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美国高中数学2-6 Analytic Geometry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...

美国高中数学2-6 Analytic Geometry
美国高中 数学 数学高考答题卡模板高考数学答题卡模板三年级数学混合运算测试卷数学作业设计案例新人教版八年级上数学教学计划 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. This page hosted by Get your own Free Home Page Let's see those answers! I need help! Take me back! 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!! This page hosted by Get your own Free Home Page 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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