MATH14 Coursewares

Fundamental Concepts

Math 14 Plane and Solid Analytic Geometry

OBJECTIVES:At the end of the lesson, the student is expected to be able to:• use the Cartesian Coordinate System efficiently and

effectively as a tool in the study of Analytic Geometry.• determine the distance between two points.• use the coordinates of the vertices of a polygon to determine

its area.

Analytic Geometry – is the branch of mathematics, which deals with the properties, behaviours, and solution of points, lines, curves, angles, surfaces and solids by means of algebraic methods in relation to a coordinate system.

DEFINITION:

FUNDAMENTAL CONCEPTS

Two Parts of Analytic Geometry

1. Plane Analytic Geometry – deals with figures on a plane surface.

2. Solid Analytic Geometry – deals with solid figures.

Directed Line – a line in which one direction is chosen as positive and the opposite direction as negative.

Directed Line Segment – consisting of any two points and the part between them.

Directed Distance – the distance between two points either positive or negative depending upon the direction of the line.

DEFINITION:

Coordinate Plane – is a plane determined by the coordinate axes.

Key Words: coordinates, abscissa, ordinate

o

y

x

P (x, y)

RECTANGULAR COORDINATE SYSTEM

DISTANCE BETWEEN TWO POINTS

The length of a horizontal line segment is the difference of the abscissas (x-coordinates) of the right endpoint and the left endpoint of the line segment. That is,

1. Horizontal Distance

Distance,

d x2 x1

d xr xl

2.VerticalThe length of a vertical line segment is the difference in the ordinates (y-coordinates) of the upper endpoint and the lower endpoint of the line segment. That is,

Distance,

d y2 y1

d yu yl

3. SlantThe distance between two points on a slant line segment is the square root of the sum of the squares of the difference of the abscissas and the difference of the ordinates. That is,

Distance ,

d a2 b2

d x2 x1 2 y2 y1 2

d xr xl 2 yu yl 2

SAMPLE PROBLEMS1. Determine the distance between a. (-2, 3) and (5, 1)b. (6, -1) and (-4, -3)2. Show that points A (3, 8), B (-11, 3) and C (-8, -2) are vertices of an isosceles triangle.3.Show that the triangle A (1, 4), B (10, 6) and C (2, 2) is a right triangle.4.Find the point on the y-axis which is equidistant from A(-5, -2) and B(3,2).

5. By addition of line segments show whether the points A(-3, 0), B(-1, -1) and C(5, -4) lie on a straight line.

6. The vertices of the base of an isosceles triangle are (1, 2) and (4, -1). Find the ordinate of the third vertex if its abscissa is 6.

7. Find the radius of a circle with center at (4, 1), if a chord of length 4 is bisected at (7, 4).

8. Show that the points A(-2, 6), B(5, 3), C(-1, -11) and D(-8, -8) are the vertices of a rectangle.

9. The ordinate of a point P is twice the abscissa. This point is equidistant from (-3, 1) and (8, -2). Find the coordinates of P.

10. Find the point on the y-axis that is equidistant from (6, 1) and (-2, -3).

AREA OF A POLYGON BY COORDINATES

Consider the triangle whose vertices are P1(x1, y1), P2(x2, y2) and P3(x3, y3) as shown below.

o

y

x

111 y,xP

222 y,xP

333 y,xP

When the vertices of the triangle are labeled in the counterclockwise direction, then the area of the triangle is determined by:

1yx

1yx

1yx

2

1A

33

22

11

Generalized formula for the area of polygon by coordinates:

1n54321

1n54321

yy..yyyyy

xx..xxxxx

2

1A

SAMPLE PROBLEMS

1.Find the area of the triangle whose vertices are (-6, -4), (-1, 3) and (5, -3).2.Find the area of a polygon whose vertices are (6, -3), (3, 4), (-6, -2), (0, 5) and (-8, 1).3.Find the area of a polygon whose vertices are (2, -3), (6, -5), (-4, -2) and (4, 0).

Reference: Analytic Geometry by Riddle

INCLINATION AND SLOPE


OBJECTIVES:At the end of the lesson, the student is expected to be

able to:• define and determine the angle of inclination and

slope of a single line, as well as of, parallel lines, perpendicular lines and intersecting lines.

• use the concept to solve common problems in Analytic geometry.

INCLINATION AND SLOPE OF A LINE

The inclination of the line, L, (not parallel to the x-axis) is defined as the smallest positive angle measured from the positive direction of the x-axis to the line L.

The slope of the line is defined as the tangent of the angle of inclination.

PARALLEL AND PERPENDICULAR LINES

If two lines are parallel their slopes are equal. If two lines are perpendicular the slope of one of the line is the negative reciprocal of the slope of the other line.

If m1 is the slope of L1 and m2 is the slope of L2 and if:

1. 2.

L1 is parallel to line L2 then m1 m2 .

L1 is perpendicular to line L2 then m1m2 1 or

m1 - 1

m2.

x

y y

Sign Conventions:

Slope is positive (+) if the line is leaning upward to the right.Slope is negative (-) if the line is leaning upward to the left.Slope is zero (0) if the line is horizontal.Slope is undefined if the line is vertical.

Examples:1. Find the slope, m, and the angle of inclination, , of the lines through each of the following pair of points.a. (8, -4) and (5, 9)b. (10, -3) and (14, -7)c. (-9, 3) and (2, -4)

2. The line segment drawn from (x, 3) to (4, 1) is perpendicular to the segment drawn from (-5, -6) to (4, 1). Find the value of x.

3. Show that the triangle whose vertices are A(8, -4), B(5, -1) and C(-2,-8) is a right triangle.

4. Show that the points A(-2, 6), B(5, 3), C(-1, -11) and D(-8, -8) are the vertices of a parallelogram. Is the parallelogram a rectangle?

5. Find y if the slope of the line segment joining (3, -2) to (4, y) is -3.

6. Show that the points A(-1, -1), B(-1, -5) and C(12, 4) lie on a straight line.

ANGLE BETWEEN TWO INTERSECTING LINES

L1

L2

21

12

mm1

mmtan

Where: m1 = slope of the initial side m2 = slope of the terminal side

The angle between two intersecting lines L1 and L2 is the least or acute counterclockwise angle.

0180:note

Examples:

1.Find the angle from the line through the points (-1, 6) and (5, -2) to the line through (4, -4) and (1, 7). 2.The angle from the line through (x, -1) and (-3, -5) to the line through (2, -5) and (4, 1) is 450 . Find x.3.Two lines passing through (2, 3) make an angle of 450 . If the slope of one of the lines is 2, find the slope of the other.4.Find the interior angles of the triangle whose vertices are A (-3, -2), B (2, 5) and C (4, 2).

REFERENCES

Analytic Geometry by Riddle

DIVISION OF LINE SEGMENT



able to:

• Determine the coordinates of a point of division of a line segment.

- Use this concept to solve common application problems in Analytic Geometry.

21

1

PP

PPr

222 y,xP

y,xP

111 y,xP 12 y,xN 1y,xM

Internal Point of Division

note : For internal point, P is always between P1 than P2 .

222 y,xP

y,xP

111 y,xP 12 y,xN 1y,xM

External Point of Division

21

1

PP

PPr

21 P than P from farther always is P point, external For :note

Since the figure is a similar triangle, the coordinates of point

P x, y is computed with the use of ratio and proportion,

wherein P1P

P1P2

P1M

P1N PM

P2N. But

P1P

P1P2r;

P1M

P1N

x x1x2 x1

;

and PMP2N

y y1y2 y1

.

Thus, x = x1 r x2 x1 and y = y1 r y2 y1 .

For internal point , r 1;

for midpoint , r 12

and;

for external point , r 1.

Alternative Formula :

If P1P

PP2

r1r2

, then x r1x2 r2x1

r1 r2 and y

r1y2 r2y1r1 r2

Examples:1.The line segment joining (-5, -3) and (3, 4) is to be divided into five equal parts. Find the point of division closest to (-5, -3).2.Find the midpoint of the segment joining (7, -2) and (-3, 5).3.The line segment from (1, 4) to (2, 1) is extended a distance equal to twice its length. Find the terminal point.4.On the line joining (4, -5) to (-4, -2), find the point which is three-seventh the distance from the first to the second point.5.Find the trisection points of the line joining (-6, 2) and (3, 8).

6. The line segment joining a vertex of a triangle and the midpoint of the opposite side is called the median of the triangle. Given a triangle whose vertices are A(4,-4), B(10, 4) and C(2, 6), find the point on each median that is two-thirds of the distance from the vertex to the midpoint of the opposite side.

REFERENCE

Analytic Geometry by Riddle

EQUATION OF LOCUS


AN EQUATION OF A LOCUS


able to:• Define the locus of the equation• Determine the equation of a locus defining line,

circle and conics

EQUATION OF A LOCUS

An equation involving the variables x and y is usually satisfied by an infinite number of pairs of values of x and y, and each pair of values corresponds to a point. These points follow a pattern according to the given equation and form a geometric figure called the locus of the equation.Since an equation of a curve is a relationship satisfied by the x and y coordinates of each point on the curve (but by no other point), we need merely to consider an arbitrary point (x,y) on the curve and give the description of the curve in terms of x and y.

Examples:Find an equation for the set of all points (x, y) satisfying the given conditions1.It is equidistant from (5, 8) and (-2, 4).2.The sum of its distances from (0, 4) and (0, -4) is 10.3.It is equidistant from (-2, 4) and the y-axis4.It is on the line having slope of 2 and containing the point (-3, -2).5.The difference of its distances from (3, 0) and (-3, 0) is 2.

REFERENCES

Analytic Geometry, 6th Edition, by Douglas F. RiddleAnalytic Geometry, 7th Edition, by Gordon Fuller/Dalton Tarwater

Analytic Geometry, by Quirino and Mijares

LINES AND FIRST DEGREE EQUATIONS



able to:• Define and determine the general equation of a line• Define and determine the different standard

equations of line• Determine the directed distance from a point to a

line• Determine the distance between parallel lines

STRAIGHT LINESA straight line is a locus of a point that

moves in a plane with constant slope. It may also be referred to simply as a line which contains at least two distinct points.

LINES PARALLEL TO A COORDINATE AXIS

If a straight line is parallel to the y-axis, its equation is x = k, where k is the directed distance of the line from the y-axis. Similarly, if a line is parallel to the x-axis, its equation is y = k, where k is the directed distance of the line from the x-axis.

DIFFERENT STANDARD FORMS OF THE EQUATION OF A STRAIGHT LINE

A. POINT-SLOPE FORM:If the line passes through the point (x1, y1), then the slope of the line is . Rewriting the equation we have which is the standard equation of the point-slope form.

1

1

xx

yym

11 xxmyy

The equation of the line through a given point P1 (x1, y1) whose slope is m.

y

x

111 y,xP

y,xP

m

form. slope

-point the as known is

which x-xmy-y or

y-yx-xm us give

willx-xby sidesboth

gmultiplyin and x-x

y-ym

formula, slopegsinU

11

11

1

1

1

EXAMPLE:Find the general equation of the line:a.through (2,-7) with slope of 2/5b.through the point (-3, 4) with slope of -2/5

B. TWO-POINT FORM:If the line passes through the points (x1, y1) and

(x2, y2), then the slope of the line is .

Substituting it in the point-slope formula, we have which the standard equation ofthe two-point form.

12

12

xx

yym

112

121 xx

xx

yyyy

The equation of the line through points P1 (x1, y1) and P2 (x2, y2)

y

x

111 y,xP

222 y,xP

m

y,xP

form. point-two

the as known is which

xxxx

yyy-y

us give willform slope

-point the to it ing substitutand

xx

yym slopethe ingUs

112

121

12

12

C. SLOPE-INTERCEPT FORM:Consider a line not parallel to either axes of the

coordinate axes. Let the slope of the line be m and intersecting the y-axis at point (0, b), then the slope of the line is . Rewriting the equation, we have

which is the standard equation of the slope-intercept form.

0x

bym

bmxy

EXAMPLE:Find the general equation of the line: a.passing through (4,-5) and (-6, 3)b.passing through (2,-3) and (-4, 5)

y

x

b,0

y,xP

m

form. intercept-slope

the as known is which

bmxy therefore

and b-ymx us give

willxby sidesboth

gmultiplyin and 0-x

b-ym

formula, slopegsinU

The equation of the line having the slope, m, and y-intercept (0, b)

EXAMPLE:a.Find the general equation of the line with slope 3 and y-intercept of 2/3. b.Express the equation 3x-4y+8=0 to the slope-intercept form and draw the line.

D. INTERCEPT FORM:Let the intercepts of the line be the points (a, 0)and (0, b). Then the slope of the line and itsequation is . Simplifying the equationwe have which is the standard equation ofthe intercept form.

a

bm

0xa

bby

1b

y

a

x

The equation of the line whose x and y intercepts are (a, 0) and (0, b) respectively.

y

x

b,0

0,a

m y,xP

form.

intercept the as known is which

1b

y

a

x us give willabby sides

both dividing and abaybx

become willpositive terms the all

make to equation the arranging

-reby Then .bxab-ay

us give willaby sidesboth

gmultiplyin and 0xa

b-b-y

formula, slope-point gsinU

EXAMPLE:Find the general equation of the line: a.with x-intercept of 2 and y-intercept of -3/4 b.through (-2, 7) with intercepts numerically equal but of opposite sign

E. NORMAL FORM:Suppose a line L, whose equation is to be found, has its distance from the origin to be equal to p. Let the angle of inclination of p be

o

b

y

x

p

L

sin

pb

b

psin

Since p is perpendicular to L, the slope of p is equal to the negative reciprocal of the slope L.

Substituting in the slope-intercept form,y = mx + b , we obtain

Simplifying, we have the normal form of the straight line

sin

cosm or ,cot

tan

1m

sin

px

sin

cosy

p y sin cos x

Reduction of the General Form to the Normal FormThe slope of the line Ax+By+C=0 is . The slope of p which is perpendicular to the line is therefore . Thus, .From Trigonometry, we obtain the values and . If we divide through the generalequation of the straight line by , we have

Transposing the constant to the right, we obtain This is of the normal form . Comparing the two equations, we note that .

B

A

A

B

A

Btan

22 BA

Bsin

22 BA

Acos

22 BA

0 BA

C y

BA

B x

BA

A222222

BA

C y

BA

B x

BA

A222222

22 BA

Cp

p y sin cos x

EXAMPLE: 1.Reduce 5x+3y-4=0 to the normal form.

2. Find the equation of a line parallel to the line 4x-y+8=0 passing at a distance ±3 from the point (-2,-4).

034

4

34

y3

34

x5 ,thus

34925BA

-4C 3B 5A

:Solution

22

4yx4424yx4

is 4,2 through gsinpas

and line given the to parallel line a of equation The

:Solution

1734yx4

or 17317

4

17

y

17

4x

be wouldform, normal the in lines, required the of equations The17

4

17

y

17

4x116

4

116

y

116

4x

have weform, normal the to ducingRe

PARALLEL AND PERPENDICULAR LINES The lines Ax+By+C=0 and Ax+By+K=0 are parallel

lines. But, the lines Ax+By+C=0 and Bx-Ay+K=0 are perpendicular lines.

EXAMPLE: Find the general equation of the line:a. through (-3, 8) parallel to the line 6x-5y+15=0 b. through (6,-1) and perpendicular to the line 4x-5y-6=0c. passing through (-1, 5) and parallel to the line through

(1 ,3) and (1,-4)

DIRECTED DISTANCE FROM A POINT TO A LINE The directed distance from the point P(x1, y1) to the

line Ax+By+C=0 is , where the sign of B isTaken into consideration for the sign of the . If

B>0, then it is and B<0, then it is . But if B=0, take the sign of A.

22

11

BA

CByAxd

22 BA 22 BA 22 BA

y

x

111 y,xP

222 y,xP

0CByAx 11

0d1

0d2

line the below is

point the 0,d if

line the above is

point the 0,d if

:note

EXAMPLE: a.Find the distance of the point (6,-3) from the line 2x-y+4=0.b.Find the equation of the bisector of the acute angle for the pair of lines L1: 11x+2y-7=0 and L2: x+2y+2=0.c.Find the distance between the lines 3x+y-12=0 and 3x+y-4=0

EXERCISES:1. Determine the equation of the line passing through (2, -3) and parallel to the line passing through (4,1) and (-2,2).2. Find the equation of the line passing through point (-2,3) and perpendicular to the line 2x – 3y + 6 = 03. Find the equation of the line, which is the perpendicular bisector of the segment connecting points (-1,-2) and (7,4).4. Find the equation of the line whose slope is 4 and passing through the point of intersection of lines x + 6y – 4 = 0 and 3x – 4y + 2 = 0

5. The points A(0, 0), B(6, 0) and C(4, 4) are vertices of triangles. Find:a. the equations of the medians and their intersection pointb. the equations of the altitude and their intersection pointc. the equation of the perpendicular bisectors of the sides and their intersection points

Exercises:1. Find the distance from the line 5x = 2y + 6 to the pointsa. (3, -5)b. (-4, 1)c. (9, 10)2. Find the equation of the bisector of the pair of acute angles formed by the lines 4x + 2y = 9 and 2x – y = 8.3. Find the equation of the bisector of the acute angles and also the bisector of the obtuse angles formed by the lines x + 2y – 3 = 0 and 2x + y – 4 = 0.

REFERENCES



CIRCLES



able to:• Determine the center and radius of the circle given

an equation.• Determine the general and standard form of

equation of the circle given some geometric conditions.

• Convert general equation of a circle to the standard form and vice-versa

• Determine the equation of a circle defining family of circles

• Determine the radical axis

CIRCLEA circle is a locus of points that moves in a plane at a constant distance from a fixed point. The fixed point is called the center and the distance from the center to any point on the circle is called the radius.Parts of a Circle:● Center - It is in the center of the circle and the distance from this point to any other point on the circumference is the same. ● Radius - The distance from the centre to any point on the circle is called the radius. A diameter is twice the distance of a radius. ● Circumference - The distance around a circle is its circumference. It is also the perimeter of the circle

● Chord - A chord is a straight line joining two points on the circumference. The longest chord in a circle is called a diameter. The diameter passed through the center.

● Segment - A segment of a circle is the region enclosed by a chord and an arc of the circle.

● Secant - A secant is a straight line cutting at two distinct points.

● Tangent - If a straight line and a circle have only one point of contact, then that line is called a tangent. A tangent is always perpendicular to the radius drawn to the point of contact.

EQUATION OF THE CIRCLE

P(x,y)

C(h,k)

y

x

x

yk

h

r

o

Let:C (h, k) - coordinates of the center of the circle r - radius of the circle P (x, y) - coordinates of any point along the circle

From the figure: Distance CP = radius ( r )Recall the distance formula:

Squaring both sides of the equation:r2 = (x – h)2 + (y – k)2

The equation is also called the center-radius form or the Standard Form. (x – h)2 + (y – k)2 = r2

If the center of the circle is at the origin (0, 0):h = 0 k = 0C (h, k) C (0, 0)

From: (x – h)2 + (y – k)2 = r2

(x – 0)2 + (y – 0)2 = r2

x2 + y2 = r2 → (Center at the origin)

From: (x – h)2 + (y – k)2 = r2 → Standard FormCenter at (h, k): (x2 – 2hx + h2) + (y2 – 2ky + k2) = r2

x2 + y2 – 2hx – 2ky + h2 + k2 - r2= 0Let: -2h = D

-2k = E CONSTANTSh2 + k2 - r2 = F

Therefore,

x2 + y2 + Dx + Ey + F = 0 →General Form

Examples:

1.If the center of the circle is at C(3, 2) and the radius is 4 units, find the equation of the circle2.Find the equation of the circle with center (-1, 7) and tangent to the line 3x – 4y + 6 = 0.3.Find the equation of the circle having (8, 1) and (4,-3) as ends of a diameter. 4.Reduce to standard form and draw the circle whose equation is 4x2 + 4y2 – 4x – 8y – 31 = 0.5.Find the equation of the circle passing through the intersection of 2x-3y+6=0 and x+3y-6=0 with center at (3,-1).

Case II: Three noncollinear points determine a circle as shown in Figure 2. The three points are the three conditions in this case, knowing them gives three conditions in D, E, and F in the general form of a circle. Note that one point (two coordinates) on a circle is a single “condition”, while each coordinate of the center is a condition. More generally, knowing that the center is on the given line can be counted on as a “condition” to determine a circle; knowing h and k is equivalent to knowing that the center is on the lines x = h and y = k.

Case III: The equation of a tangent line, the point of tangency, and another point on the circle as shown in the Figure 3. The center is on the perpendicular to the tangent at the point of tangency. It is also on the perpendicular bisector of the segment joining any two points of the circle. These two lines determine the center of the circle; the radius is now easily found.

Case IV: Tangent line and a pair of points on a circle determine two circles as shown in the Figure 4.

Figure 1 Figure 2

Figure 3 Figure 4

Example: 1. Find the equation of the circle if the circle is

tangent to the line 4x – 3y + 12=0 at (-3, 0) and also tangent to the line 3x + 4y –16 = 0 at (4, 1).

2. Find the equation of the circle which passes through the points (1, -2), (5, 4) and (10, 5).

3. Find the equation of the circle which passes through the points (2, 3) and (-1, 1) and has its center on the line x – 3y – 11 = 0.

4. Find the equation(s) of the circle(s) tangent to 3x-4y-4=0 at (0,-1) and containing the point (-1,-8)

Exercises: 1. Find the equation of the circle passing through (7,

5) and (3, 7) and with center on x-3y+3=0.2. Find the points of intersection of the circles

x2 + y2 – 4x – 4y + 4 = 0 and x2 + y2 + 2x – 4y - 8 = 0. Draw the circles.

3 Find the equation of the circle if it is tangent to the line x + y = 2 at point (4 -2) and the center is at the x-axis.

4 A triangle has its sides on the lines x + 2y – 5 = 0, 2x – y – 10 = 0 and 2x + y + 2 = 0. Find the equation of the circle inscribed in the triangle.

5 Determine the equation of the circle circumscribing the triangle determined by the lines x + y = 8, 2x + y = 14 and 3x + y = 22.

6 A triangle has its sides on the lines x + 2y – 5 = 0, 2x – y – 10 = 0 and 2x + y + 2 = 0. Find the equation of the circle inscribed in the triangle.

FAMILIES OF CIRCLES

Let x2+y2+D1x+E1y+F1=0 and x2+y2+D2x+E2y+F2=0 be the equation of two circles and taking “k” as the parameter, then the equation of the families of circles passing through the intersection of two circles is (x2+y2+D1x+E1y+F1) + k(x2+y2+D2x+E2y+F2) =0. Except for k=-1, it would become a linear equation (D1-D2)x + (E1-E2)y + (F1-F2) = 0, which is called a “radical axis” of the two given circles.

Example: 1. Write the equation of the family of circles C3 all

members of which pass through the intersection of the circles C1 and C2 represented by the equations C1: x2+y2-6x+2y+5=0 and C2: x2+y2-12x-2y+29=0. find the member of the family C3 that passes through the point (7, 0).

2. Graph the circles C1 and C2 whose equations are C1: x2+y2-12x-9y+50=0 and C2: x2+y2-25=0. also graph the member C3 of the family of circles for which k=1.

3. Draw the graph of the equations x2+y2-4x-6y-3=0 and x2+y2-12x-14y+65=0. Then find the equation of the radical axis and draw the axis.

REFERENCES

Analytic Geometry, 6th Edition, by Douglas F. RiddleAnalytic Geometry, 7th Edition, by Gordon Fuller/Dalton

TarwaterAnalytic Geometry, by Quirino and Mijares

PARABOLA


OBJECTIVES: At the end of the lesson, the student is expected to

be able to:

• define conic section• identify the different conic section• describe parabola• convert general form to standard form of equation of parabola and vice versa.• give the different properties of a parabola and sketch its graph

Conic Section or a Conic is a path of point that moves so that its distance from a fixed point is in constant ratio to its distance from a fixed line.

Focus is the fixed point Directrix is the fixed line Eccentricity is the constant ratio usually represented by (e)

The conic section falls into three (3) classes, which varies in form and in certain properties. These classes are distinguished by the value of the eccentricity (e).

If e = 1, a conic section which is a parabolaIf e < 1, a conic section which is an ellipseIf e > 1, a conic section which is a hyperbola

THE PARABOLA (e = 1)

A parabola is the set of all points in a plane, which are equidistant from a fixed point and a fixed line of the plane. The fixed point called the focus (F) and the fixed line the directrix (D). The point midway between the focus and the directrix is called the vertex (V). The chord drawn through the focus and perpendicular to the axis of the parabola is called the latus rectum (LR).

PARABOLA WITH VERTEX AT THE ORIGIN, V (0, 0)

(-a, y)

Let: D - Directrix F - Focus 2a - Distance from F to D LR - Latus Rectum = 4a (a, 0) - Coordinates of F

Choose any point along the parabolaSo that,

or

Squaring both side,

Equations of parabola with vertex at the origin V (0, 0)

Examples

1. Determine the focus, the length of the latus rectum and the equation of the directrix for the parabola 3y2 – 8x = 0 and sketch the graph.2. Write the equation of the parabola with vertex V at (0, 0) which satisfies the given conditions:a. axis on the y-axis and passes through (6, -3)b. F(0, 4/3) and the equation of the directrix is y + 4/3 = 0c. Directrix is x – 4 = 0d. Focus at (0, 2)e. Latus rectum is 6 units and the parabola opens to the leftf. Focus on the x-axis and passes through (4, 3)

PARABOLA WITH VERTEX AT V (h, k)

We consider a parabola whose axis is parallel to, but not on, a coordinate axis. In the figure, the vertex is at (h, k) and the focus at (h+a, k). We introduce another pair of axes by a translation to the point (h, k). Since the distance from the vertex to the focus is a, we have at once the equation

y’2 = 4ax’ Therefore the equation of a parabola with vertex at (h, k) and focus at (h+a, k) is

(y – k)2 = 4a (x – h)

Equations of parabola with vertex at V (h, k)

Standard Form General Form

(y – k)2 = 4a (x – h) y2 + Dy + Ex + F = 0

(y – k)2 = - 4a (x – h)

(x – h)2 = 4a (y – k) x2 + Dx + Ey + F = 0

(x – h)2 = - 4a (y – k)

Examples

1. Draw the graph of the parabola y2 + 8x – 6y + 25 = 02. Express x2 – 12x + 16y – 60 = 0 to standard form and construct the parabola.3. Determine the equation of the parabola in the standard form, which satisfies the given conditions.a. V (3, 2) and F (5, 2)b. V (2, 3) and axis parallel to y axis and passing through (4, 5)c. V (2, 1), Latus rectum at (-1, -5) & (-1, 7)d. V (2, -3) and directrix is y = -74. Find the equation of parabola with vertex at (-1, -2), axis is vertical and passes through (3, 6).

5. A parabola whose axis is parallel to the y-axis passes through the points (1, 1), (2, 2) and (-1, 5). Find the equation and construct the parabola.6. A parabola whose axis is parallel to the x-axis passes through (0, 4), (0, -1) and (6, 1). Find the equation and construct the parabola.7. A parkway 20 meters wide is spanned by a parabolic arc 30 meters long along the horizontal. If the parkway is centered, how high must the vertex of the arch be in order to give a minimum clearance of 5 meters over the parkway.8. A parabolic suspension bridge cable is hung between two supporting towers 120 meters apart and 35 meters above the bridge deck. The lowest point of the cable is 5 meters above the deck. Determine the lengths (h1 & h2) of the tension members 20 meters and 40 meters from the bridge center.

9. Water spouts from a horizontal pipe 12 meters above the ground and 3 meters below the line of the pipe, the water trajectory is at a horizontal distance of 5 meters. How far from the vertical line will the stream of the water hit the ground? 10. A parabolic trough 10 meters long, 4 meters wide across the top and 3 meters deep is filled with water at a depth of 2 meters. Find the volume of water in the trough.

REFERENCES



ELLIPSE


OBJECTIVES: At the end of the lesson, the student is expected to be able to:

• define ellipse• give the different properties of an ellipse with center at ( 0,0)• identify the coordinates of the different properties of an ellipse with center at ( 0, 0)• sketch the graph of an ellipse

THE ELLIPSE (e < 1)

An ellipse is the set of all points P in a plane such that the sum of the distances of P from two fixed points F’ and F of the plane is constant. The constant sum is equal to the length of the major axis (2a). Each of the fixed points is called a focus (plural foci).

The following terms are important in drawing the graph of an ellipse:

Eccentricity measure the degree of flatness of an ellipse. The eccentricity of an ellipse should be less than 1.Focal chord is any chord of the ellipse passing through the focus. Major axis is the segment cut by the ellipse on the line containing the foci a segment joining the vertices of an ellipseVertices are the endpoints of the major axis and denoted by 2a.Latus rectum or latera recta in plural form is the segment cut by the ellipse passing through the foci and perpendicular to the major axis. Each of the latus rectum can be determined by:

Properties of an Ellipse:1.The curve of an ellipse intersects the major-axis at two points called the vertices. It is usually denoted by V and V’.2.The length of the segment VV’ is equal to 2a where a is the length of the semi- major axis.3.The length of the segment BB’ is equal to 2b where b is the length of the semi-minor axis.4.The length of the segment FF’ is equal to 2c where c is the distance from the center to the foci. 5. The midpoint of the segment VV’ is called the center of an ellipse denoted by C. 6. The line segments through F1 and F2 perpendicular to the major – axis are the latera recta and each has a length of 2b2/a.7. The relationship of a, b and c is given by; a2 = b2 + c2 where, a > b.8. c = a e

ELLIPSE WITH CENTER AT ORIGIN C (0, 0)

B

B’

ELLIPSE WITH CENTER AT ORIGIN C (0, 0)d1 + d2 = 2a

Considering triangle F’PFd3 + d4 = 2a

d3 = 2a – d4

Equations of ellipse with center at the origin C (0, 0)

ELLIPSE WITH CENTER AT C (h, k)

ELLIPSE WITH CENTER AT (h, k)

If the axes of an ellipse are parallel to the coordinate axes and the center is at (h,k), we can obtain its equation by applying translation formulas. We draw a new pair of coordinate axes along the axes of the ellipse. The equation of the ellipse referred to the new axes is

The substitutions x’ = x – h and y’ = y – k yield

ELLIPSE WITH CENTER AT (h, k)

Examples:

1. Find the equation of the ellipse which satisfies the given conditions a. foci at (0, 4) and (0, -4) and a vertex at (0,6)b. center (0, 0), one vertex (0, -7), one end of minor axis (5, 0)c. foci (-5, 0), and (5, 0) length of minor axis is 8d. foci (0, -8), and (0, 8) length of major axis is 34e. vertices (-5, 0) and (5, 0), length of latus rectum is 8/5f. center (2, -2), vertex (6, -2), one end of minor axis (2, 0)g. foci (-4, 2) and (4, 2), major axis 10h. center (5, 4), major axis 16, minor axis 10 with major axis parallel to x-axis.

2. Sketch the ellipse 9x2 + 25y2 = 2253. Find the coordinates of the foci, the end of the

major and minor axes, and the ends of each latus rectum. Sketch the curve.

4. Reduce the equations to standard form. Find the coordinates of the center, the foci, and the ends of the minor and major axes. Sketch the graph.a. x2 + 4y2 – 6x –16y – 32 = 0b. 16x2 + 25y2 – 160x – 200y + 400 = 0c. 3x2 +2y2 – 24x + 12y + 60 = 0d. 4x2 + 8y2 + 4x + 24y – 13 = 05. The arch of an underpass is a semi-ellipse 6m wide and 2m high. Find the clearance at the edge of a lane if the edge is 2m from the middle.

a. b.

6 m.

2 m.

x

y

2 m.

●(2, y)

C(0, 0)

1.49m. or .m3

52y

920

y 9

20y

209y 36y916

3614

y94

14

y94

12y

32

equation working 1by

ax

2

22

22

2

2

2

2

2

2

2

2

6. The earth’s orbit is an ellipse with the sun at one focus. The length of the major axis is 186,000,000 miles and the eccentricity is 0.0167. Find the distances from the ends of the major axis to the sun. These are the greatest and least distances from the earth to the sun.

●F

sun

Least Distance

Greatest Distance

C(0, 0)

miles 100,553,94100,553,1000,000,93ca :distance greatest the for solveTo

miles 900,446,91100,553,1000,000,93ca :distance least the for solveTo

miles 100,553,10167.093,000,000aec miles 93,000,000a 0186,000,002a

7. A hall that is 10 feet wide has a ceiling that is a semi-ellipse. The ceiling is 10 feet high at the sides and 12 feet high in the center. Find its equation with the x-axis horizontal and the origin at the center of the ellipse.

x

12 ft.

10 ft.

10 ft.

y

C(0, 0)

0100y25x4 is equation the

100y25x4

10014

y25x

14

y25x

:equation working the in b and a ngSubstituti

4b 21012b

25a 5a 102a

:figure the From

equation working 1by

ax

22

22

2222

22

2

2

2

2

2

2

REFERENCES



HYPERBOLA



• give the properties of hyperbola.• write the standard and general equation of a hyperbola.• sketch the graph of hyperbola accurately.

THE HYPERBOLA (e > 1)

A hyperbola is the set of points in a plane such that the difference of the distances of each point of the set from two fixed points (foci) in the plane is constant.

The equations of hyperbolas resemble those of ellipses but the properties of these two kinds of conics differ considerably in some aspects.

To derive the equation of a hyperbola, we take the origin midway between the foci and a coordinate axis on the line through the foci.

The following terms are important in drawing the graph of a hyperbola

Transverse axis is a line segment joining the two vertices of the hyperbola.Conjugate axis is the perpendicular bisector of the transverse axis.

General Equations of a Hyperbola1. Horizontal Transverse Axis : Ax2 – Cy2 + Dx + Ey + F = 0

2. Vertical Transverse Axis: Cy2 – Ax2 + Dx + Ey + F = 0

HYPERBOLA WITH CENTER AT THE ORIGIN C(0,0)

DIRECTRIX

Then letting b2 = c2 – a2 and dividing by a2b2, we have

if foci are on the x-axis

if foci are on the y-axis

The generalized equations of hyperbolas with axes parallel to the coordinate axes and center at (h, k) are

if foci are on a axis parallel to the x-axis

if foci are on a axis parallel to the y-axis

Examples:

1. Find the equation of the hyperbola which satisfies the given conditionsa.Center (0,0), transverse axis along the x-axis, a focus at (8,0), a vertex at (4,0).b.Center (0, 0), conjugate axis on x-axis, one focus at , equation of one directrix is . c.Center (0,0), transverse axis along the x-axis, a focus at (5,0), transverse axis = 6d.Center (0,0), transverse axis along y-axis, passing through the points (5,3) and (-3,2).e.Center (1, -2), transverse axis parallel to the y-axis, transverse axis = 6 conjugate axis = 10

13,013139y

f.Center (-3,2), transverse axis parallel to the y-axis, passing through (1,7), the asymptotes are perpendicular to each other.g.Center (0,6), conjugate axis along the y-axis, asymptotes are 6x – 5y + 30 = 0 and 6x + 5y – 30 = 0.h.With transverse axis parallel to the x-axis, center at (2,-2), passing throughi.Center at (2,-5), conjugate axis parallel to the y-axis, slopes of asymptotes numerically one-sixteenth times the length of the latus rectum, and distance between foci is . j.Center (1,-1), TA // to x-axis, LR=9, DD’= .k.Center (4,-1), TA // to y-axis, FF’=10, LR=9/2.l.CA // to x-axis, C (3, 6), FF’= , DD’= .m.C (-7,-2), TA // to x-axis, eccentricity= , LR=4/3.

4 ,1032 and 0 ,232

145213138

56 5524311

2. Reduce each equation to its standard form. Find the coordinates of the center, the vertices and the foci. Draw the asymptotes and the graph of each equation.a. 9x2 –4y2 –36x + 16y – 16 = 0b. 49y2 – 4x2 + 48x – 98y - 291 = 03. Determine the equation of the hyperbola if its center is at (-4,2) if its vertex is at (-4, 7) and the slope of an asymptote is 5/2.

REFERENCES



TRANSLATION OF AXIS


Translation of Axes


• translate coordinate axes.• find new coordinates after translation.• find new equation if the origin is translated to a given point..• simplify the equations by translation of axes.• graph the transformed equation.

Definition: Translation of Axes

Consider a transformation in which the new axes are parallel to the original axes and similarly directed. Translation of axes is related to performing two geometric transformations: a horizontal shift and a vertical shift. Hence the new axes can be obtained by shifting the old axes h units horizontally and k units vertically while keeping their directions unchanged. Let x and y stand for the coordinates of any point P when referred to the old axes, and x’ and y’ the coordinates of P with respect to the new axes. Then x = x’ + h and y = y’ + k.

Examples:1.Find the new coordinates of the point P(4,-2) if the origin is moved to (-2, 3) by a translation.2.Find the new equation of the circle x2+y2-6x+4y-3=0 after a translation that moves the origin to the point (3,-2).3.Translate the axes so that no first-degree term will appear in the transformed equation.

a. x2+y2+6x-10y+12=0b. 2x2+3y2+10x-18y+26=0c. x2-6x-6y-15=0

REFERENCES

Analytic Geometry

ROTATION OF AXIS



• transform an equation by rotation of axes.• simplify the equations by rotation of axes.• identify a conic given an equation.• graph the transformed equation.

2θ

REFERENCES

Analytic Geometry

ALGEBRAIC CURVES


OBJECTIVES:

At the end of the lesson, the student is expected to be able to:

• define and describe the properties of algebraic curves• identify the intercepts of a curve• test the equation of a curve for symmetry• identify the vertical and horizontal asymptotes• sketch algebraic curves

ALGEBRAIC CURVES

An equation involving the variables x and y is satisfied by an infinite number of values of x and y, and each pair of values corresponds to a point. When plotted on the Cartesian plane, these points follow a pattern according to the given equation and form a definite geometric figure called the CURVE or LOCUS OF THE EQUATION.

The method of drawing curves by point-plotting is a tedious process and usually difficult. The general appearance of a curve may be developed by examining some of the properties of curves.

PROPERTIES OF CURVES The following are some properties of an algebraic curve:1. Extent 2. Symmetry3.Intercepts4.Asymptotes

1. EXTENT The extent of the graph of an algebraic curve involves its domain and range. The domain is the set of permissible values for x and the range is the set of permissible values for y. Regions on which the curve lies and which is bounded by broken or light vertical lines through the intersection of the curve with the x-axis. To determine whether the curve lies above and/or below the x-axis, solve for the equation of y or y2 and note the changes of the sign of the right hand member of the equation.

2. SYMMETRY Symmetry with respect to the coordinate axes exists on one side of the axis if for every point of the curve on one side of the axis, there is a corresponding image on the opposite side of the axis. Symmetry with respect to the origin exists if every point on the curve, there is a corresponding image point directly opposite to and at equal distance from the origin.

Test for Symmetry

1. Substitute –y for y, if the equation is unchanged then the curve is symmetrical with respect to the x-axis.2. Substitute –x for x, if the equation is unchanged the curve is symmetrical with respect to the y- axis.3. Substitute – x for x and –y for y, if the equation is unchanged then the curve is symmetrical with respect to the origin.

Simplified Test for Symmetry

1. If all y terms have even exponents therefore the curve is symmetrical with respect to the x-axis.2. If all x terms have even exponents therefore the curve is symmetrical with respect to the y-axis.3. If all terms have even exponents therefore the curve is symmetrical with respect to the origin.

3. INTERCEPTS

These are the points which the curve crosses the coordinate axes.a. x-intercepts – abscissa of the points at which the curve crosses the x-axis.b. y-intercepts – ordinate of the points at which the curve crosses the y-axis.

Determination of the InterceptsFor the x-intercepta. Set y = 0b. Factor the equation.c. Solve for the values of x.

For the y-intercepta.Set x = 0b.Solve for the values of y

4. Asymptotes

A straight line is said to be an asymptote of a curve if the curve approaches such a line more and more closely but never really touches it except as a limiting position at infinity. Not all curves have asymptotes.

Types of Asymptotes1.Vertical Asymptote2.Horizontal Asymptote3.Slant/Diagonal Asymptote

Steps in Curve Tracing1. If the equation is given in the form of f( x, y) = 0, solve for y (or y2) to express the equation in a form identical with the one of the four general types of the equation.2. Subject the equation to the test of symmetry.3. Determine the x and y intercepts.4. Determine the asymptotes if any. Also determine the intersection of the curve with the horizontal asymptotes.Note: The curve may intercept the horizontal asymptotes but not the vertical asymptotes.

5. Divide the plane into regions by drawing light vertical lines through the intersection on the x-axis.

Note: All vertical asymptotes must be considered as dividing lines.

6. Find the sign of y on each region using the factored form of the equation to determine whether the curve lies above and/or below the x-axis.

7. Trace the curve. Plot a few points if necessary.

REFERENCES

Analytic Geometry

POLAR CURVES



• define polar coordinates.• plot points of polar coordinates.• write and sketch the graphs of polar equations.• convert polar to rectangular and vice versa.

Examples: Plot the given points on a polar coordinate system.a.P (3, 45°)

b. P (3, –60°)

c. P (–3, 220°)

d. P (-3, –75°)

RELATIONS BETWEEN RECTANGULAR AND POLAR COORDINATES

The transformation formulas that express the relationship between rectangular coordinates and polar coordinates of a point are as follows:

and

Also, ;

;

2. Express the rectangular coordinates (-4, 3) in terms of polar coordinates.

Therefore, P(5, 143.10)

3. Find the polar coordinate equation corresponding to x2 + 2y2 = 6.

substituting: x = r cos θ and y = r sin θ, we have

6sin2cosr

6sinr2cosr

6sinr2cosr

6y2x

222

2222

22

22

4. Transform the equation r = 8 cos θ to rectangular coordinate equation.

0yx8x

x8yx

yx

x8yx

r

x8 8cosr

have we,r

x cos and yxr:ngSubstituti

22

22

22

22

22

cos-12

r d. 4 cos r c.

4 b. 4r a.

equation. coordinate r rectangula to equations following the Transform .7

4y2x d. y4x c. 0y3x b. 4y a.

equation. coordinate polar to equations following the Transform .605y2x 05x2y

5rx

ry

2r

5cos2sinr cossin2

5r

:have we ,ry

sin and rx

cos : ngSubstituti

equation. coordinate r rectangula to cossin2

5r Transform .5

222

SpiralcLogarithmi e r

SpiralHyperbolic or Reciprocal a r

Archimedes of l Spira a r

Spirals4.

2 sina r or 2 cos a r

Lemniscate 3.

loop. inner an has graph the ,a b if origin; the

ng surroundicurve a is graph the ,a b if cardioid; a called is

limacon the b, a if ,cos a b r or sina b r

Limacon .2

a

2222

Cardioid Limacon

Eight – leaf Rose Three – leaf Rose

Spiral of Archimedes Logarithmic Spiral

Lemniscate

θ 00 300 600 900 1200 1500 1800 2100 2400 2700 3000 3300 3600

cosθ 1 0.87 0.5 0 -0.5 -0.87 -1 -0.87 -0.5 0 0.5 0.87 1

2cosθ 2 1.73 1 0 -1 -1.73 -2 -1.73 -1 0 1 1.73 2

r 3 2.73 2 1 0 -0.73 -1 -0.73 0 1 2 2.73 3

Example 1: Trace r = 1 + 2 cosθ

θ 00 300 600 900 1200 1500 1800 2100 2400 2700 3000 3300 3600

3θ 00 900 1800 2700 3600 4500 5400 6300 7200 8100 9000 9900 10800

cos3θ 1 0 -1 0 1 0 -1 0 1 0 -1 0 1

Example 2: Construct the graph r = cos3θ.

θ 00 300 600 900 1200 1500 1800 2100 2400 2700 3000 3300 3600

2θ 00 600 1200 1800 2400 3000 3600 4200 4800 5400 6000 6600 7200

cos2θ 1 0.5 -0.5 -1 -0.5 0.5 1 0.5 -0.5 -1 -0.5 0.5 1

4cos2θ 4 2 -2 -4 -2 2 4 2 -2 -4 -2 2 4

r 2 1.4 i i i 1.4 2 1.4 i i i 1.4 2

Example 3: Construct the graph r2 = 4 cos 2θ.

REFERENCES

Analytic Geometry

PARAMETRIC EQUATIONS



• define parametric equation.• sketch the graphs of parametric equations by point plotting.• find the equation by eliminating the parameter of the parametric equation and sketch the graph.

DEFINITION: PARAMETRIC EQUATIONS If there are functions f and g with a common domain T, the equations x = f(t) and y = g(t), for t in T, are parametric equations of the curve consisting of all points ( f(t), g(t) ), for t in T. The variable t is the parameter. The equations x = t + 2 and y = 3t – 1 for example are parametric equations and t is the parameter. The equations define a graph. If t is assigned a value, corresponding values are determined for x and y. The pair of values for x and y constitute the coordinates of a point of the graph. The complete graph consists of the set of all points determined in this way

as t varies through all its chosen values. We can eliminate t between the equations and obtain an equation involving x and y. Thus, solving either equation for t and substituting in the other, we get 3x – y = 7 The graph of this equation, which also the graph of the parametric equations, is a straight line.

Example 1: Sketch the graph of the parametric equations x = 2 + t and y = 3 – t2 .

t -3 -2 -1 0 1 2 3

x -1 0 1 2 3 4 5

y -6 -1 2 3 2 -1 -6

y

x

●●

● ●●

●●

t=-1

t=2t=-2

t=-3 t=3

t=0

t=1

Example 2: Eliminate the parameter between x = t + 1 and y = t2 + 3t + 2 and sketch the graph.Solution:Solving x = t + 1 for t, we have t = x – 1. Substitute into y = t2 + 3t + 2, then

y = (x – 1)2 + 3(x – 1) + 2 y = x2 – 2x + 1 + 3x – 3 + 2 y = x2 + x

Reducing to the standard form, y + ¼ = x2 + x + ¼ y + ¼ = (x + ½)2 , a parabola with V(-½,-¼)

opening upward

y

x-1-2 210

-2

-1

2

1

Example 3: Eliminate the parameter between x = sin t and y = cos t and sketch the graph.Solution:Squaring both sides of the parametric equations, we have

x2 = sin2 t and y2 = cos2 tAnd adding the two equations will give us x2 + y2 = sin2 t + cos2 t But

sin2 t + cos2 t = 1 Therefore

x2 + y2 = 1 , a circle with C(0, 0) and r = 1

y

x-1-2 210

-2

-1

2

1

Example 4: Find the parametric representation for the line through (1, 5) and (-2, 3).

Solution:Letting (1, 5) and (-2, 3) be the first and second points,

respectively, of x = x1 + r(x2 – x1)

andy = y1 + r(y2 – y1)

We then havex = 1 + r(-2 – 1) and y = 5 + r(3 – 5)x = 1 – 3r y = 5 – 2r

Example 5: Eliminate the parameter between x = sin t + cos t and y = sin t.

Solution:Solving sin2 t + cos2 t = 1 for cos t, we have

Substitute into x = sin t + cos t , then x = sin t +

But y = sin t and y2 = sin2 tTherefore x = y +

x – y = Squaring both sides

(x – y)2 = 1 – y2

tsin1tcos 2

tsin1 2

2y1 2y1

Exercises: Eliminate the parameter and sketch the curve.1. x = t2 + 1, y = t + 1 2. x = t2 + t – 2 , y = t + 2 3. x = cos θ , y = cos2 θ + 8 cos θ4. x = 4 cos θ , y = 7 sin θ5. x = cos θ , y = sin 2θ6. x = 1 + cos 2θ , y = 1 – sin θ

REFERENCES

Analytic Geometry

SPACE COORDINATES AND SURFACES


OBJECTIVES: At the end of the lesson, the student is expected to be able to:• define space coordinates.• plot points of space coordinates.• write and sketch the graphs of space coordinate equations.• know the different kinds planes and surfaces.

Let OX, OY, and OZ be three mutually perpendicular lines. These lines constitute the x-axis, the y-axis, and the z-axis of a three-dimensional rectangular coordinate system. The axes, in pairs, determine three mutually perpendicular planes called coordinate planes. The planes are designated as the XOY-plane, the XOZ-plane, and the YOZ-plane or, more simply, the xy-plane, the xz-plane, and the yz-plane. The coordinate planes divide space into eight regions called octants. The distance of P from the yz-plane is called the x-coordinate, the distance from the xz-plane the y-coordinate, and the distance from the xy-plane the z-coordinate. The coordinates of a point are written in the form (x, y, z), in this order, x first, y second, and z third.

z

x y

o

xy-plane

yz-planexz-plane

Example:Plot the given points in a three-dimensional

coordinate system.

1. (3, 0, 0)2. (0, 3, 0)3. (0, 0, 3)4. (1.5,-1, 2)5. (0, 2, -2)6. (2, 2.5, 3)

(0, 0, 3)

(0, 3, 0)

(3, 0, 0)

(2, 2.5, 3)(1.5,-1, 2)

(0, 2, -2)

z

x

yo

-z

-x

-y

THEOREMS: Let P1(x1, y1, z1) and P2(x2, y2, z2) be the coordinates of two points in a three-dimensional coordinate system. Then the distance d between P1 and P2 is given by

The coordinates P(x, y, z) of the midpoint of the line segment joining P1(x1, y1, z1) and P2(x2, y2, z2) are given by the equations

This theorem may be generalized by letting P(x, y, z) be any division point of the line through P1 and P2. If the ratio of P1P to P1P2 is a number r, then

2

12

2

12

2

12 zzyyxxd

212121

2zz

z 2

yyy

2xx

x

121121121 zzrzz yyryy xxrxx

EXAMPLES: 1. Find the distance between the points P1(-4, 4, 1) and P2(-3, 5,-4).

2. Find the coordinates of the midpoint of the line segment that joins A(3,-2, 4) and B(-6, 5, 8). 3. Find the coordinates of the point P(x, y, z), which is one-third of the way from A(1, 3, 5) to B(5, 7, 9). 4. Given: A(1, 4, 7) and B(5,-1, 11), find the point P so that the ratio of AP to PB is equal to 4 to 7.

SURFACESA.PLANE: Ax + By + Cz + D = 0

a) x = k, plane parallel to yz-planeb) y = k, plane parallel to xz-planec) z = k, plane parallel to xy-planed) Ax + By + D = 0, plane parallel to z-axise) By + Cz + D = 0, plane parallel to x-axisf) Ax + Cz + D = 0, plane parallel to y-axisg) Ax + By + Cz = 0, plane

EXAMPLE: 1. x = 3 2. y = 3

y

x

z

y

x

z

3. z = 3

y

x

z

y

x

z

(0, 4, 0)

(6, 0, 0)

4. 2x + 3y = 12

5. 2x + 3z = 12

y

x

z

y

x

z

6. 2y + 3z = 12

(0, 0, 4)

(6, 0, 0)

(0, 6, 0)

(0, 0, 4)

6. 2x + 3y + 4z = 12

y

x

z

(0, 4, 0)

(0, 0, 3)

(6, 0, 0)

B. CYLINDERS and SPHERE:1. x2 + y2 = 4

y

x

z

(-2, 0, 0)

(0, 2, 0)

(2, 0, 0)

(0,-2, 0)

CIRCULAR CYLINDER

2. 4x2 + y2 = 4

y

x

z

(-1, 0, 0)

(0, 2, 0)

(1, 0, 0)

(0,-2, 0)

ELLIPTICAL CYLINDER

3. x2 = y

y

x

z

4. y2 = x

y

x

z

5. z2 = y

y

x

z

5. z2 = y – 1

y

x

z

V(0, 1, 0)

SPHERE: (x – h)2 + (y – k)2 + (z – l)2 = r2 Ax2 + Ay2 + Az2 + Gx + Hy + Iz = J

r = 0 (point)r = - (no locus)r = + (sphere)

y

x

z

EXAMPLE: Describe the locus of x2 + y2 + z2 + 2x – 4y – 8z + 5 = 0.

Sketch the graph.SOLUTION: x2 + 2x + 1 + y2 – 4y + 4 + z2 – 8z + 16 = –5 + 1 + 4 + 16

(x + 1)2 + (y – 2)2 + (z – 4)2 = 16C(–1, 2, 4) and r = 4

y

x

z

y’

x’

z’

(-1, 6, 4)

(-5, 2, 4)

(-1, 2, 0)(3, 2, 4)

(-1,- 2, 4)

(-1, 2, 8)

C(-1, 2, 4)

QUADRIC SURFACESSecond-Degree Equation in x, y, zForm: Ax2+By2+Cz2+Dxy+Exz+Fyz+Gx+Hy+Iz+J=0 The graph of such equation is called quadric surface or simply quadric.Six Common Types of Quadric Surfaces: 1.Ellipsoid2.Hyperboloid of One Sheet3.Hyperboloid of Two Sheets4.Elliptic Paraboloid5.Hyperbolic Paraboloid6.Elliptic Cone

QUADRIC SURFACES

ELLIPSOID

1cz

by

ax

2

2

2

2

2

2

x

y

QUADRIC SURFACES

HYPERBOLOID OF ONE SHEET

1cz

by

ax

2

2

2

2

2

2

QUADRIC SURFACES

HYPERBOLOID OF TWO SHEETS

1by

ax

cz

2

2

2

2

2

2

QUADRIC SURFACES

ELLIPTIC PARABOLOID

2

2

2

2

by

ax

z

QUADRIC SURFACES

HYPERBOLIC PARABOLOID

2

2

2

2

ax

by

z

x’

y’

z’

QUADRIC SURFACES

ELLIPTIC CONE

2

2

2

22

by

ax

z

EXAMPLES: Sketch the quadric surface.1. 36x2+9y2+4z2=36

Solution:

19z

4y

1x

361

36z4y9x36

222

222

x y z

x ±1 0 0

y 0 ±2 0

z 0 0 ±3

:Intercepts .I

(ellipse)

14y

1x

:0z let

plane-xy i):Traces .II

22

(ellipse)

19z

4y

:0x let

plane-yz iii)22

(ellipse)

19z

1x

:0y let

plane-xz ii)22

y

x

z

(-1,0,0)

(1,0,0)

(0,-2,0) (0,2,0)

(0,0,-3)

(0,0,3)

14y

1x 22

19z

4y 22

19z

1x 22

2. 16x2+36y2-9z2=144:Solution

116z

4y

9x

1441

144z9y36x16

222

222

:Intercepts .I

x y z

x ±3 0 0

y 0 ±2 0

z 0 0 ±4i

(ellipse)

14y

9x

:0z let


22

)(hyperbola

116z

4y

:0x let

plane-yz ii)22

)(hyperbola

116z

9x

:0y let

plane-xz iii)22

(ellipse)

18y

18x

21

24y

9x

114y

9x

116

)4(4y

9x

4z let :plane-xy to parallel Sections III.

2222

22222

y

x

z

y”

y’

x”

x’

(-4.2,0,-4)

(3,0,0)

(0,2,0)

(0,-2,0)

(-3,0,0)

(0,2.8,-4)

(4.2,0,4)

(0,2.8,4)(0,-2.8,4)

(-4.2,0,4)

(0,-2.8,-4)

(4.2,0,-4)

z=4

z=-4

18y

18x 22

14y

9x 22

116z

9x 22

116z

4y 22

3. 4z2-4x2-y2=4:Solution

14y

1x

1z

41

4yx4z4

222

222

:Intercepts .I

x y z

x ±i 0 0

y 0 ±2i 0

z 0 0 ±1

trace) (no

14y

1x

:0z let


22

)(hyperbola

14y

1z

:0x let

plane-yz ii)22

)(hyperbola

11x

1z

:0y let

plane-xz iii)22

(ellipse)

132y

8x

81

84y

x

14y

1x

9 14y

1x

13)(

3z let :plane-xy to parallel Sections III.

2222

22222

y

x

z

y”

y’

x”

x’

(-2.8,0,-3)

(0,0,-1)

(0,0,1)

(0,-5.7,-3)

(0,5.7,3)

(0,5.7,-3)

(2.8,0,3)

(-2.8,0,3)

(0,-5.7,3)

(2.8,0,-3)

z=3

z=-3

132y

8x 22

14y

1z 22

1xz 22

REFERENCES

Analytic Geometry

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