas
A parabola is the set of all points P in the plane that are
equidistant from a fixed point F (focus) and a fixed line
d (directrix).
Demonstration of Focus Point for a Parabola (Manipula Math)
Drawing a Parabola (Manipula Math)
The equations of the parabola are as follows:
For parabolas opening up/down, the directrix is a
horizontal line in the form y = + p
For parabolas opening right/left, the directrix is a
vertical line in the form
x = + p
The vertex point for all of the above is (0, 0)
Sample Problems
1) Find the focus point and directrix and graph the
2parabola: y = x/8
Solution: The parabola opens up. 1/4p =
1/8 means 4p = 8 and
p = 2. This is the distance from the vertex to the
directrix or to the focus point. The focus point is 2 units up so it is (0, 2). The directrix is a horizontal line 2 units down from the vertex. The equation is y = -2
To determine how wide the parabola opens, the distance |4p| is the distance of the chord connecting the two sides of the parabola through the focus point perpendicular
to the axis of symmetry. In this case 4p = 8, so the parabola is 4 units from the focus point both right and
left. The two points are given on the graph.
2) Find the focus point and the directrix and graph the
2parabola: x = -2y
Solution: This parabola opens to the left. 1/4p = -2
-8p = 1
p = -1/8
The focus point is at (-1/8, 0) and the directrix is a vertical line at x = 1/8
The distance across the parabola through the focus is 1/2, so the parabola is one-fourth
unit up and down from the focus point.
3) Find the equation of the parabola with vertex at (0,
0) and directrix
y = 2.
Solution: Since the directrix is a horizontal
line and is above the vertex, the parabola opens down. p = 2 (distance from directrix to vertex), so 4p =
28. Thus the equation is y = -(1/8)x
4) Find the equation of a parabola with focus at (2, 0)
and directrix at x = -2
Solution: The vertex for this parabola is
inbetween the directrix and focus. So the vertex is (0,
0). The parabola opens to the right with p = 2.
2So 4p = 8. Thus the equation is x = (1/8)y
Translations of the parabola
The equations of the parabola with vertex (h, k) are:
5) Find the vertex, focus and directrix and graph the
parabola
2y = 2x - 8x + 1
Solution:
Put the equation in the correct form.
2y - 1 = 2(x - 4x ) Complete the square
2y - 1 + 8 = 2(x - 4x + 4) added 8 to both sides!
2y + 7 = 2(x - 2)
The parabola opens up with vertex at (2, -7)
1/4p = 2
8p = 1
p = 1/8
Focus point at (2, -6 7/8)
directrix at y = -7 1/8
6) Find the equation of the parabola with focus ( 1, 3)
and directrix x = -3.
Solution:
The parabola opens to the right. The vertex is midway between the focus and directrix. The vertex is at (-1,
3). p = 2 so 4p = 8
The equation is:
2(x + 1) = (1/8)(y - 3)
On to the last section:
Back to the previous section:
Section 6-6: Systems of Second-Degree
Equations
Try the quiz at the bottom of the page!
go to quiz
There are many methods for solving a system of second-degree equations in two variables. In this section we will concentrate on the algebraic approach using substitution and/or elimination. We have talked
about solving them using a graphing caluculator.
22221) Solve the system x + y = 20 and (x - 5) + (y - 5) =
10.
Solution: Write both in expanded form:
22 x + y= 20
22x - 10x + y - 10y = -40
Subtract the two equations to get:
-10x - 10y = -60
Divide by -10
x + y = 6
This line represents the line containing any intersection
points of the two circles. Isolate for either x or y.
y = 6 - x
now substitute back into one of the original
equations. Use the top one
22x + (6 - x) = 20
22x + 36 - 12x + x = 20
22x - 12x + 16 = 0
2x - 6x + 8 = 0
(x - 4)(x - 2) = 0
x = 4 or x = 2
To find y, use the red equation above.
When x = 4, y = 2
When x = 2, y = 4
The intersection points are (4, 2) and (2, 4)
22222) Find the intersection of 3x + y = 15 and x - y = 1.
Solution: The first equation is an ellipse
and the second is a hyperbola. Add the two equations
to get:
2 4x= 16
2x = 4
x = 2 or x = -2
Now replace these answers in one of the above
equations to find the y values.
24 - y = 1
2-y = -3
2y = 3
y = 1.7 or y = -1.7
The intersection points are (2, 1.7), (2, -1.7), (-2, 1.7),
(-2, -1.7)
22223) Find the intersection of x + y = 1 and x + 4y =
13.
Solution: Subtract these two equations to
get:
23y = 12
2y = 4
y = 2 or y = -2
Now put these into one of the original equations. Use
the first one.
2x + 4 = 1
2x = -3
This means that the answers are imaginary. What does that mean about the intersection? You are right! No
intersection. Here is the graph:
Bring on the sample test:
Let me restudy:
Current quizaroo # 6
221) Find the center point and radius for the circle: x + 4x + y - 6y - 23 =
0
a) (-2, 3), with radius 36
b) (-2, 3) with radius 6
c) (2, -3) with radius 36
d) (2, -3) with radius 6
e) (2, 3) with radius 6
22) Which of the following is a vertex point for the ellipse 4(x - 1) +
225(y - 2) = 100
a) (3, 2)
b) (1, 4)
c) (1, 7)
d) (6, 2)
e) (6, 4)
3) Which one is an equation of an asymptote for the hyperbola: (x -
221) - (y - 3) = 36
a) y - 3 = -1(x - 1)
b) y = x
c) y = -x
d) y - 3 = -6(x - 1)
e) y - 3 = 6(x - 1)
4) A parabola is the set of all points equdistant from a fixed point to a
fixed line. The fixed line
is called?
a) latus rectum
b) chord
c) directrix
d) focus
e) major axis
5) What is the most number of times a hyperbola can intersect a
circle?
a) 2
b) 3
c) 4
d) 5
e) infinite number
click here for answers!!
Sample Test
1) Graph each of the following conics and give all appropriate information.
2) Find the equation of each:
a) Ellipse with vertex at (2, 8); foci at (2, -4) and (2, 4).
b) Parabola: vertex at (3, 1); focus at (3, 3).
c) Hyperbola: center at (2, -3); vertex at (7, -3); end of minor
axis at (2, 1).
3) Find the intersection points if any for each:
2222 a) x + y = 25 and x + 10y = 169.
2222 b) y - x = 64 and x + y = 25.
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Answer Page
1)
2)
3)
4)
5)
6) V(2,8) foci at(2, -4), (2, 4)
Ellipse with center at (2, 0) Half way between the two foci's. 2a = 8 , a = 64 distance from center to vertex 2c = 4, c = 16 distance from center to focus. 222b = 48 (a - c)
22(x - 2) + y
-------- ---- = 1
48 64
7) V(3,1) F(3, 3)
p = 2 Distance from vertex to focus. Parabola opens up.
4p = 8
2y - 1 = (1/8)(x - 3)
8) C(2, -3) V(7, -3) E(2, 1)
The transverse axis is horizontal with: 2a = 5, a = 25 (distance from center to vertex) 2b = 4, b = 16 (distance from center to end of conjugate axis)
22(x - 2) (y + 3)
-------- - --------- = 1
25 16
9) 2222x + y = 25 and x + 10y = 169 2Solve for x in the first equation 221) x = -y + 25
Substitute in the second equation 22 = 169 (-y + 25) + 10y
Combine like terms 29y + 25 = 169
Solve for y 29y = 144 2y = 144/9 = 16
y = + 4
Substitute this back into equation 1 above to find the x values 22x = -(+ 4) + 25 2x = 9
x = + 3
There are four intersection points: ( 3, 4), (3, -4), (-3, 4) and (-3, -4)
10)
2222y - x = 64 and x + y = 25
Isolate for y2 in the first equation 22 + 64 1) y = x
Substitute into the second equation 22x + x + 64 = 25 22x + 64 = 25
Solve for x 22x = - 39 2x = -39/2
This yields imaginary answers, since we would be taking the square root
of a negative number. This tells us the graphs do not intersect!!
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That should do it for another chapter!!
It's time to head toward trig!!
If you need to study somemore, hit the
home button!!
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