Descartes

MA109 Workbook Lesson 6
Geometric Transformations


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1. Transformations

A transformation of the plane is simply a function from the plane into itself, i.e. it is a rule which assigns to each point P = a, b) in the plane another well defined point T(P) in the plane. We say that T(P) is the value of the transformation T at the point P. Some types of transformations:

  1. T is one-to-one or injective if T(P) = T(Q) is never true for any two different points P and Q.
  2. T is said to be onto or surjective if every point Q in the plane is the value of the transformation at some point of the plane.
  3. T is said to be a one-to-one correspondence or bijective if it is both one-to-one and onto.

Some examples of one-to-one correspondences of the plane:

  1. A translation or shift is a transformation of the type T(x, y) = (x + a, y + b) for all points (x, y). This shift simply moves every point a units to the right and b units upwards. Note that if a is negative, this means that the point is moved |a| units to the left and similarly for b. If b (respectively a) is zero, then the transformation is called a horizontal shift (respectively vertical shift).
  2. A dilation or expansion is a transformation of the form T(x, y) = (ax, by) where a and b are fixed positive numbers. If b = 1 (respectively a = 1), then the transformation is called a horizontal (respectively vertical) expansion. Note that if, for example, a is smaller than 1, then the horizontal expansion actually is a contraction.
  3. Reflection about the x-axis (respectively reflection about the y-axis) is the transformation T(x, y) = (x, -y) (respectively T(x, y) = (-x, y)).
  4. A rotation about the origin is best understood by thinking of the plane of complex numbers. A rotation is a transformation of the form T(z) = cz where c is a complex number with |c| = 1. Such a number c is of the form c = (cos(θ), sin(θ)) for some angle θ. If z = x + iy, then we can describe the rotation as T(x, y) = (x cos(θ) - y sin(θ), x sin(θ) + y cos(θ)).

Exercise 1. What are the equations that define the following transformations:

  1. Shift one unit to the right, left, up, and down.
  2. Expand so that the unit circle becomes a circle of radius 2 or 1/3.
  3. Rotate 45 degrees clockwise, counterclockwise.

Exercise 2. One can obtain more transformations by composing the basic ones given above.

  1. Show that every translation is the composition of a horizontal shift and a vertical shift. Does the order matter?
  2. What is the analogue for dilations?
  3. Show that T(x, y) = (2x + 1, 3y - 2) can be expressed as the composition of a shift and an expansion. Does the order matter?
  4. Show that a rotation through 180 degrees can be expressed as a composition of reflections.

The inverse of a one-to-one transformation T is a transformation S such that S(T(P)) = P for all points P in the plane. The inverse undoes whatever T does; one also has T(S(P)) = P for all points P.

Exercise 3. Find the inverses of each of the following transformations:

  1. The translation T(x, y) = (x + a, y + b)
  2. The dilation T(x, y) = (ax, by) where a and b are positive numbers.
  3. Reflections about the x-axis and y-axis.
  4. Rotation of θ degrees about the origin.
  5. The inverse of an arbitrary one-to-one transformation T.
  6. If T is the composition of R and S (in this order) and if U and V are the inverses of R and S respectively, what is the inverse of T?

Curves in the plane are often defined by some equation. For example, the unit circle is the set of points (x, y) which satisfy f(x, y) = 0 where f(x, y) = x2 + y2 - 1. One can think of the equation f(x, y) as a function f from the plane into the line. Now suppose we apply a transformation T to the curve. What is the equation of the new curve? If T has an inverse S, then the equation is simply the composition g of f and S. To see this, if Q is a point on the transformed curve, then it is of the form Q = T(P) where P is a point on the original curve. So, P = S(Q) because S is the inverse of T. But then g(Q) = f(S(Q)) = f(P) = 0. Similarly, if g(Q) = 0 for some point Q, then let P = S(Q). We have f(P) = f(S(Q)) = g(Q) = 0 and so P is a point on the original curve and Q is a point of the transformed curve.

For example, consider the unit circle defined by f(x, y) = x2 + y2 - 1 = 0. Apply a horizontal expansion T(x, y) = (2x, y). The transformed curve looks like a stretched circle; it is called an ellipse. The inverse of T is S where S(x, y) = (x/2, y). The equation of the ellipse is obtained by composing f with S. That is x2/4 + y2 - 1 = 0. For example, (2, 0) is on the ellipse and satisfies this equation.

2. Lines

Let's start with the x-axis. Its equation is y = 0. If we rotate it θ degrees counterclockwise, this is the same as applying the transformation T where

T(x, y) = (x cos(θ) - y sin(θ), x sin(θ) + y cos(θ)). The inverse is a rotation through an angle of -θ S(x, y) = (x cos(θ) + y sin(θ), -x sin(θ) + y cos(θ)). Composing with the equation y = 0 gives -x sin(θ) + y cos(θ) = 0. We can rewrite the equation as y = tan(θ) x (what happens if θ is ±90o?). This is the equation of an arbitrary line through the origin. One usually writes m = tan(θ) and y = mx; the quantity m is called the slope of the line. To obtain an arbitrary line, one need only translate the line vertically b units.

Exercise 4.

  1. Show that the equation of the line obtained by translating the line y = mx upward a distance of b units is given by y = mx + b. This is called the slope-intercept form of the equation of the line and (0, b) is called the y-intercept of the line; it is the point of intersection of the line and the y-axis. The x-intercept of the line is the point where the line intersects the x-axis. How would you calculate the x-intercept of this line? What about the exceptional cases where the line might not intersect one axis or the other?
  2. In your high school algebra course, you probably learned the formula m = (y2 - y1)/(x2 - x1) for the slope of a line through the points (x1, y1) and (x2, y2). Explain why this formula is valid for lines which are not vertical.
  3. The point-slope form of the equation of the line is y - y1 = m(x - x1) Explain why this formula is valid for non-vertical lines with slope m which pass through the point (x1, y1).

Exercise 5. Find the equations of the lines which form the sides of an equilateral triangle in the first quadrant one of whose sides is the line segment from the origin to the point (3, 0).

Exercise 6. Explain why two parallel lines have the same slope. Explain why the product of the slopes of two non-vertical perpendicular lines is always -1.

3. Parabolas

A parabola is the set of points which are equidistant from a given point F and a given line D. The point is called the focus of the parabola and the line is called its directrix.

To obtain an equation for the parabola, choose a set of coordinates such that the focus is the origin and the line has equation y = -d. In polar coordinates, the distance from the focus to the point P on the parabola is just the polar coordinate r. The distance to the line is d + r cos(θ) and so the equation of the parabola is

r = d + r sin(θ)

We can also convert this to cartesian coordinates. Just square both sides to get x2 + y2 = (d + y)2. Multiplying this out and collecting terms, one gets the equation

y = x2/(2d) + d/2 By applying reflections, expansions, and translations, we see that the graph of any quadratic function is a parabola.

Exercise 7. Explain why y = 2x2 + 12 x - 6 is a parabola. Where is the focus and the directrix? Repeat the exercise with y + 4x2 - 3 = 0.

Exercise 8. Find the equation of the parabola with directrix x = 6 and focus (1, 2).

4. Ellipses and Hyperbolas

Fix a point F (called the focus), a line d (called the directrix), and a positive number e > 0. Consider the locus of points whose distance to the focus is e times the distance from the point to the directrix. This locus is called an ellipse when e < 1 and a hyperbola when e > 1. When e = 1, the locus is a parabola. In all cases, the equation is easy to obtain in polar coordinates. Just take the focus to the origin and the directrix to be y = -d. Then the equation of the locus is

r = e(d + r sin(θ))

As with the case of the parabola, it is easy to convert to cartesian coordinates. Squaring both sides, one gets

x2 + y2 = e2 (d + y)2 Multiplying this out and completing the square gives after some calculation ±x2/b2 + (y - c)2/a2 = 1 where b = ed/(|1-e2|)1/2, a = ed/(|1 - e2|), c = ae, and the sign is plus when e < 1 or -1 when e > 1.

Translate c units downward to obtain the equation

y2/a2 ± + x2/b2 = 1 From this equation, there is an obviously a symmetry with respect to reflections across both axis. After an expansion, the equation becomes y2 ± x2 = 1 In the case of an ellipse, this is our familiar unit circle.

Exercise 9. In the case of a hyperbola, rotate through an angle of 45 degrees to see that the graph is essentially y = 1/x.

Exercise 10. Because of the symmetry, we see that we could have defined the same figures using the reflection of the focus and the directrix. For example, for an ellipse, the other focus would be at (0, 2c). You may be familiar with the definition of an ellipse as being the set of points such that the sum of the distances from the point to each of two foci is be equal to 2a. Show that this ellipse satisfies this same condition. (Hint: The square of the distance from the second focus to the point is given by

(2c - r sin(θ))2 + (r cos(θ))2 Using the equation, you should be able to reduce this to (r - 2a)2.)

What do you conclude in case the curve is a hyperbola?