Embedded Files
MODULE 5 - ANALYTIC GEOMETRY
OVERVIEW
This module contains the necessary descriptions, equations and examples that will help you learn and understand Analytic Geometry. This booklet contains an overview to the following topics: The Distance and Midpoint Formulas, Lines, Circles, Conic Sections and Polar Coordinates.
This module contains the necessary descriptions, equations and examples that will help you learn and understand Analytic Geometry. This booklet contains an overview to the following topics: The Distance and Midpoint Formulas, Lines, Circles, Conic Sections and Polar Coordinates.
LEARNING OBJECTIVES:
At the end of this module, you should be able to:1. Solve problems involving lengths and distances in the plane, including midpoint and point of division formula.2. Recognize the relationship between equation in two variables and graphs in the plane and use the pertinent information such as points of intersection and intercepts.3. Sketch graphs and discuss relevant features of lines, circles and other conic sections.4. Performs translations and rotations of the coordinate axes to eliminate certain terms from equation.5. Determine the relationship between polar coordinates & rectangular coordinates.
At the end of this module, you should be able to:1. Solve problems involving lengths and distances in the plane, including midpoint and point of division formula.2. Recognize the relationship between equation in two variables and graphs in the plane and use the pertinent information such as points of intersection and intercepts.3. Sketch graphs and discuss relevant features of lines, circles and other conic sections.4. Performs translations and rotations of the coordinate axes to eliminate certain terms from equation.5. Determine the relationship between polar coordinates & rectangular coordinates.
CARTESIAN COORDINATE PLANE
The Cartesian Coordinate system is also known as the rectangular coordinate system. Origin is the intersection of the coordinate axes: the horizontal axis (x- axis) and the vertical axis (y-axis). Cartesian plane has four quadrants whose points are defined as P(x,y) where x is the x-coordinate or also known as abscissa and y as the y-coordinate or also known as ordinate. Intercepts are the points that lie on the coordinate axes, y-intercept if found in the y axis and x-intercept in the x-axis.
The Cartesian Coordinate system is also known as the rectangular coordinate system. Origin is the intersection of the coordinate axes: the horizontal axis (x- axis) and the vertical axis (y-axis). Cartesian plane has four quadrants whose points are defined as P(x,y) where x is the x-coordinate or also known as abscissa and y as the y-coordinate or also known as ordinate. Intercepts are the points that lie on the coordinate axes, y-intercept if found in the y axis and x-intercept in the x-axis.
DISTANCE BETWEEN TWO POINTS
The distance formula, derived from the Pythagorean theorem, is used to find the distance between two points in the plane. The Pythagorean Theorem, a2 + b2 = c2 , is based on a right triangle where a and b are the lengths of the legs adjacent to the right angle, and c is the length of the hypotenuse. Given endpoints (x1, y1) and ( x2, y2), the distance between two points is given by: d=√[(x2 − x1)2 + (y2 − y1)2].
I got an example from quiz 5 shown at the left. What I first did is mark which is the x1 , x2 , y1 , and y2. Then, just substitute the values to the distance formula to get the distance.
The distance formula, derived from the Pythagorean theorem, is used to find the distance between two points in the plane. The Pythagorean Theorem, a2 + b2 = c2 , is based on a right triangle where a and b are the lengths of the legs adjacent to the right angle, and c is the length of the hypotenuse. Given endpoints (x1, y1) and ( x2, y2), the distance between two points is given by: d=√[(x2 − x1)2 + (y2 − y1)2].
I got an example from quiz 5 shown at the left. What I first did is mark which is the x1 , x2 , y1 , and y2. Then, just substitute the values to the distance formula to get the distance.
MIDPOINT BETWEEN TWO POINTS
The average of a segment's endpoints can be used to calculate the segment's midpoint. This notion aids in recalling a formula for determining the midpoint of a segment given its endpoint coordinates. Remember that you may find the average of two numbers by dividing their total by two. Given endpoints (x1, y1) and ( x2, y2), the midpoint between two points is given by:
The average of a segment's endpoints can be used to calculate the segment's midpoint. This notion aids in recalling a formula for determining the midpoint of a segment given its endpoint coordinates. Remember that you may find the average of two numbers by dividing their total by two. Given endpoints (x1, y1) and ( x2, y2), the midpoint between two points is given by:
SLOPE OF A LINE
The slope of a line is a number that measures its "steepness", usually denoted by the letter m. It is the change in y for a unit change in x along the line (see figure at the left). Aside from the given formula slope is also m = tan α,where α is the angle of inclination.
Note that m is positive if the line is incline right, m is negative if the line is incline left, m is zero if the line is horizontal, and m is undefined if the line is vertical. Further, when the equation of the line is in general from Ax + By + C = 0, then slope is just -A/B.
As a generalization of this practical description, the mathematics of differential calculus defines the slope of a curve at a point as the slope of the tangent line at that point. When the curve is given by a series of points in a diagram or in a list of the coordinates of points, the slope may be calculated not at a point but between any two given points.
The slope of a line is a number that measures its "steepness", usually denoted by the letter m. It is the change in y for a unit change in x along the line (see figure at the left). Aside from the given formula slope is also m = tan α,where α is the angle of inclination.
Note that m is positive if the line is incline right, m is negative if the line is incline left, m is zero if the line is horizontal, and m is undefined if the line is vertical. Further, when the equation of the line is in general from Ax + By + C = 0, then slope is just -A/B.
As a generalization of this practical description, the mathematics of differential calculus defines the slope of a curve at a point as the slope of the tangent line at that point. When the curve is given by a series of points in a diagram or in a list of the coordinates of points, the slope may be calculated not at a point but between any two given points.
DIVISION OF LINE SEGMENT
A line segment can be divided into n number of equal parts where “n” is determined as any natural number. To find the coordinates of the point dividing the line segment, the formula is shown at the left.
A line segment can be divided into n number of equal parts where “n” is determined as any natural number. To find the coordinates of the point dividing the line segment, the formula is shown at the left.
Example from quiz 5: Determine the coordinates of the point which is 3/5 of the way from the point (2,-5) to the point (-3,5). The first thing I did is to find the second ratio. If the ratio 1 is 3/5 then ratio 2 must be 2/5. Then, I just input the values to the given formulas to get the value of x and the value of y. The final answer is (-1,1).
STRAIGHT LINES
Line is series of infinite number of points. Straight Line is the line of one or uniform slope. Line is a locus of a moving point so that it is always equidistant to two fixed points. The slope of the line is m , the y-intercept is b, and the x-intercept is a. The equation of the line can be represented in several forms, and the procedure for finding the equation depends on the form chosen to represent the line. The general equation of straight lines is: Ax + By + C = 0.
Line is series of infinite number of points. Straight Line is the line of one or uniform slope. Line is a locus of a moving point so that it is always equidistant to two fixed points. The slope of the line is m , the y-intercept is b, and the x-intercept is a. The equation of the line can be represented in several forms, and the procedure for finding the equation depends on the form chosen to represent the line. The general equation of straight lines is: Ax + By + C = 0.
VARIOUS FORMS OF A LINE
1. Point-slope form: y − y1 = m(x− x1) where m is the slope and (x1 , y1) is a point on the line
2. Two-point form: y − y1 = (y2 − y1 / x2 − x1)(x − x1) where (x1 , y1) and (x2 , y2) are the points on the line
3. Slope intercept form: y = mx + b where m is the slope and b is the y-intercept
4. Intercept form: (x/a) + (y/b) = 1 where a is the x-intercept and b is the y-intercept
1. Point-slope form: y − y1 = m(x− x1) where m is the slope and (x1 , y1) is a point on the line
2. Two-point form: y − y1 = (y2 − y1 / x2 − x1)(x − x1) where (x1 , y1) and (x2 , y2) are the points on the line
3. Slope intercept form: y = mx + b where m is the slope and b is the y-intercept
4. Intercept form: (x/a) + (y/b) = 1 where a is the x-intercept and b is the y-intercept
Example from problem set 5: Find the equation of the line passing through (5,-5) and has a slope of -5.
From the forms above, we can say that the form will be a point-slope form since a point and a slope were given. By following the format and just substitute the slope or m, x1 and y1, we can find the equation y = -5x + 20.
From the forms above, we can say that the form will be a point-slope form since a point and a slope were given. By following the format and just substitute the slope or m, x1 and y1, we can find the equation y = -5x + 20.
PROPERTIES OF A LINE
PROPERTIES OF A LINE
Example from problem set 5: Find the distance between the two parallel line y = 3x - 2 and line 2y -6x + 18 = 0.I first identified from the equations which is the line 1 and the line 2. Note that I simplified the line 2 since it can be expressed in simplest form. Then, we all know that the general form of a line is Ax + By + C = 0. Since parallel lines have the same A and the same B and seeing the two equations, we can now substitute the values. Then, I just input them in a calculator to get 2.21 units.
PROPERTIES OF A LINE
PROPERTIES OF A LINE
Example from quiz 5: Find the equation of the line passing through the origin and parallel to the line 5x - 3y + 8 = 0.
I find first the slope of the equation by transforming it into a slope-intercept form which is equal to 5/3. Since in parallel lines slopes are equal, then the equation to be found has a slope of 5/3 also. Then, the point of the line given was the origin or (0,0). Through the point-slope formula, we can arrive at the equation by substituting the values. Finally, the equation is 5x - 3y = 0.
I find first the slope of the equation by transforming it into a slope-intercept form which is equal to 5/3. Since in parallel lines slopes are equal, then the equation to be found has a slope of 5/3 also. Then, the point of the line given was the origin or (0,0). Through the point-slope formula, we can arrive at the equation by substituting the values. Finally, the equation is 5x - 3y = 0.
PROPERTIES OF A LINE
Example from quiz 5: Find the equation of the line through the point (3,-2) and perpendicular to the line 2x + 3y + 4 = 0.
Just like the previous example, I first get the slope of the given equation by transforming it into a slope-intercept form. We all know that if two lines are perpendicular, the slope of one is the negative reciprocal of the other. We can get the slope of the missing equation. If the slope of the first equation is -2/3 then the slope of the second equation must be 3/2. Finally, we have now a slope and a point, thus we can now give the equation through the point-slope form of the line. The missing equation is 3x - 2y - 13 = 0.
Just like the previous example, I first get the slope of the given equation by transforming it into a slope-intercept form. We all know that if two lines are perpendicular, the slope of one is the negative reciprocal of the other. We can get the slope of the missing equation. If the slope of the first equation is -2/3 then the slope of the second equation must be 3/2. Finally, we have now a slope and a point, thus we can now give the equation through the point-slope form of the line. The missing equation is 3x - 2y - 13 = 0.
PROPERTIES OF A LINE
CONIC SECTIONS
A conic section is any one of several curves produced by passing a plane through a cone in different ways.
However, there are other ways for a plane and the cones to intersect to form what are referred to as degenerate conics: a point, one line, and two lines. The general equation of conic sections is: Ax2 + Bxy+ Cy2 + Dx + Ey + F = 0.
A conic section is any one of several curves produced by passing a plane through a cone in different ways.
- Circle - When the plane is horizontal
- Ellipse - When the (tilted) plane intersects only one cone to form a bounded curve
- Parabola - When the plane intersects only one cone to form an unbounded curve
- Hyperbola - When the plane (not necessarily vertical) intersects both cones to form two unbounded curves (each called a branch of the hyperbola)
However, there are other ways for a plane and the cones to intersect to form what are referred to as degenerate conics: a point, one line, and two lines. The general equation of conic sections is: Ax2 + Bxy+ Cy2 + Dx + Ey + F = 0.
CIRCLES
Circle is a locus of points equidistant to a fixed point. The constant distance is the radius and the fixed point is the center. The general equation of a circle is Ax2 + Cy2 + Dx + Ey + F = 0 where A = C. Its standard equations are:
Circle is a locus of points equidistant to a fixed point. The constant distance is the radius and the fixed point is the center. The general equation of a circle is Ax2 + Cy2 + Dx + Ey + F = 0 where A = C. Its standard equations are:
- x2 + y2 = 0 (center origin)
- (x - h)2 = (y - k)2 = r2 (center h,k)
Example from quiz 5: Find the equation of a circle whose center is at (3,-5) and whose radius is 4.
Since the center and the radius are already given, we simply substitute them to the standard formula where (h,k) is the center and r is the radius. Then, perform the square of binomial, combine like terms, and simplify. The general equation of the circle is x2 + y2 - 6x + 10y + 18 = 0
Since the center and the radius are already given, we simply substitute them to the standard formula where (h,k) is the center and r is the radius. Then, perform the square of binomial, combine like terms, and simplify. The general equation of the circle is x2 + y2 - 6x + 10y + 18 = 0
PARABOLAS
Parabola is a locus of points equidistant to a fixed point and a fixed line. The fixed point is the focus and the fixed line is the directrix. The general equation of a parabola is Ax2 + Dx + Ey + F = 0 or Cy2 + Dx + Ey + F = 0. Its standard equations are:
Note that c is the unit distance from the vertex to the focus and the unit distance from the vertex to the directrix.
Parabola is a locus of points equidistant to a fixed point and a fixed line. The fixed point is the focus and the fixed line is the directrix. The general equation of a parabola is Ax2 + Dx + Ey + F = 0 or Cy2 + Dx + Ey + F = 0. Its standard equations are:
- Vertex at origin
- y2 = -4cx (opens left)
- y2 = 4cx (opens right)
- x2 = -4cy (opens downward)
- x2 = 4cy (opens upward)
- Vertex at (h,k)
- (y - k)2 = -4c(x - h) (opens left)
- (y - k)2 = 4c(x - h) (opens right)
- (x - h)2 = -4c(y - k) (opens downward)
- (x - h)2 = 4c(y - k) (opens upward)
Note that c is the unit distance from the vertex to the focus and the unit distance from the vertex to the directrix.
Example from problem set 5; Find the directrix of the parabola y2 - 8x - 4y + 20 = 0.
To find the directrix of the parabola, the first thing I did is transform the equation in general form into standard form. Then, from the standard form, we can already identify the vertex of the parabola, the orientation of the parabola, and the value of c. For instance, the vertex is at (2,2), c is equal to 2, and the parabola opens rightward since we are dealing with y2 and 4c is positive. With that information, we can already identify where the directrix is even without the formula. We all know that c is the distance from the vertex to the directrix, so 2 units from (2,2) going to the left (since it opens rightwards) will be directrix. Thus, the directrix is at x = 0.
To find the directrix of the parabola, the first thing I did is transform the equation in general form into standard form. Then, from the standard form, we can already identify the vertex of the parabola, the orientation of the parabola, and the value of c. For instance, the vertex is at (2,2), c is equal to 2, and the parabola opens rightward since we are dealing with y2 and 4c is positive. With that information, we can already identify where the directrix is even without the formula. We all know that c is the distance from the vertex to the directrix, so 2 units from (2,2) going to the left (since it opens rightwards) will be directrix. Thus, the directrix is at x = 0.
ELLIPSES
Ellipse is a locus of points such that the sum of its distance from two fixed points is constant and is equal to the length of the major axis. The two fixed points is called the foci and the length of the major axis is 2a. The general equation of an ellipse is Ax2 + Cy2 + Dx + Ey + F = 0 where A ≠ C.
The standard form of an ellipse is on the figure at the right. Note that it is for an ellipse which the center is at (h,k). Meanwhile, for an ellipse which center is at the origin, we can just change the numerators of the equation into x2 and y2. Also, note that a is the semi-major axis and b is the semi-minor axis. Further, c = √(a2 - b2).
Ellipse is a locus of points such that the sum of its distance from two fixed points is constant and is equal to the length of the major axis. The two fixed points is called the foci and the length of the major axis is 2a. The general equation of an ellipse is Ax2 + Cy2 + Dx + Ey + F = 0 where A ≠ C.
The standard form of an ellipse is on the figure at the right. Note that it is for an ellipse which the center is at (h,k). Meanwhile, for an ellipse which center is at the origin, we can just change the numerators of the equation into x2 and y2. Also, note that a is the semi-major axis and b is the semi-minor axis. Further, c = √(a2 - b2).
Example from quiz 5: Find the equation of the ellipse whose foci (-3,0) and (3,0) with corresponding directrices x = -6 and x = 6.
To better understand the problem, I first illustrated it to determine the orientation of the ellipse. We had also identified the center which is at the origin. To arrive to the equation, we should find the value of a and b. From the formula of foci of an ellipse (±ae , 0), we therefore say that ae is equal to 3. Also, from the formula of the directrices of an ellipse (±a/e), we therefore say that a/e is equal to 6/1. We have now two equations. In solving systems of equations, we can use the substitution method. As can be seen in my solution, I derived the value of e which is equal to 3/a. Then, I substitute this 3/a to the e of the second equation to get rid of e; and so that only letter a will be the left variable. From this, we can get the value of a by performing the necessary operations. I got a = √18. Then, to get the value of b, I used the formula for e. I substitute first the value of e by 3/a and since a is equal to √18, I input it in the equation so that the only variable left is b. From there, I got the value of b which is 3. Finally, using the standard equation for a horizontal ellipse, I arrived to the equation of the ellipse shown in the figure.
To better understand the problem, I first illustrated it to determine the orientation of the ellipse. We had also identified the center which is at the origin. To arrive to the equation, we should find the value of a and b. From the formula of foci of an ellipse (±ae , 0), we therefore say that ae is equal to 3. Also, from the formula of the directrices of an ellipse (±a/e), we therefore say that a/e is equal to 6/1. We have now two equations. In solving systems of equations, we can use the substitution method. As can be seen in my solution, I derived the value of e which is equal to 3/a. Then, I substitute this 3/a to the e of the second equation to get rid of e; and so that only letter a will be the left variable. From this, we can get the value of a by performing the necessary operations. I got a = √18. Then, to get the value of b, I used the formula for e. I substitute first the value of e by 3/a and since a is equal to √18, I input it in the equation so that the only variable left is b. From there, I got the value of b which is 3. Finally, using the standard equation for a horizontal ellipse, I arrived to the equation of the ellipse shown in the figure.
HYPERBOLAS
Hyperbola is a locus of points whose distances from two fixed points differ by a certain constant and is equal to the length of the transverse axis. The two fixed points is the foci and the length of the transverse axis is equal to 2a. The general equation of a hyperbola is Ax2 - Cy2 + Dx + Ey + F = 0 or Cy2 - Ax2 + Dx + Ey + F = 0.
Shown at the right is the standard form of hyperbola. Note that it is only for those which the center is at (h,k). If the center is at the origin then just change the numerators into x2 and y2. Also, remember a2 is always the denominator of the positive term. Same as in ellipse, c = √(a2 - b2).
Hyperbola is a locus of points whose distances from two fixed points differ by a certain constant and is equal to the length of the transverse axis. The two fixed points is the foci and the length of the transverse axis is equal to 2a. The general equation of a hyperbola is Ax2 - Cy2 + Dx + Ey + F = 0 or Cy2 - Ax2 + Dx + Ey + F = 0.
Shown at the right is the standard form of hyperbola. Note that it is only for those which the center is at (h,k). If the center is at the origin then just change the numerators into x2 and y2. Also, remember a2 is always the denominator of the positive term. Same as in ellipse, c = √(a2 - b2).
Example from quiz 5: Find the coordinates of one of the foci of the hyperbola (x2/16) - (y2/9) = 1.
The first thing I did is find the value of a and b. Since a2 is always the denominator of the positive term, then a = 4 and b = 3. To get the value of c, we will use the formula given above; c = 5. Given that the center of the hyperbola is at the origin and c is the unit distance from the center to the focus, we can already tell the coordinates of the foci even without the formula. Five units to the left of the origin is (-5,0) and five units to the right is (5,0). Thus, the said coordinates must be the foci.
The first thing I did is find the value of a and b. Since a2 is always the denominator of the positive term, then a = 4 and b = 3. To get the value of c, we will use the formula given above; c = 5. Given that the center of the hyperbola is at the origin and c is the unit distance from the center to the focus, we can already tell the coordinates of the foci even without the formula. Five units to the left of the origin is (-5,0) and five units to the right is (5,0). Thus, the said coordinates must be the foci.
SUMMARY
The conic sections namely, circle, parabola, ellipse and hyperbola are unique that they have different purposes, advantages or uses. They are extremely helpful especially in the field of engineering. Let me just add this information that one can already identify the kind of conic section just by looking at its general form. It is a circle when x2 and y2 are present with equal coefficients. It is a parabola when either x2 or y2 is present but not both. It is an ellipse when x2 and y2 are present and positive but the coefficients are different. Lastly, it is a hyperbola when x2 and y2 are present and exactly one of the coefficients is negative and the other is positive.
The conic sections namely, circle, parabola, ellipse and hyperbola are unique that they have different purposes, advantages or uses. They are extremely helpful especially in the field of engineering. Let me just add this information that one can already identify the kind of conic section just by looking at its general form. It is a circle when x2 and y2 are present with equal coefficients. It is a parabola when either x2 or y2 is present but not both. It is an ellipse when x2 and y2 are present and positive but the coefficients are different. Lastly, it is a hyperbola when x2 and y2 are present and exactly one of the coefficients is negative and the other is positive.
Jess Marck Salazar -PS5- BSCE1B.pdfPROBLEM SET #5 (45/45)
Salazar_Quiz5Solutions_BSCE1B.pdfQUIZ #5 SOLUTIONS (65/65)
REFLECTION
In this module, there were new things I learned. Most of these things are about the slope of a line since our teacher in high school (grade 8) who was supposed to teach this topic in detail fail to discuss because of some reasons. I am happy that through this module, I was able to fill the gap with the knowledge that I supposed to learn back then. Specifically, I have learned that the slope of parallel lines are equal and the slope of perpendicular lines is the negative reciprocal of the other. I have also recalled many formulas that was taught to us before. Further, I was also able to relearn the conic sections which was taught to us in grade 11. When it comes to the activities given to us, I really enjoyed answering the problem set as well as the quiz. Though we had limited time, our teacher was still able to discuss the topics precisely and in a way that everybody can understand. Our teacher is doing an excellent job.
In this module, there were new things I learned. Most of these things are about the slope of a line since our teacher in high school (grade 8) who was supposed to teach this topic in detail fail to discuss because of some reasons. I am happy that through this module, I was able to fill the gap with the knowledge that I supposed to learn back then. Specifically, I have learned that the slope of parallel lines are equal and the slope of perpendicular lines is the negative reciprocal of the other. I have also recalled many formulas that was taught to us before. Further, I was also able to relearn the conic sections which was taught to us in grade 11. When it comes to the activities given to us, I really enjoyed answering the problem set as well as the quiz. Though we had limited time, our teacher was still able to discuss the topics precisely and in a way that everybody can understand. Our teacher is doing an excellent job.
REFERENCES:
https://wallpaperaccess.com/full/3280804.jpghttps://res.cloudinary.com/jerrick/image/upload/v1581025680/5e3c898fc935ad001e25fc42.jpghttps://cutewallpaper.org/21/color-background/Free-photo-color-background-Abstract-Bright-Closeup-.pnghttps://i.pinimg.com/564x/24/b7/31/24b7317b0cbfb62e564ee191fa24e887.jpghttps://24wallpapers.com/app-gateway/wallpaper-uploads/wallpapers/legacyUploads/wi89124833a91-caf7-4c3e-aad5-42da99b7dba92079665946.jpghttps://kinetikgroup.co/wp-content/uploads/2020/05/Beulah-International-MelbourneCBD.jpghttps://www.swedishnomad.com/wp-content/images/2018/10/Burj-Khalifa-Dubai.jpghttps://www.mathwarehouse.com/algebra/linear_equation/slope-of-a-line.php#:~:text=The%20slope%20of%20a%20line%20characterizes%20the%20direction%20of%20a,of%20those%20same%202%20points.https://en.wikipedia.org/wiki/Slopehttps://www.onlinemathlearning.com/image-files/forms-equation-of-line.pnghttps://www.onlinemathlearning.com/image-files/equation-circle.pnghttps://encrypted-tbn0.gstatic.com/images?q=tbn:ANd9GcTvJqrOBHSGj0wXKvvBeiMCtawmnjySp_J4cA&usqp=CAUhttps://useruploads.socratic.org/B6ETKfBRxqwsZ9PQdJUg_hyperbola-advanced-algebra-8-728.jpghttp://ellipsesconicsections.weebly.com/uploads/2/2/5/5/22554190/6074916_orig.png
Page updated
Report abuse