Analytic Geometry - iiw.kuleuven.be · Ananlytic Geometry 7 Notice that ( , ) (0,0)a b = is...

Analytic Geometry

Team Mathematics

Dr. Caroline Danneels

1 Vectors and lines ................................................................................................................... 3

1.1 Vectors ........................................................................................................................................ 3

1.2 Lines ............................................................................................................................................ 5

1.3 Parallel lines ................................................................................................................................ 8

1.4 The euclidian vectorspace ........................................................................................................... 9

1.5 Angles ....................................................................................................................................... 11

1.6 Orthogonality of 2 lines ............................................................................................................ 12

1.7 Distance from a point to a line (O.N.B.) ................................................................................... 13 1.8 Exercises ................................................................................................................................... 14

2 Conic sections ...................................................................................................................... 19

2.1 Introduction ............................................................................................................................... 19

2.2 The circle .................................................................................................................................. 20

2.3 The parabola.............................................................................................................................. 21

2.4 Exercises ................................................................................................................................... 23

_____________________________________________________________________________ Ananlytic Geometry 3

1 Vectors and lines

1.1 Vectors

1.1.1 The concept of a vector

π is the set of all points of the plane. Take in the plane π a fixed origin O. A translation t in the plane is now completely determined by the image point P of O by this translation: t(O) = P. The translation is determined by point P or also by any couple of points (A,B) so that t(A)=B.

With a translation t, we associate a free vector. The vector AB��

is the set of couples of points

(A,B) such that t(A)=B. AB CD=��

if there is a translation t such that t(A)=B and t(C)=D.

A free vector can now be represented by means of an arrow with origin in O and with terminus

point t(O)=P. Now each point P in the plane can be seen as the terminus point of P OP=��

. P��

is called a bound vector.

The vector O��

is called the zero vector.

The plane in which we consider each point a terminus point of bound vectors is called πo.

1.1.2 Basis and coordinate system

Let’s choose in πo xE��

en yE��

which together with O��

form a triangle.

For each P��

in πo there is only one 2( , )x y ∈ℝ such that x yP xE yE= +��

.

{ },x yE E��

is called a basis of πo.

(x,y) are called the cartesian coordinates of P��

with respect to the basis { },x yE E��

. We emphasize

the notion “coordinate” is related to a basis.

OEx is the x-axis; OEy is the y-axis; together they represent the coordinate axes of the xy-coordinate system.

If (a1,a2) are the coordinates of A��

we also write this vector as A(a1,a2) or 1 2( , )A OA a a= =��

. This

means that we will use the word vector (with starting point the origin) and the word point in the xy-plane for the same concept.


1.1.3 Operations with vectors

In πo we define the following operations:

1. The addition of 2 vectors A��

and B��

is performed by adding the 2 associated vectors OA��

and OB��

using the parallelogram rule. The result C��

is determined by its terminus point.

From the figure one can easily see that, if ( )1 2,A a a=��

and ( )1 2,B b b=��

, then the sum

vector is: ( )1 2 1 2,C A B a a b b= + = + +��

.

A

B

C

x

y

1a

2a

1b

2b

O

2. The multiplication of a vector A��

of πo and a scalar k ∈R is a multiple of A��

: k A��

.

Furthermore, if k > 0, k A��

has the same sense as A��

, and if k < 0, k A��

has the opposite sense.

If ( )1 2,A a a=��

, then ( )1 2,k A ka ka=��

.

1.1.4 Midpoint of a line segment

The midpoint of a line segment [AB] with ( )1 2,A a a=��

and ( )1 2,B b b=��

is determined by

1 1 2 2,2 2 2

a b a bA BM

+ ++ = =

��


1.2 Lines

1.2.1 Vector equation

Take a line e through the origin. Every other point S on that line is the terminus point of a

vector S��

that determines the direction of e in a unique way. Such a vector is called a direction

vector of e. If S��

is a direction vector of e, then every kS��

with 0k ∈R also is.

O

S

e

Take a line 1e parallel to e and two different points P��

and 0P��

on 1e .

0P�

P�

S�

0P P−� �

1e

e

If S�

is a direction vector of e , then 0P P kS− =��

or 0P P kS= +��

for a certain k ∈R .

Otherwise, for every k , the terminus point of the vector 0P kS+��

is on the line 1e .

The vector equation 0P P kS= +��

is therefore a necessary and sufficient condition for a point to

lie on the line 1e (given by a point 0P and the direction S�

).

0P P kS= +��

with k ∈R

is called the vector equation of the line.


1.2.2 The parameter equations

Replacing the vectors by their coordinates in 2R with ( , )P x y=

�, 0 0 0( , )P x y=�

and ( , )S a b=�

gives

0 0( , ) ( , ) ( , )x y x y k a b= + with k ∈R

Furthermore

0 0( , ) ( , )x y x ka y kb= + + with k ∈R

which is equivalent with

0

0

x x ka

y y kb

= + = +

with k ∈R

These equations are called parameter equations of a line through a given point 0 0( , )x y with

direction numbers ( , )a b , the coordinates of the direction S�

. It is important to see that the above

equations are not unique since both 0P�

and S�

may be chosen.

1.2.3 Cartesian equation(s)

So we have two ways to give the equation of a line: the vector equation and the parameter equations. A third possibility, although less general, is the Cartesian equation(s). They may be obtained starting from the parameter equations. From the parameter equations, k may be eliminated:

0 0x x y y

a b

− −= if 0ab ≠

or

0 0( )b

y y x xa

− = − if 0≠a

Both equations are called the Cartesian equation of the line. The ratio b

a (in front of x) is called

the slope m.

If 0a = , the parameter equations are

0

0

x x

y y kb

= = +

with k ∈R

The second of these equations simply says that y may be any real number. Since there is no limitation on y, this equation is superfluous. The other equation then becomes the Cartesian equation:

0x x=

This line is parallel to the y-axis. The simplest direction numbers are (0,1). The slope m is not defined, since 0a = .

In the same way, if 0b = , the cartesian equation becomes 0y y= . This line is parallel to the

x-axis. The simplest direction numbers are (1,0). Its slope is m = 0.


Notice that ( , ) (0,0)a b = is impossible. The zero vector is excluded as direction vector of a line.

If the line is given by two points 0 0 0( , )P x y en 1 1 1( , )P x y then a direction vector of the line is

given by 1 0−� �P P , so we obtain:

0 1 0( )P P k P P= + −� � � �

with k ∈R

as vector equations,

0 1 0

0 1 0

( )

( )

x x k x x

y y k y y

= + − = + −

met k ∈R

as parameter equations,

0 0

1 0 1 0

x x y y

x x y y

− −=− −

as cartesian equation if 01 xx ≠ and .

If 01 xx = the equation is similar to a = 0; if 01 yy = the equation is similar to b= 0.

So, in general, the Cartesian equation of a line in 2ℝ takes the form

0Ax By C+ + =

with A and B not both zero at the same time.

Otherwise it can be proven that each equation of this form in 2ℝ respresents a line.

What is the slope of this line?

Find a set of direction numbers.

01 yy ≠


1.3 Parallel lines

If is given :

line e1 with direction vector 1S , direction numbers (a1,b1), slope m1, equation A1x+B1y+C1=0,

line e2 with direction vector 2S , direction numbers (a2,b2), slope m2, equation A2x+B2y+C2=0

e e 1

2 O S S

1 2

then

e1 // e2

12 SkS =⇔ with 0∈k R

==

⇔12

12

kbb

kaa with 0∈k R

12 mm =⇔

2 1

2 1

A kA

B kB

=⇔ =

with 0∈k R

Use:

Given line e with equation Ax+By+C=0 and point P0(x0,y0)

Determine the equation of line f // e and through P0

⇒ f has equation: A(x- x0)+B(y- y0)=0


1.4 The euclidian vectorspace

1.4.1 Definitions

If 1 2( , )A a a��

and 1 2( , )B b b��

in are given, then the scalar product of these two vectors is

1 1 2 2A B a b a b⋅ = + ∈��

ℝ

Then these vectors are perpendicular if

A B A . B 0⊥ ⇔ =��

Then the norm of the vector is

2 21 2A A A a a 0= ⋅ = + >

��

Then A��

is a normalized vector if

1=A

Then the distance between 1 2( , )A a a and ( )1 2,B b b is

( ) ( )222

211),( babaBABAd −+−=−=

A

B

A - B

O

π 0


1.4.2 Normalizing a vector

If OA ≠��

then

a

AE

A=��

��

has length 1 and thus is a unit vector in the same sense as A��

.

This process is called normalizing a vector.

1.4.3 Orthonormal basis (O.N.B.)

{ },x yE E��

is an orthonormal basis

==

⊥⇔

1yx

yx

EE

EE


1.5 Angles

First we want to mention that angles are positive angles if the terminal ray rotates counterclockwise around the vertex from the initial ray.

1.5.1 The angle of a line with the x-axis

(analytical expressions hold in an orthonormal basis)

In the following figure, we see in the right triangle that

mm ==1

tanα

In other words, the slope of the line is equal to the tangent of the angle between the x-axis and the line.

α

E y

E x

y

x

e

0

1

m

1.5.2 The angle between two lines

If line e1 has direction vector 1S��

, direction numbers (a1,b1) and slope m1

and line e2 has direction vector 2S��

, direction numbers (a2,b2) and slope m2

then the angle α between these lines is equal to the acute angle determined by the direction vectors of the lines:

1 2 1 2 1 2

2 2 2 21 2 1 2 1 2

cosS S a a b b

S S a a b bα

⋅ += =

+ ⋅ +

��

��


1.6 Orthogonality of 2 lines

If line e1 has direction vector 1S , direction numbers (a1,b1), slope m1, equation

1 1 1 0A x B y C+ + =

and line e2 has direction vector2S , direction numbers (a2,b2), slope m2 , equation 2 2 2 0A x B y C+ + =

then (the analytical expression with respect to an orthonormal basis is):

e

e

1

2 O

S

S

1

2

e1 ⊥ e2

2 1S S⇔ ⊥��

02121 =+⇔ bbaa

12

1

mm −=⇔

02121 =+⇔ BBAA

Use:

Find the equation of line f ⊥ e with equation 0Ax By C+ + = and which passes through point

( )0 0 0,P x y .

⇒ f has equation: ( ) ( )0 0 0B x x A y y− − − =


1.7 Distance from a point to a line (O.N.B.)

Consider in a point 0 0 0( , )P x y and a line : 0e Ax By C+ + = .

To find the (orthogonal) distance from 0P��

to the line e , one may proceed in the following way:

• Construct the line r through 0P��

perpendicular to e .

• Find the intersection S��

of r and e .

• The distance between 0P��

and e is then 0( , )d P S��

.

So:

r may be given by means of:

0

0

x x kAk

y y kB

= +∈ = +ℝ

To find the intersection S��

of r and e, the parameter equations of r is substituted in the equation of : 0e Ax By C+ + = . Then this equation is solved for k:

( ) ( )0 0 0A x kA B y kB C+ + + + =

gives the value of the parameter kS which corresponds to the intersection S��

:

0 02 2S

Ax By Ck

A B

+ += −+

So the distance is:

( ) ( )( ) ( )( )2 20 0 0 0 0 0, S Sd P S P S x x k A y y k B= − = − + + − +��

( ) 2 20, Sd P S k A B= +��

and finally

0 00 0 2 2

( , ) ( , )Ax By C

d P e d P SA B

+ += =

+

��

π 0


1.8 Exercises

1. Given are A(5,-1), B(-1,5), C(-7,2). a. Determine the coordinates of the midpoints of the sides of the triangle ABC.

( ) 7 12,2 , 4, , 1,

2 2 − −

b. Determine the coordinates of the centroid (or geometrical center or center of area) of the triangle ABC.

( )1,2−

2. Determine the direction numbers and the slope of the lines determined as follows: a. through (-2, 7) and (1, -8) (3, -15) and m = -5 b. line x (k,0) and m = 0 c. line y (0,k) and m does not exist! d. the line with equation: 2 4 0x y− + = (1, 2) and m = 2

e. the line with equation: 3

4y x= (4, 3) and

3

4m =

f. the line with equation: 2 3y x= + (1, 2) and m = 2

3. Make a drawing of the lines with the following equations: a. 4y x= b. 2 3 0x y+ = c. 4 2 5 0x y+ + =

4. Determine the angle which the following lines make with the x-axis in an orthonormal coordinate system. a. 5 0y x− + = α = 45°;α = 135°

b. 3 5 0y x− − = α = 60°;α = 120°

5. Proof that the figure formed by A(-2,1), B(-1,4), C(5,6), D(4,3) is a parallelogram.

6. Given line e with equation 2 4x y+ = a. does A(4,1) belong to e ? No b. does B(4,0) belong to e ? Yes c. determine the abscissa of the point on e with ordinate -5 x = 14

d. determine the ordinate of the point C with abscissa 1 3

2y =

e. draw line e f. find the intersection points with the x-axis and the y-axis (4, 0) and (0, 2)


7. Given line e with equation: 3 2 0ax y+ + = with a∈ℝ . Determine, if possible, a such that:

a. e passes through (2, 0) a = -1 b. e passes through (0,0) impossible c. e is parallel to line f with equation 3 5 0x y− − = a = -9 d. e is parallel to the x-axis a = 0 e. e is parallel to the y-axis impossible

f. the x-intercept of e is +3 2

3a = −

g. the y-intercept of e is +5 impossible

h. the y-intercept of e is a∈ℝ

8. Find the equation of the line : a. whose slope is m = -2 and passes through the point (3, 4) 2 10 0y x+ − = b. which passes through the points (2, 3) and (5, 1) 3 2 13 0y x+ − = c. which passes through the origin and the point (2, 6) 3 0y x− = d. which passes through the points (3, 5) and (7, 5) 5y = e. which passes through the points (-2, 4) and (-2, 1) 2x = − f. which passes through the points (1, 3) and (-2, -6) 3 0y x− = g. with slope m = 2 and whose y-intercept is +4 2 4 0y x− − = h. whose y-intercept is +3 and x-intercept is +2 3 2 6 0x y+ − = i. whose y-intercept is -6 and which passes through the point (2, 4) 5 6 0y x− + = j. which passes through the point (-2, 6 ) and is parallel to the line e: 3 2 5 0x y+ − =

2 3 6 0y x+ − =

k. which passes through the point (0, 3) and is parallel to the line through the points (2,0) and (5, 2) 3 2 9 0y x− − =

l. which passes through the point (2, 3) and is parallel to the y-axis x = 2 m. which passes through the point (-2, 4) and is parallel to the x-axis y = 4

9. Given line e with equation : ( 2) ( )y a x a b= − + + with ,a b ∈ℝ . Determine a and b such that: a. e passes through the points (1, 3) and (3, 5) a = 3 and b = -1 b. e passes through the point (2, -3) and is parallel to line f: 5 0x y+ + = a = 1 and b = -2 c. the x-intercept of e is -2 and e has (2, -10) as direction numbers a = -3 and b = -7 d. e passes through (0, 0) en (-2, -4) a = 4 and b = -4

10. Determine the equation of line e (in an O.N.B.): a. which passes through the point (4, -3) and forms an angle of 45° with the x-axis

7 0; 1 0y x y x− + = + − = b. whose y-intercept is +3 and forms an angle of 30° with the positive x-axis

3 3 9 0;3 3 9 0y x y x− − = + − =

−2

3


11. Determine the equation of line e which passes through the point (2, 4): a. whose positive x-intercept is double the positive y-intercept 2 10 0x y+ − = b. whose negative x-intercept and positive y-intercept are equal 2 0x y− + =

12. Determine a such that e // f:

: ( 1) ( 2) 1 0

: ( 1) (2 4 ) 0

e a x a y

f a x a y a

− − + + =+ + − + =

a = 0 or a = 3

13. Examine if the following points are collinear:

a. (8, 3), (-6, -3), (15, 6) Yes b. (1, 1), (4, -1), (-5, 5) Yes

14. Determine a such that the points (2, 3), (a, 2) and (a + 2, a - 3) are collinear. Determine the equation of the line passing through these points.

3 and 5 0

4and 2 8 0

a y x

a y x

= + − == + − =

15. Determine the points of intersection of the following lines:

a. : 2 4

:3 7

e x y

f x y

+ = + =

(2, 1)

b. : 5 3 1 0

: 2 8 0

e x y

f x

+ − = + =

(-4, 7)

c. : 6 3 4

3: 2 3 0

2

e y x

f x y

− = − + =

coinciding lines

16. Give the general equation of all lines passing through (2, 1). 1 ( 2)y m x− = −

17. Give the general equation of all lines with slope 2. 2y x q= +

18. Proof that the 3 medians of a triangle are concurrent. 19. For A(2,0), B(1,1), C(1,2) given in an O.N.B., determine A B⋅��

2

A C⋅��

2

C B⋅��

3 A A⋅��

4 B B⋅��

2 C C⋅��

5

20. Proof that the points A(4,6), B(2,-4), C(-2,2) form a right triangle.


21. Determine the equation of the line perpendicular to

a. : 5 6 0e x y+ − = and passing through (0, 0) 15 3

5 0 and ,13 13

x y − =

b. : 2 5 0f x y− + = and passing through (2, -6) 2 10 0 and ( 4, 3)x y+ + = − − Determine also the intersection points.

22. Determine the equation of the line perpendicular to the line with equation 2 5 4 0x y+ + = in the intersection point with the x-axis. 5 2 10 0x y− + =

23. A line has slope 2. Determine the equation of the line perpendicular to that line and passing through the point (4,1). 2 6 0y x+ − =

24. Determine the equation of the perpendicular bisector of the line segment [AB] with A(3,-1)and B(1,3). 2 0y x− =

25. The equations of the sides of a triangle are: 3 4 14 0;4 13 0; 5 8 0x y x y x y− + = + − = + − = . Determine the equations of the altitudes and the coordinates of the orthocenter.

5 5 0

3 4 15 0

4 10 0

30 55,

19 19

y x

y x

y x

− + =+ − =− − =

26. Determine the length of the sides of the triangle formed by A(5, -3), B(-1, 5), C(-7, 2). 10;13;3 5

27. Determine the distance from the origin to A(-2, 4) and to the line : 3 5e x y− =

102 5;

2

28. Determine the distances from A(-3, 4), B(2, 1), C(-3, 2) to the line : 5 0e x y− + =

2;3 2;0

29. Determine the distances from A(5, 1) to B(1, -2), C(-2, 2) and to the line BC.

5;5 2;5


30. Determine the distance from A(-4, 4) to the line : 2 3 0e x y+ − = using a. the formula b. the construction method

7 5

5

31. Determine the distance from A(3, -5) to the line perpendicular to 022: =+− yxe and

through B(1, 1). 2 5

32. Determine the distance from the origin to the perpendicular bisector of the line segment [AB] with A(-9, 7), B(15, -3).

2

33. Determine the distance between the parallel lines with equations

2 3 0 and 2 2 0x y x y+ + = + − = . 5 34. If : 2 1 0; : 2 3 0e x y f x y+ + = + − = , determine p such that the distances from A(1,p) to the

lines e and f are equal.

1 and 5

3p p= − =

35. Determine the equation of the line through A(2, 3) and at a distance 3 from the origin.

3 and 5 12 39 0y y x= + − =

36. Determine the equation of the line through A(-5, 1) such that the distance from B(3, -1) and C(-3, 2) to that line is equal.

2 3 0 and 10 5 0y x y x+ + = + − =

37. Determine the equation of the line parallel to line : 5 12 11 0e x y+ − = and at a distance 1 from A(-2, 1).

5 12 15 0

5 12 11 0

x y

x y

+ − =+ + =


2 Conic sections

2.1 Introduction

Parabolas, ellipses and hyperbolas are conic sections. A conic section is a curve which is obtained by the intersection of a cone with a plane. Circles are also conic sections, since they are special cases of ellipses. The type of a conic section depends on the angle between the intersecting plane and the cone.

Figure 1 snijding van een kegel door een vlak

Left: to obtain an ellipse, the angle between the plane and the axis is larger than the angle between the axis of the cone and a line on the cone.

Middle: to obtain a parabola, the angle between the plane and the axis of the cone is equal to the angle between the axis and a line on the cone.

Right: to obtain an hyperbola the angle between the plane and the axis of the cone is smaller than the angle between the plane and a line on the cone.

In the following paragraphs we give the Cartesian equations in standard form of a circle and a parabola in an orthonormal coordinate system. The equations represent the condition which points P(x,y) must satisfy to be on the graph of the conic section.


2.2 The circle

Although the circle is only a special case of an ellipse, yet we will first pay attention to its definition and its equation.

Definition:

A circle C is the set of points P(x,y) which are on a constant distance R from a fixed point M(x0,y0). This fixed distance R is called the radius, the fixed point M(x0,y0) the center of the circle.

2 2 20 0

( , )

( ) ( )

P d P M R

x x y y R

∈ ⇔ =⇔ − + − =

C

If the center M is in the origin, the equation of the cirlce is 2 2 2x y R+ =

The general equation of the circle is:

022 =++++ CByAxyx

Be aware that not all equations of this form represent a circle (see exercise 3).


2.3 The parabola

Definition:

A parabola P is the set of points P(x,y) for which the distance to a fixed line d is equal to the distance to a fixed point F which is not on d. The point F is called the focal point (focus) of the parabola, the line d the directrix.

( , ) ( , )Q d Q d d Q F∈ ⇔ =P

The equation of the parabola P with focal point F(p,0) and directrix :d x p= − is:

2 4y p x=

This is easily found from the definition:

Q

F

d y

x T

Figure 2 parabool

2 2

2 2 2

2

( , ) ( , )

( ) ( 0)

( ) ( 0) ( )

4

Q d Q d d Q F

x p y x p

x p y x p

y p x

∈ ⇔ =

⇔ − + − = +

⇔ − + − = +⇔ =

P

So the equation of the parabola is 2 4y p x= .

The x-axis is the symmetry axis of the parabola. The point F is the focal point, the line d is the directrix and the point halfway between the focal point and the directrix is the top of the parabola.

It is important to see that the above equation is only valid for this specific choice of the coordinate system and of the position of the focal point and the directrix. If the parameter p is negative, the focal point is at the left of the top and the directrix at the right side. The parabola will open to the negative x-axis. The equation is the same.

If we select a horizontal directrix, the parabola will be oriented vertically. Again, depending on the sign of p, the parabola is open to the positive y-axis if p > 0 and is open to the negative y-axis if p < 0.


So there are four possibilities for a parabola with its top in the origin and one of the coordinate axes as symmetry axis.

2 4y p x= with 0p > 2 4y p x= with 0p <

2 4x p y= with 0p > 2 4x p y= with 0p <

If the parabola is shifted to an arbitrary position of its top T(x0,y0) one finds two possible equations:

Horizontal symmetry axis: 2

0 0( ) 4 ( )y y p x x− = − with 0 0( , )F x p y+ and 0:d x x p= −

Vertical symmetry axis: 2

0 0( ) 4 ( )x x p y y− = − with 0 0( , )F x y p+ and 0:d y y p= −

The general equation of a parabola with symmetry axis parallel to a coordinate axis is:

Horizontal symmetry axis:

CByAyx ++= 2

Vertical symmetry axis:

CBxAxy ++= 2


2.4 Exercises

(we work in an orthonormal coordinate system)

1. Determine the equation of the circle with center M(a, b) and radius R. a. 0, 0, 5a b R= = = 2 2 5x y+ =

b. 4, 1, 2a b R= = = 2 2( 4) ( 1) 4x y− + − =

2. Determine the center and radius of the following circles: a. 2 2 8 6 0x y x y+ − − = ( )4,3 5M and R =

b. 2 23 3 2 3 1 0x y x y+ − + + = 1 1 1

,3 2 6

M and R − =

c. 2 216 16 8 15 0x y x+ − − = 1

,0 14

M and R =

d. ( )2 236 48 36 227 0x y x y+ − + − = 2 1

, 73 2

M and R − =

3. Examine if following equations represent circles. If so, determine the center and the radius. a. 2 2 6 14 59 0x y x y+ − + + = no circle

b. 2 216 16 8 64 335 0x y x y+ + − − = 1

,2 ,54

C −

c. 2 24 4 12 40 109 0x y x y+ − + + = 3

, 5 ,02

C −

4. Find the equation of the circle which passes through (3, 3) and (5, 7) and has its center on the line : 5a x y− =

2 2 16 6 48 0x y x y+ − − + =

5. Find the equation of the circle which has the line segment joining the points A(5,6) and B(-1,0) as diameter.

2 2 4 6 5 0x y x y+ − − − =

6. Find the equation of the circle through P(-3,4) and concentric with the circle2 2: 3 4 1 0c x y x y+ + − − =

2 2 3 4 0x y x y+ + − =

7. Find the equation of the circle circumscribing the triangle ABC with A(2, 2), B(6, -2), C(-3, -5).

2 22 2 5 11 28 0x y x y+ − + − =


8. A circle has its center in M(3, 0) and passes through P(1, 1). Determine: a. the equation of the circle. 2 2( 3) 5x y− + =

b. the equation of the circle with the same center and double area. ( )2 23 10x y− + =

9. Determine the focal point and the directrix of the following parabolas and make a drawing. a. 2 8 0y x− = (2,0); 2x = −

b. 2 6 0y x+ = 3 3

,0 ;2 2

x − =

c. 22 0x y+ = 1 1

0, ;8 8

y − =

d. 22 0x y− = 1 1

0, ;8 8

y = −

10. Find the equation of the parabola with the x-axis as the symmetry axis and with the top in the origin and through the point A(-1, 3).

2 9y x= −

11. Find the equation of the parabola with focal point 1

0,2

−

and directrix 1

:2

d y = .

2 2 0x y+ =

12. Find the equation of the parabola with focal point (7, -2) and directrix : 3d x = . 2 4 8 44 0y y x+ − + =

13. Find the equation of the parabola with focal point (7, -2) and as directrix d the bisector of the second quadrant.

2 2 2 28 8 106 0x y xy x y+ − − + + =

14. Make a drawing of:

a. 23 24 2 50 0x x y− − + = ( ) ( )2 24 1

3x y− = −

b. 24 40 3 100 0y y x+ − + = ( )2 35

4y x+ =

Analytic Geometry - iiw.kuleuven.be · Ananlytic Geometry 7 Notice that ( , ) (0,0)a b = is...

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Transcript of Analytic Geometry - iiw.kuleuven.be · Ananlytic Geometry 7 Notice that ( , ) (0,0)a b = is...