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x Analytic

• Analytic geometry, usually called coordinate geometry or analytical geometry, is the study of geometry using the principles of • The link between algebra and geometry was made possible by the development of a which allowed geometric ideas, such as and , to be described in algebraic terms like real numbers and . • Central idea of analytic geometry – relate geometric points to real numbers.

Dimensions Affine

a 0 b V – 1D  – nonempty set of points y 2D x

By defining each point with a z unique set of real numbers, geometric figures such as lines, , and conics can be 3D y described with algebraic equations.

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"An affine space is nothing more than a vector space whose origin we try to forget about, by adding translations to the linear maps“ Marcel Berger V – Vector space  – nonempty set of points

PPV v B A There exists point O , such that P V: a  A  O is one-to-onecorrespondence

Affine subspace

Let us consider an affine space A and its associated vector space V. Affine subspaces of A are the of A of the form O V  O  w; w  W where O is a point of A, and V a linear subspace of W. The linear subspace associated with an affine subspace is often called its direction, and two subspaces that share the same direction are said .

2 Vectors in R2

• Magnitude of the vector is equal to the of head and tail points. A = (1,3) B = (3,1) u = A  vector v = B w2 = B – A w = u + v dotprod = u*v

mu = [u] Magnitude of the (with sgn) crossprod = u ⊗ w

Vectors in R3

Linear combination, linear dependence Vector subspace

A = (1,3,2) B = (3,1,0) O = (0,0,0) u = A v = B w = u + 2v a = [A,B,O] b = PerpendicularLine[O,a] C = Point[b] n = C dotprod = n*u

3 Linear combination Let V is a vector space over a field R. x  x,,,, x x x  Vector x V is given by ordered . 1 2 3 n

Vector x is a linear combination of set ( e1, … , en ), iff there exists n-tuple ( α1, … , αn ) real numbers which yields n x ai e i  a1 e 1  a 2 e 2   a n e n i1 Affine combination

Take an arbitrary point A in affine space (O,V) n A O  x  O  ai e i; where e i  E i  O i1

AOaEOaEO 1 1   2 2    aEOnn   n A a1 E 1  a 2 E 2 ...  an E n  O 1   a i i1

Affine combination of points O, E1,…, En Arbitrary point A in affine space (O,V) could be expressed as an affine combination n A a1 E 1  a 2 E 2 ...  an E n  O 1   a i i1 n A a1 E 1  a 2 E 2 ...  an E n  a 0 O , where a i  1 i0

Convex combination of points O, E1,…, En - linear combination of points where all coefficients are non- negative and sum to 1

n AaEaE1 1  2 2 ...  aEaOn n  0 , where a i  0 and  a i  1 i0

4 Convex combination of points O, E1,…, En

1 4 AaEaEaEaEaO 1 1  2 2  3 3  4 4  0  aii, where a  0 i0

Straight line in two-dimensional space

A straight line is unambiquously determined by two different points. A straight line can be analytically expressed in – form y mx q – Parametric form X  A  t  u – General ax by  cz  d  0

5 Parametric equations of a line • All points X = A + t.u where t  R form a line and vice versa -- all points on that line have the form X = A + t.u for some t. • u – direction vector p

X3

u X 2 X1 = A + u

X2 = A + 2.u X1 X3 = A + 3.u X4 X4 = A + 1/2 . u A X5 = A + (-1) . u X6 X6 = A + (- 3/2) . u X5  X  p  X  A t u ; t R

Task Příklad Determine the parametric form for the line AB.

u = B – A

C = A+t*u GeoGebra-primka.ggb

6 y = mx + q

The slope of a line m = rise over run.

Calculating Slope

• Slope (m) = rise (change in y) / run (change in x) • Rise is the vertical change and run is the horizontal change

M = y/x M = 3/3 Rise (3) M = 1 The slope is 1. This means that for Run every increase of 1 on the x axis, (3) there will be an increase of 1 on the y axis.

7 Parametric form for the plane in 3D space

X2 X3

3v 2u + 3v C v

X A u B 2u 1

X1 = A + 2u

u = B – A X2 = A + 3v v = C – A X3 = A + 2u + 3v X = A+t*u+s*v

Parametric form for the plane in 3D space

• All points X = A + t.u + s·v, where t, s  R form a plane and vice versa -- all points on that plane have the form X = A + t.u + s·v for some real numbers t,s. • u, v – direction vectors

X3  v

A u

  X    X  A t  u  sv; t, s R

Vektorový součin

8 General equation of the hyperplane in 2D space X  p  ax by  c  0 p

X[x; y] (X  P)  n(a;b)

X Arbitrary point on the line p P[ p ; p ]  1 2 n () vector of p  X  p  X  P n (for X≠P)  X  Pn  0

x  p1 a  y  p2 b  0 ax by   ap1 bp2  0 label:  ap1 bp2   c

General equation of the hyperplane in 3D space  n(a,b,c) X   ax by  cz  d  0

X[x, y, z] (X  P)

P[ p , p , p ] 1 2 3 Perpendicular (normal) vector of p  X   X  P n (for X≠P)  X  Pn  0

x  p1 a  y  p2 b  z  p3 c  0

ax by  cz   ap1 bp2 cp3  0

label:  ap1 bp2 cp3   d

9 Conic Sections

Where do you see conics in real life?

10 Circles

A is a set of points in a plane that are equidistant from a fixed point. The distance is called the radius of the circle, and the fixed point is called the center.

• A circle with center (h, k) and radius r has length r  (x  h)2  (y  k ) 2 to some point (x, y) on the circle. • Squaring both sides yields the center-radius form of the equation of a circle.

r 2  (x  h)2  ( y  k)2

Center-Radius Form of the Equation of a Circle

The center-radius form of the equation of a circle with center (h, k) and radius r is

222 ()()x h  y  k  r .

Notice that a circle is the graph of a relation that is not a function, since it does not pass the vertical line test.

11 Finding the Equation of a Circle Example Find the center-radius form of the equation of a circle with radius 6 and center (–3, 4). Graph the circle and give the domain and range of the relation.

Solution Substitute h = –3, k = 4, and r = 6 into the equation of a circle.

62  (x  (3))2  (y  4)2 36  (x  3)2  (y  4)2

General Form of the Equation of a Circle

For real numbers c, d, and e, the equation x2  y2  cx  dy  e  0 can have a graph that is a circle, a point, or is empty.

12 Parametric equations for the circle

xy221

Parametric equations for the circle x rcos t y rsin t ; t  0,2 

x2 y 2 r 2 r2cos 2 t r 2 sin 2 t r 2

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http://tube.geogebra.org/

Equations and Graphs of

A parabola is a set of points in a plane equidistant from a fixed point and a fixed line. The fixed point is called the focus, and the fixed line the directrix, of the parabola. • For example, let the directrix be the line y = –c and the focus be the point F with coordinates (0, c).

14 Equations and Graphs of Parabolas

• To get the equation of the set of points that are the same distance from the line y = –c and the point (0, c), choose a point P(x, y) on the parabola. The distance from the focus, F, to P, and the point on the directrix, D, to P, must have the same length. d(P,F)  d(P,D) (x  0)2  (y  c)2  (x  x)2  (y  (c))2 x2  y2  2yc  c2  y2  2yc  c2 x2  y2  2yc  c2  y2  2yc  c2 x2  4cy

Parabola with a Vertical Axis and (0, 0)

The parabola with focus (0, c) and directrix y = –c has equation x2 = 4cy. The parabola has vertex (0, 0), vertical axis x = 0, and opens upward if c > 0 or downward if c < 0.

• The focal chord through the focus and perpendicular to the axis of of a parabola has length |4c|. – Let y = c and solve for x. x2  4cy x2  4c2  x  2c or 2c

The endpoints of the chord are ( x, c), so the length is |4c|.

15 Determining Information about Parabolas from Equations Example Find the focus, directrix, vertex, and axis of each parabola. (a) x2  8y (b) y2  28x

Solution (a) 4c  8 c  2

Since the x-term is squared, the parabola is vertical, with focus at (0, c) = (0, 2) and directrix y = –2. The vertex is (0, 0), and the axis is the y-axis.

Determining Information about Parabolas from Equations (b) 4c  28 c  7

The parabola is horizontal, with focus (–7, 0), directrix x = 7, vertex (0, 0), and x-axis as axis of the parabola. Since c is negative, the graph opens to the left.

16 An Application of Parabolas

Example The Parkes radio tele- scope has a parabolic dish shape with 210 feet and depth 32 feet. Because of this parabolic shape, distant rays hitting the dish are reflected directly toward the focus.

An Application of Parabolas

(a) Determine an equation describing the cross section. (b) The receiver must be placed at the focus of the parabola. How far from the vertex of the parabolic dish should the receiver be placed?

Solution (a) The parabola will have the form y = ax2 (vertex at the 210 origin) and pass through the point  2 ,32  (105,32). 32 a (105)2 32 32 a  The cross section can be described by 1052 11,025 32 yx 2. 11,025

17 An Application of Parabolas

32 (b) Since y  x2 , 11,025

1 4c  a 11,025 4c  32 11,025 c   86.1. 128

The receiver should be placed at (0, 86.1), or 86.1 feet above the vertex.

Trajectory of a projectile

path that a thrown or launched projectile or missile

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GeoGebra-kuzelosecky.ggb

Parametric equations of the ellipse

y

X[x;y] x = a · cos t + m y t y = b · sin t + n n × S[m;n] where t is a polar m x between radius vector of X 0 x and x axis  x m22 y n t<0;2π). 221 ab a.cos t m  m22 b .sin t  n  n 1 ab22 at.22 cos bt.22 sin 1 ab22

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