Chapter 6

Transformations in

Coordinate Geometry

Transformations

As Gregor Samsa awoke one morning from uneasy dreams, he found himself transformed…

• Franz Kafka, “The Metamorphosis”

Our main theme for this chapter will be understanding relationships between geometric transformations [such as shifting a graph to the right by 3 units] and algebraic transformations [such as replacing every x in an equation by $ (x - 3) $ ]. To keep matters simple, we’ll consider just three geometric transformations: reflections, shifts, and stretches.

Reflections

Here sad self-lovers saw in tragic error

Some lovely other or another sky;

In your reversing yet unlying mirror

I saw I was I.

• John Hollander, “At a Forest Pool”

Apart from reveling in Hollander’s narcissistic palindrome, our main object in this brief section is to understand what happens on a point-by-point basis when we reflect a geometric object across a line. In the figure at right, a possibly recognizable personage is depicted, along with his reflection across a line.

When we reflect a point across a line, it ends up just as far

from the line as it had originally been, but on the opposite side. The segment joining a point and its reflected image (segment WC, for example) is always perpendicular to the line of reflection. This is the case, you may observe, with the segment joining the jolly red noses of Mr. Fields and his doppelgänger. The nearer a point lies to the reflecting line, the shorter the distance it will move when reflected. Points on the line don’t move at all.

Since we will be concerned almost exclusively with reflections over the axes, we shall use the phrase vertical reflection to refer specifically to a reflection over the x-axis (such a reflection is “vertical” because it moves points vertically), and horizontal reflection for a reflection over the y-axis.

Exercises

1. Draw the graph of $ y = x^{2} $ and its vertical reflection.

2. Draw the graph of $ y = x^{2} $ and its horizontal reflection.

3. Draw the graph of $ y = x^{2} $ and its reflection across the line y = 1.

4. Draw the graph of $ (x-1)^{2}+(y-1){2}=4 $ and its horizontal reflection.

Stretches

What, will the line stretch out to the crack of doom?

• Macbeth (Act 4, Scene 1)

When we stretch, we stretch relative to some fixed line. Once we’ve specified the line, a “stretch factor” is needed to specify the stretch’s strength: When the stretch factor is k, the stretch sends each point in the plane to a new location exactly k times as far from the fixed line as it had been. Points on the fixed line itself do not move.

In the figure at right, I’ve superimposed the “before” state (in solid black) and the “after” state (in dashed grey) of a horizontal stretch relative to the depicted vertical line by a factor of 3. Scrutinize it carefully until you thoroughly understand it. Observe that a stretch, in contrast to a reflection, changes a figure’s shape. A reflected circle remains circular; a stretched circle does not.

If we stretch a graph by a factor of 1/2, the distance of each of its points to the fixed line is halved. Such a “stretch” is really a compression. This will happen whenever we “stretch” by a factor less than 1. If we view the previous figure with new eyes, seeing the dashed grey figures as the “before” state and the solid black figures as the “after” state, then it represents a horizontal stretch by a factor of 1/3.

Since we will be concerned almost exclusively with stretches relative to the axes, we’ll use the phrase vertical stretch to refer specifically to a stretch relative to the x-axis (we call it “vertical” because it moves points vertically), and horizontal stretch for a stretch relative to the y-axis.

Exercises

5. The figure at right shows a horizontally stretched ellipse.

What stretch factor will transform the black ellipse into the grey one? What stretch factor will transform the grey ellipse into the black one?

6. Draw the graph of $ y = $ and then stretch it vertically by a factor of 4. Label some points (with their coordinates) on both graphs.

7. Draw the graph of $ y = x^{2} $ and then stretch it horizontally by a factor of 1/2. Label some points on both graphs. Could the same effect have been achieved by stretching the original graph vertically? If so, then by what factor? If not, why not?

8. Draw the graph of $ x^{2} + y^{2} = 1 $ and then stretch it horizontally by a factor of 3. Label some points on both graphs. Could the same effect have been achieved by stretching the original graph vertically? If so, by what factor? If not, why not?

9. Draw the graph of $ (x - 2)^{2} + y^{2} = 1 $ and then stretch it horizontally by a factor of 1/2. Find the coordinates of the compressed circle’s topmost point, and those of its leftmost point.

Shifts

Shift that fat ass, Harry.

But slowly, or you’ll swamp the damned boat.

• George Washington. $ ^{*} $

Last (and least) are shifts, which are easy to understand: To shift a graph, we simply move each of its points a specified distance in a specified direction. For example, in the figure at right, if we shift the solid black circle vertically by 3 units, it will occupy the dashed grey circle’s position.

We will concern ourselves exclusively with horizontal and vertical shifts, since any shift can be broken into horizontal and vertical components. We’ll often let the algebraic sign (+ or –) do the talking for us when we wish to distinguish between right and left, or up and down. For instance, a vertical shift by 5 units will refer to a shift up, whereas a vertical shift by -5 units will signify a shift down. (Similarly, a negative horizontal shift is a shift to the left.)

Exercises

10. Draw the graph of $ y = |x| $ , and then shift it vertically by -2 units. Label some points on both graphs.

11. Draw the graph of $ y = $ , then shift it horizontally by 3 units. Label some points on both graphs.

12. In general, geometric transformations are non-commutative. That is, the order in which we carry them out often matters a great deal. Convince yourself of this by comparing the results of the following transformations:

1. Begin with the graph of $ y = x^{2} $ . Shift it horizontally by 2 units, then reflect it over the y-axis.

2. Begin with the graph of $ y = x^{2} $ . Reflect it over the y-axis, then shift it horizontally by 2 units.

13. Geometric transformations sometimes do commute. Convince yourself of this by comparing the results of the following transformations:

1. Begin with the graph of $ y = x^{2} $ . Shift it horizontally by 2 units, then reflect it over the x-axis.

2. Begin with the graph of $ y = x^{2} $ . Reflect it over the x-axis, then shift it horizontally by 2 units.

14. Suppose you must do the following three things to the graph of $ y = x^{3} $ , but the order in which to do them isn’t specified: shift right by 1, reflect over the y-axis, stretch horizontally by a factor of 2.

1. In how many different orders can these three transformations be applied?

2. How many different graphs result from the different orders?

The Transformation Table

And it came to pass, as soon as he came nigh unto the camp, that he saw the calf, and the dancing: and Moses’ anger waxed hot, and he cast the tables out of his hands, and brake them beneath the mount.

• Exodus 32:19

Now that you understand what reflections, stretches, and shifts are, we can discuss how to use them in coordinate geometry. Because students sometimes find this material difficult, I shall begin at the end. I shall declaim. I shall tell you – with the grave and irrefutable voice of authority, accompanied by the solemn majesty of a table – the precise correspondences between these geometric transformations and their algebraic analogues. Provided you will pause in your revels about the golden calf, I shall deign to show you how without telling you why.

Of course, the table that I shall show you was not actually handed down to Moses on Mt. Sinai, and you should not, in the long-run, accept it as though it had been. You should demand an explanation. And I will provide one after you’ve developed a feel for using the table. As you’ll see, the explanation is not terribly difficult, but it does require some new notation and a slightly new way of thinking about graphs. All this in due time. For now, here is the holy transformation table.

	Horizontal	Vertical
Reflection	Substitute $ (-x) $ for each $ x $	Substitute $ (-y) $ for each $ y $
Stretch by a factor of $ k $	Substitute $ (x) $ for each $ x $	Substitute $ (y) $ for each $ y $
Shift by $ k $ units	Substitute $ (x-k) $ for each $ x $	Substitute $ (y-k) $ for each $ y $

The geometric transformations are on the table’s margins; the algebraic transformations are in its body. Horizontal and vertical transformations correspond, respectively, to substitutions for x and y. There are only three types of substitution, and all are easily memorized. Just as you shouldn’t need to consult Moses’ tables to remind yourself of their proclamations (Oh, I can’t recall… Was it thou shalt commit adultery?’), you shouldn’t need to consult the transformation table to know what it says about horizontal shifts. Memorize the table right away.

Our first example will reaffirm a result we’ve established by other means, and should therefore give you some confidence that the transformation rules do indeed work as advertised.

Example 1. If we shift the unit circle up by 5 units, what will its new equation be?

Solution. According to our transformation table, a vertical shift by 5 units corresponds to a substitution of $ (y - 5) $ for y. Making this substitution in the equation of the unit circle, $ x^{2} + y^{2} = 1 $ , we obtain the shifted circle’s equation, $ x^{2} + (y - 5)^{2} = 1 $ .

Recall that we use + and - to distinguish between right and left shifts (or between up and down shifts). For example, a shift down by 5 units is a vertical shift by -5 units. To shift a graph this way, the transformation table tells to substitute $ (y - (-5)) $ , that is, $ (y + 5) $ , for each y in its equation.

Example 2. If we shift the unit circle 7 units to the left, what will its new equation be?

Solution. The table tells us that shifting left by 7 units corresponds to substituting $ (x+7) $ for x. Putting this into the unit circle’s equation yields the shifted circle’s equation: $ (x+7)^{2}+y{2}=1 $ .

Now let’s try some stretches.

Example 3. Find the equation of the graph that results from stretching the unit circle vertically by a factor of 2.

Solution. Vertically stretching by a factor of 2 corresponds to substituting $ (y/2) $ for y. Substituting this into the unit circle’s equation yields the equation of the stretched circle (called an ellipse): $ x^{2} + (y^{2}/4) = 1 $ .

Suppose we stretch a circle vertically by some factor (as in the previous example), and then stretch the resulting ellipse horizontally by the same factor. Would the result be a circle again? This seems plausible, but perhaps it isn’t so. We need not wonder for long. Coordinate geometry to the rescue!

Example 4. If we stretch the graph we obtained in the previous example horizontally by a factor of 2, will the result be a circle?

Solution. We already found the ellipse’s equation in the previous example. If we stretch the ellipse horizontally, its equation changes; the transformation table tells us that we’ll get its new equation if we substitute x/2 for x in the ellipse’s equation. Doing so yields $ (x^{2}/4) + (y^{2}/4) = 1 $ .

Clearing fractions, this becomes $ x^{2} + y^{2} = 4 $ , which we recognize. This is indeed the equation a circle: the circle of radius 2 centered at the origin.

This last example suggests that we can transform the unit circle into any circle in four “moves” (or less): two stretches to attain the desired radius, and two shifts to move its center to the desired location.

Example 5. If we reflect the graph of $ y = $ over the x-axis, then shift the result up by 1 unit, what will the final graph’s equation be?

Solution. Vertical reflection corresponds to substituting -y for y.

Thus, the equation of the reflected graph (dashed, grey at right) is

\[ -y=\sqrt{x}. \]

Isolating y gives us an equivalent form,

\[ y=-\sqrt{x}. \]

Next, to shift this up 1 unit, we substitute $ (y - 1) $ for y, obtaining

\[ y-1=-\sqrt{x}. \]

Isolating y again, we obtain the final graph’s equation:

\[ \boldsymbol{y}=\mathbf{1}-\sqrt{\boldsymbol{x}}. \]

Exercises

15. State the algebraic substitution corresponding to each of the following geometric transformations:

1. shift right by 8 b) shift left by 8 c) shift down by 8 d) shift up by 8 e) shift left by 1/2

2. vertical stretch by a factor of 6 g) vertical stretch by a factor of 1/3 h) shift right by 4

3. horizontal stretch by a factor of 10 j) horizontal stretch by a factor of 1/10

4. horizontal stretch by a factor of 17 l) shift up by 2 m) vertical stretch by a factor of 3/2

5. vertical stretch by a factor of 3/16 o) reflection over the x-axis p) shift right by $ $

6. vertical stretch by a factor of 2/17 r) reflection over the y-axis s) shift down by 1

16. Draw the graph of y = 1/x, then shift it horizontally by 3 units. Label some points on each graph. What is the shifted graph’s equation?

17. Draw the ellipse $ x^{2} + (y^{2}/4) = 1 $ (which we met in example 3), then shift it down by 3 units. Label some points on each graph. What is the shifted ellipse’s equation?

18. Draw the graph of $ (x-1)^{2}+(y-1){2}=2 $ and its reflection over the y-axis. Find the equation of the reflected graph. Finally, find the coordinates of the points where the reflected graph crosses the axes.

[When simplifying the equation, remember that $ (-a-b)^{2}=(-(a+b)){2}=(a+b)^{2} $ .]

19. The graph at right, called the Folium of Descartes, played a small but important role in the history of coordinate geometry – and calculus. (You can read about this online if you are curious.) Its equation is $ x^{3} + y^{3} = 3xy $ .

1. The dotted line is not part of the Folium; it is the Folium’s asymptote: the line it approaches but never touches. Find the asymptote’s equation.

2. Draw the reflection of the Folium over the x-axis. What is its equation?

3. Find the coordinates of point D.

[Hint: It is the intersection of the Folium and a certain line through the origin…]

4. If we were to shift the Folium horizontally until D lies on the y-axis, what would its equation in this new position be?

5. If we were to shift the Folium so as to put D at the origin, what would its equation in its new position be?

20. The graph at right is called a cocked hat (or a bicorn). Its equation is

\[ y^{2}(1-x^{2})=(x^{2}+2y-1)^{2}. \]

1. Where does the cocked hat cross the y-axis?

2. The Mad Hatter wants a cocked hat with “corners” lying at $ (,0) $ , but whose peak still lies at $ (0,1) $ . Draw it and find its equation.

3. Next, the Mad Hatter creates a cocked hat with corners at $ (2,0) $ and $ (3,0) $

and peak at $ (5/2, 1) $ . Draw it and find its equation.

[Hint: This will require two successive transformations.]

4. Finally, being a truly mad hatter, he creates an inverted cocked hat, whose corners are at $ (-1,0) $ and $ (1,0) $ and whose peak – or trough, as the case may be here – is at $ (0,-5000) $ . Find its equation, and determine precisely where this crazy hat crosses the y-axis.

[Hint: You need not solve an equation to find the points of intersection. Just keep track of the coordinates of the original intersection points as you transform the graph of the original cocked hat.]

21. The equation of the graph shown at right, a pear-shaped quartic, is

\[ x^{4}-2x^{3}+4y^{2}=0. \]

1. Observe that if we substitute -y for y in the curve’s equation, the equation itself does not change. This algebraic fact corresponds to a geometric property of the curve. What is it?

2. Mr. Knickerbocker is writing a computer program in which the pear

will appear onscreen with its tip at a given point to alert the user to something important happening there. If the pear is to appear on screen with its tip at $ (x_{0}, y_{0}) $ , what will its equation be? (The pear always remains the same size and always has its tip on the left.)

3. After using the equation you found for him in part (b), Mr. Knickerbocker runs his program and finds a few bugs. For instance, when $ (x_{0}, y_{0}) $ is near the right hand edge of the screen, the pear is cut off by the screen’s edge, and is thus invisible as far as the user is concerned. To fix this, Mr. Knickerbocker wants to revise his program so that pear will sometimes point in the opposite direction (i.e. with its tip on the right). If the pear is to appear on screen tip rightmost and pointing at $ (x_{0}, y_{0}) $ , what will its equation be?

22. The graph at right is an example of a Devil’s curve, a class of curves first studied in 1750 by Gabriel Cramer. The equation of this particular specimen is

\[ -x^{4}+10x^{2}+y^{4}-9y^{2}=0. \]

1. Find the points at which this Devil’s curve crosses the axes.

2. If we wished to compress the graph vertically so that it crossed the y-axis at $ (0, ) $ , what substitution would we have to make in its equation? What would the new equation be?

3. If we wished to compress the result of part (b) horizontally so that it crossed the x-axis at $ (,0) $ , what substitution would we have to make? What would the new equation be?

23. The equation of the fish curve at right is

\[ (2x^{2}+y^{2})^{2}-2\sqrt{2}x(2x^{2}-3y^{2})+2(y^{2}-x^{2})=0. \]

1. Find the points at which the fish curve crosses the x-axis.

2. If we were to replace every y in the equation with $ (5y) $ , what would happen to the fish?

3. If we were to replace every x in the fish’s equation with $ (-x/2) $ , what would happen to the fish? [Hint: You can achieve the same net algebraic effect in two steps – replace each x by x/2, then replace each x by -x.]

24. We can compress the transformation table by packing the stretches and reflections together:

	Horizontal	Vertical
Stretch by a factor of k (with a reflection if k < 0)	Substitute $ ( x) $ for each x	Substitute $ ( y) $ for each y
Shift by k units	Substitute $ (x - k) $ for each x	Substitute $ (y - k) $ for each y

Explain why this is valid. After doing so, feel free to use the table in this compressed form.

Preliminaries to the Proof

Now that you know how the transformation rules work, it’s high time you understand why they work. I will explain this on the next page – after introducing a few preliminary ideas here.

A function of two variables is a rule that unambiguously turns ordered pairs of numbers into numbers. The notation for a function of two variables is exactly what you’d expect.

Example 1. If $ f(x,y)=2x^{2}+y $ , find the values of $ f(3,-7) $ and $ f(-7,3) $ .

Solution. $ f(3,-7)=2(3)^{2}-7=11. $

\[ f(-7,3)=2(-7)^{2}+3=101. \]

Functions of two variables provide the right language for discussing equations in two variables, because we can write any equation in two variables in the form $ f(x, y) = 0 $ simply by pushing all its terms over to the left-hand side.

Example 2. Rewrite the equation $ x^{2} + (y - 1)^{2} = 4 $ in the form $ f(x, y) = 0 $ .

Solution. Pushing all terms to the left, we obtain an equivalent equation with the required form:

\[ \overbrace{x^{2}+(y-1)^{2}-4}^{f(x,y)}=0 \]

In the previous exercise set, I presented the equations of the fish curve, Devil’s curve, and pear-shaped quartic in this $ f(x,y)=0 $ form.

There’s no real advantage, except arguably a kind of tidiness, gained by putting any particular equation in the form $ f(x,y)=0 $ . However, this form is extremely useful whenever we wish to discuss two-variable equations in general. Thus, whenever we wish to discuss the abstract idea of a two-variable equation, or prove a theorem that holds for all graphs, we often begin by carelessly drawing a squiggle (such as the one at right) meant to represent any old curve, which we then think of as the graph of some equation $ f(x,y)=0 $ , with the function f left unspecified.

The Explanation at Last

We need to justify all six entries in the transformation table. If you understand the proof of the first entry, you’ll understand all the others, as they are all cut from the same logical pattern. This being the case, I’ll leave most of the six as exercises for you.

Claim 1. If we reflect a graph over the y-axis, we can obtain the reflected graph’s equation by replacing each x the original graph’s equation by -x.

Proof. Consider an arbitrary graph, such as the black one at right, and reflect it over the y-axis (so that it produces the grey curve). Let $ f(x,y)=0 $ be the original graph’s equation. We must now find the reflected graph’s equation: an equation that is satisfied by the coordinates of all points on the reflected graph.

Let $ (x,y) $ be a variable point on the grey reflected graph.

(Its variability allows it to represent every point on the graph.)

“Undoing” the reflection would send this point back to $ (-x, y) $ ,

and this latter point must lie, of course, on the original black graph. Therefore $ (-x, y) $ satisfies that original graph’s equation.

That is, for every point $ (x, y) $ on the reflected graph, we know that $ f(-x, y) = 0 $ .

It follows that this last equation is the reflected graph’s equation. Happily, we can obtain it from the original graph’s equation, $ f(x, y) = 0 $ , by simply substituting -x for x.

Claim 2. If we stretch a graph horizontally by a factor of k, we can obtain the reflected graph’s equation by replacing each x the original graph’s equation by x/k.

Proof. Consider an arbitrary graph, such as the black one in the figure at right, and stretch it horizontally (to produce the grey curve). Let $ f(x, y) = 0 $ be the original graph’s equation. We now seek the stretched graph’s equation: an equation that is satisfied by the coordinates of all points on the stretched graph.

Let $ (x,y) $ be a variable point on the stretched graph. “Undoing” the stretch would send this point back to $ (x/k,y) $ , and this latter point must, of course, lie on the original black graph. Therefore, $ (x/k,y) $ satisfies that original graph’s equation.

That is, for every point $ (x, y) $ on the stretched graph, we know that $ f(x/k, y) = 0 $ .

It follows that this last equation is the stretched graph’s equation. We can thus obtain it from the original graph’s equation, $ f(x, y) = 0 $ , by merely substituting x/k for x.

If you understood these two proofs, you will have no trouble constructing the other four on your own. Essentially, we obtain the transformed curve’s equation by “undoing” the transformation algebraically, then substituting the “un-transformed” coordinates back into the original curve’s equation.

Exercises

25. Prove that if we shift a graph horizontally by k units, we can obtain the shifted graph’s equation by substituting $ (x - k) $ for each x in the original graph’s equation.

26. Prove that if we reflect a graph over the x-axis, we can obtain the reflected graph’s equation by replacing each y the original graph’s equation by -y.

27. Prove that if we stretch a graph vertically by a factor of k, then we can obtain the reflected graph’s equation by replacing each y the original graph’s equation by y/k.

28. Prove that if we shift a graph vertically by k units, we can obtain the shifted graph’s equation by replacing each y the original graph’s equation by $ (y - k) $ .

29. In the proof of Claim 1 above, we showed that $ f(-x,y)=0 $ is satisfied by each point on the reflected graph. Strictly speaking, we should also have shown that $ f(-x,y)=0 $ is not satisfied by any point that isn’t on the reflected graph. Fill this gap.

[Hint: Let $ (a, b) $ be a variable point not on the reflected graph. Now “undo” the transformation, etc.]

Right-hand Side Shortcuts for Functions

When working specifically with graphs of functions (i.e. equations of the form $ y = f(x) $ ), we can carry out transformations by operating directly on the equation’s right-hand side. The following “RHS Shortcuts” are equivalent to, but more convenient than, the usual substitutions. $ ^{*} $

Vertical Stretch by a factor of $ k^{} $	Multiply RHS by k
Vertical Shift by k units	Add k to the RHS

To see that these “new” operations are really just shortcuts for familiar substitutions, consider a vertical stretch of the graph of $ y = f(x) $ . Substituting $ (y/k) $ for y gives us $ y/k = f(x) $ . Algebraic massage turns this into $ y = kf(x) $ . And voilà! The net algebraic effect is to multiply the original RHS by k, as claimed. $ ^{} $ The same sort of argument justifies the vertical shift shortcut, as you should verify.

Example 1. Roughly speaking, what does the graph of $ y = 2x^{2} + 3 $ look like?

Solution. The graph of $ y = x^{2} $ is the familiar U-shape. From $ y = x^{2} $ , we can reach $ y = 2x^{2} + 3 $ in two algebraic “moves”: First, we multiply the RHS by 2 (yielding $ y = 2x^{2} $ ), then add 3 to the RHS. The corresponding geometric transformations: Stretch vertically by a factor of 2, then shift up by

3. Hence, the graph we seek is a slightly skinnier U-shape opening up with its vertex is at $ (0,3) $ .

Never forget: The RHS shortcuts apply only to functions $ y = f(x) $ , not to equations in general!

\[ x^{2}+y^{2}=1 \]

\[ y=f(x) \]

The RHS shortcuts are gratifyingly direct: To stretch vertically by 5, we just multiply the RHS by 5; to shift up by 3 units, we just add 3 to the RHS. A pleasant contrast to the topsy-turvy world of substitutions, where we must remember to take reciprocals and reverse algebraic signs! The RHS shortcuts are easy to use, easy to remember, and easy to understand. But once again, with feeling: They apply only to functions.

Let’s practice the shortcuts by applying them to some sheep in wolves’ clothing – harmless functions dressed up in enough transformational garb to make them look fearsome to the uninitiated.

Example 2. Graph the function $ y = -3 $

Solution. This is just a transformed version of a familiar function, $ y = $ , as this analysis shows:

\[ y=\sqrt{x}\quad\xrightarrow{sub(x+2)for x}\quad y=\sqrt{x+2}\quad\xrightarrow{-3(RHS)}\quad y=-3\sqrt{x+2} \]

The corresponding geometry: Shift right by 2, then stretch vertically by a factor of 3 (accompanied by a vertical reflection).

Applying those successive geometric transformations to the graph of $ y = $ will bring us to the graph at right, as you should verify. Hence, the figure at right must be the given function’s graph. To provide more detail, we could find and label its intersection with the y-axis, which (as you should also verify) is $ (0, -3) $ .

In the previous example, we could have carried out the transformations in any order and obtained the same graph. (Try it and see.) In general, however, we must take care with the order of transformations.

Example 3. Graph the function $ y = x^{2} - 3 $ .

Solution. We recognize that this function is just a transformed version of $ y = x^{2} $ :

\[ y=x^{2}\quad\xrightarrow{(1/2)\cdot RHS}\quad y=\frac{1}{2}x^{2}\quad\xrightarrow{RHS-3}\quad y=\frac{1}{2}x^{2}-3 \]

The corresponding geometric transformations: a vertical stretch by 1/2, then a shift down by 3.

Applying these two geometric transformations (in the specified order!) to the familiar U-shaped graph of $ y = x^{2} $ yields a graph whose basic shape is shown at right. Should we desire more detail, we could find and label its intersections with the x-axis in the usual way.

[These are $ (- , 0) $ and $ (, 0) $ , as you should verify.]

As you saw in exercise 12, it is essential that you do geometric transformations in the correct order. How can you tell whether a given order will yield the correct graph? This is easy: Just check whether the corresponding algebraic transformations, applied in the same order, yield the equation whose graph you want. If they do, then all’s well. If they don’t, then there’s a problem.

In the previous example, suppose you had wondered if you could have first shifted down by 3 units, and then stretched vertically by a factor of 1/2. Would that work? Well, consider the corresponding algebraic transformations: First subtract 3 from the original RHS, then multiply the resulting RHS by 1/2. A little calculation shows that doing these to the original equation $ (y = x^{2}) $ yields $ y = (1/2)x^{2} - 3/2 $ , which, misses our algebraic target: $ y = (1/2)x^{2} - 3 $ . Hence, the corresponding geometric operations

would have missed the geometric target, too; they would have led us to the graph of the wrong equation. Order is important. First you open the window, then you put your head through.

Here’s a more challenging example – which we’ll solve in two different ways.

Example 4. Graph the function $ y = $

First Solution. This is a transformed version of $ y = $ , whose graph is the top half of the unit circle. Here’s one sequence of algebraic steps that accomplishes this transformation:

\[ y=\sqrt{1-x^{2}}\quad\xrightarrow{sub\left(x+9\right)for x}\quad y=\sqrt{1-(x+9)^{2}}\quad\xrightarrow{sub3x for x}\quad y=\sqrt{1-(3x+9)^{2}} \]

The corresponding geometric transformations are:

Shift left by 9 units, then stretch horizontally by 1/3.

Applying these geometric transformations in the specified order to the unit circle’s top half gives us the graph of $ y = $ , which, as you can verify by carrying the transformations out yourself, looks like the graph at right.*

Second Solution. If we do some preliminary algebra (factoring out a 3 inside the parentheses), we can rewrite the given function in a new form,

\[ y=\sqrt{1-(3(x+3))^{2}}, \]

which suggests a different sequence of steps:

\[ y=\sqrt{1-x^{2}}\quad\xrightarrow{sub3x for x}\quad y=\sqrt{1-(3x)^{2}}\quad\xrightarrow{sub(x+3)for x}\quad y=\sqrt{1-(3(x+3))^{2}} \]

The corresponding geometric transformations are:

Stretch horizontally by 1/3, then shift left by 3

By drawing pictures, you should verify that the graph resulting from this alternate sequence of geometric transformations will be the same that we found in the first solution. Notice that the shift here was by only 3 units (as opposed to 9 units in the first solution); because we compressed the graph first, we didn’t have to shift it as far. $ ^{†} $

Exercises

30. Give the geometric transformation corresponding to the following algebraic transformations of $ y = f(x) $ :

1. Multiply the RHS by 5

2. Substitute $ (x + 2) $ for x

3. Multiply the RHS by -1

4. Add 1 to the RHS

5. Multiply the RHS by 7/9

6. Substitute 6x for x

7. Substitute $ (2/3)x $ for x

8. Add 5 to the RHS

9. Multiply the RHS by -3

10. Multiply the RHS by $ (-3/5) $

11. Substitute -4x for x

12. Substitute $ (y + 2) $ for y.

31. True or false (and explain why the false answers are false):

1. Multiplying the RHS of $ x^{2} + y^{2} = 1 $ by 4 stretches its graph vertically by a factor of 4.

2. Adding 3 to the RHS of $ 2x^{2} + 2y^{2} = 2 $ shifts its graph up by 3 units.

3. Subtracting 7 from the RHS of $ 10xy = x^{2} + 3y^{3} - 5 $ shifts its graph down by 7 units.

4. Multiplying the RHS of $ y = 14x^{8} + $ by 3 stretches its graph vertically by a factor of 3.

32. Give the algebraic transformation corresponding to the following geometric transformations of $ y = f(x) $ :

1. Shift right by 8 b) Shift left by 8 c) Shift down by 8 d) Shift up by 8 e) V-stretch by a factor of 6

2. V-stretch by a factor of 1/3 g) H-stretch by a factor of 10 h) H-stretch by a factor of 1/10

3. V-stretch by a factor of 3/16 j) Reflection over the x-axis k) Reflection over the y-axis

4. V-stretch by a factor of 5 and a v-reflection m) V-stretch by a factor of 3/7 and a vertical reflection

33. Graph the following functions, and label key points with their coordinates. (These will include intersections with the axes, and possibly endpoints or turning points when these exist).

1. $ y = 5(x - 1)^{3} $

\[ y=\sqrt{x+3}+1 \]

\[ y=\frac{4}{x-2} \]

3. $ y = -2|x| + 3 $

\[ \mathsf{f})y=-3(x-1)^{2}+2 \]

\[ {\bf g})y=-{\frac{3}{x}}+7 \]

\[ \mathsf{d})y={\textstyle{\frac{2}{3}}}(x+2)^{2}-1 \]

\[ h)y={\sqrt[3]{8x-8}} \]

1. $ y = 2 - 3 $

10. $ y = 1 - $

\[ \begin{array}{r}{\mathsf{k})y=-\frac{1}{2}(x+1)^{3}}\end{array} \]

34. Find the functions corresponding to the following graphs, which are transformed versions of $ y = |x| $ .

a)

b)

c)

35. Using only shifts, stretches, and reflections, is it possible to transform line y = x into the line through $ (x_{0}, y_{0}) $ with slope m? If not, why not? If so, give a sequence of transformations that will do it.

Graphs of Quadratic Functions

I’m very well acquainted too with matters mathematical, I understand equations, both the simple and quadrational, About binomial theorem I am teemin’ with a lot o’ news – With many cheerful facts about the square of the hypotenuse! - Major General Stanley (Pirates of Penzance, Act I)

We’ve already proven that the graphs of all linear functions (that is, functions of the form $ y = ax + b $ ) are straight lines. In this section we’ll prove that graphs of all quadratic functions (that is, functions of the form $ y = ax^{2} + bx + c $ ) are U-shapes. Moreover, we’ll see later that they are not just any old U-shapes: They are parabolas. Before proving that all quadratic functions have U-shaped graphs, we’ll consider one specific quadratic. This will contain, in a very tangible form, all the key ideas we’ll need for the abstract universal proof. One idea we’ll need is “completing the square”, so you may wish to review that technique before reading on.

Example. Graph the quadratic function $ y = 2x^{2} + 12x + 13 $ .

Solution. We begin by rewriting the equation in an equivalent form by completing the square.

\[ \begin{aligned}y&=2x^{2}+12x+13\\&=2(x^{2}+6x)+13\\&=2(x^{2}+6x+\mathbf{9}-\mathbf{9})+13\\&=2[(x+3)^{2}-9]+13\\&=2(x+3)^{2}-5.\end{aligned} \]

In this form, we see that our quadratic is just a transformed version of $ y = x^{2} $ , whose graph is the familiar U-shape. Specifically, it is related to $ y = x^{2} $ by the following sequence of transformations:

\[ y=x^{2}\quad\xrightarrow{2(RHS)}\quad y=2x^{2}\quad\xrightarrow{sub\ (x+3)\ for\ x}\quad y=2(x+3)^{2}\quad\xrightarrow{RHS-5}\quad y=2(x+3)^{2}-5 \]

The corresponding sequence of geometric transformations is:

Stretch vertically by 2, shift left by 3, shift down by 5

Following these transformations of the graph of $ y = x^{2} $ in your mind’s eye, you’ll see the rough shape of the graph we seek: a stretched out U, which bottoms out at $ (-3, -5) $ . For a bit more detail, we note that the equation tells us that when x is 0, we have y = 13. The graph we want is thus a U whose lowest point is $ (-3, -5) $ and which crosses the y-axis at $ (0, 13) $ . This much information is already enough to produce the graph at right.

To include more detail, if we desire it, we could find the points where the curve crosses the x-axis. [These, as you can verify, are $ (-3 , 0) $ .]

The method we employed in the preceding example can be used to graph any quadratic function: Complete the square, then apply the appropriate transformations to the graph of $ y = x^{2} $ . Let us now tackle the problem in the abstract.

Claim. Every quadratic function has a U-shaped graph.

Proof. Consider the general quadratic function $ y = ax^{2} + bx + c $ .

Rewriting the equation in an equivalent form by completing the square, we obtain

\[ \begin{aligned}y&=ax^{2}+bx+c\\&=a\left(x^{2}+\frac{b}{a}x\right)+c\\&=a\left(x^{2}+\frac{b}{a}x+\frac{b^{2}}{4a^{2}}-\frac{b^{2}}{4a^{2}}\right)+c\\&=a\left[\left(x+\frac{b}{2a}\right)^{2}-\frac{b^{2}}{4a^{2}}\right]+c\\&=a\left(x+\frac{b}{2a}\right)^{2}+\left(c-\frac{b^{2}}{4a}\right).\\ \end{aligned} \]

This reveals that the general quadratic function is an algebraically transformed version of $ y = x^{2} $ . Consequently, its graph can be obtained from the U-shaped graph of $ y = x^{2} $ by a sequence of geometric transformations: a vertical stretch by a factor of a (with a vertical reflection if a < 0), then a horizontal shift, and finally, a vertical shift. Since a stretched U is still a U (albeit a skinnier or a fatter one), and a shifted U obviously is still a U, we may therefore conclude that the graph of $ y = ax^{2} + bx + c $ must always be U-shaped, as claimed.

One further observation: In the proof, we saw that if a, the quadratic’s leading coefficient, is negative, then the U will be reflected vertically. If this happens, it will, of course, open downwards. Otherwise (if the leading coefficient is positive) the U will open upwards. To sum up what we’ve discovered,

The graph of every quadratic function (i.e. function of the form $ y = ax^{2} + bx + c $ ) is U-shaped.

• The U opens upwards if the leading coefficient is positive, and downwards if it is negative.

• To graph a quadratic, we complete the square and apply the transformations revealed thereby.

Finally, a little terminology: The turning point in a U-shape is called its vertex. In the graph of a quadratic function, the vertex lies either where the function attains its maximum output value (if the graph opens downwards) or its minimum value (if it opens upwards).

Exercises

36. Graph the following quadratics. Find the graphs’ vertices and their intersections with the axes.

\[ \begin{aligned}&a)y=x^{2}+8x+1\quad&b)y=2x^{2}+4x+2\quad&c)y=5x^{2}-3\quad&d)y=-x^{2}+10x-7\\&e)y=3x^{2}+4x+5\quad&f)y=-5x^{2}-12x+2\quad&g)y=\frac{3}{2}x^{2}+6x\quad&h)y=-\frac{3}{14}x^{2}+\frac{2}{7}x+1\end{aligned} \]

37. Can the graph of $ y = x^{2} $ be contained between two vertical lines? If so, which ones? If not, why not?

38. Many textbooks state that in the graph of $ y = ax^{2} + bx + c $ , the vertex’s x-coordinate will be -b/2a.

1. Explain why this is so.

2. Use this result to find the coordinates of the vertex of $ y = 5x^{2} + 4x - 1 $ .

3. Although this result will allow you to solve certain homework problems (like those in exercise 36) quickly, memorizing it is counterproductive. It will not help you learn mathematics. In contrast, each time that you graph a quadratic by completing the square, you reinforce two important mathematical techniques: completing the square and transformations. (Besides, if you want a shortcut, why not just use a computer?) The second reason for eschewing this formula is that once you’ve learned a little calculus, you’ll be able to find a quadratic’s vertex in seconds without having to rummage in your memory for anything.

39. If two rectangles have the same perimeter, must they have the same area? If so, explain why. If not, provide a counterexample.

40. Mr. Square plans to fence off a rectangular area in the middle of field. He has 100 feet of fencing. He dimly remembers learning that rectangles with the same perimeter can have different areas, and he wants his rectangle’s area as large as possible. Being Mr. Square, he’s pretty sure that the way to maximize the area is to make a square, but then, he’s been wrong before. Is he right this time? If not, why not? If so, prove it.

[Hint: Consider a rectangle with perimeter 100. Let x be the length of one of its sides. Express the rectangle’s area as a function of x. The function will be quadratic. Graph it, and think about its vertex in the context of Mr. Square’s problem.]

41. Mr. Square’s neighbor, Lana Evitneter, also plans to fence off a rectangular area with 100 feet of fencing. However, since a wall of her house will serve as one side of the rectangle, she actually needs fencing for just three sides. She asks Mr. Square to help her maximize the area of her rectangle. Naturally, Mr. Square recommends making a square. Is he right this time? If so, prove it. If not, find the dimensions of the rectangle that will actually maximize the enclosed area.

42. Mortified by his mistake, Mr. Square flees the neighborhood in his hot-air balloon. He takes off from the base of a very long hill, which has a constant slope of 1/3 (see the figure). The path that Mr. Square’s balloon takes happens to be the graph of $ y = -x^{2} + 4x $ . Alas, as you can see from the figure, his balloon doesn’t make it over the hill. Assuming that each unit on the axes represents 1000 feet,

1. What are the coordinates of the point at which the balloon lands?

2. How far (in feet) is the launch point from the landing point?

3. If the launch point is at sea level, then what is the balloon’s maximum altitude relative to sea level?

4. What is the balloon’s maximum altitude relative to the ground?

Parabolas

It has been observed that missiles and projectiles

describe a curved path of some sort. However, no

one has yet pointed out that this path is a parabola.

This… I have succeeded in proving.

• Galileo Galilei, Dialogues Concerning Two New Sciences, 3^{rd} Day, Introduction

The distance from a point to a line is, by definition, the length of the shortest path joining them: a straight path meeting the line at right angles. Thus, in the figure at right, the distance from point P to line AB is the length of segment PQ.

Draw a line and a point on a piece of paper. Call the point F and the line d. Locate a point equidistant from F and d, and mark it on the page. Then, find and mark as many other points as you can that are equidistant from F and d. After a while, you’ll have a picture that looks something like the one at right. In your mind’s eye, picture the curve that passes through each and every one of the infinitely many points equidistant from F and d. This curve is called a parabola.

Definition. A parabola is the set of all points equidistant from some fixed point (called the parabola’s focus) and some fixed line (called the parabola’s directrix).

While all parabolas are U-shapes, very few U-shapes are parabolas. If you draw a random U-shape on a page, it will almost certainly not be a parabola. Try it: First draw a random U-shape on a page, then try to guess where its focus and directrix would be if it were a parabola. Now start checking points on the U (preferably with the aid of a ruler). If you can find even one point on the U that is not exactly the same distance from your prospective focus and directrix, then your U-shape is not a parabola – at least not with those choices of focus and directrix. Difficult as it is to draw a reasonably accurate circle freehand (without the aid of a compass), it is still more difficult to draw a reasonably accurate parabola.

Mathematicians have studied parabolas for well over two thousand years on account of their remarkable geometric properties. Amazingly, in the early 17 $ ^{th} $ century, Galileo proved that parabolas have physical significance as well: A projectile moving under the influence of gravity alone will always follow a parabolic path. When you toss a ball to your dog, the path that the ball follows when it leaves your hand is not merely U-shaped, but parabolic: As it moves through the air, the ball remains equidistant from an invisible focus and directrix. Take a physics class, and you’ll learn why.

We can discover a parabola’s equation by translating its definition into algebraic terms. This will be especially easy if we place the axes so that the parabola’s vertex is at the origin and its focus lies on the positive y-axis. This setup ensures that the focus’s coordinates will be $ (0, p) $ for some positive number p. Moreover, since the vertex is p units from the focus, it must also (by the parabola’s definition) lie p units up from the directrix; hence, the equation of the directrix must be y = -p.

Problem. Derive the equation of the parabola with vertex $ (0,0) $ and focus $ (0, p) $ .

Solution. We seek an equation satisfied by the coordinates of all the parabola’s points. Let $ (x, y) $ be a variable point on the parabola. Its distance to the focus $ (0, p) $ is, by the distance formula,

\[ \sqrt{(\Delta x)^{2}+(\Delta y)^{2}}=\sqrt{\boldsymbol{x}^{2}+(\boldsymbol{y}-\boldsymbol{p})^{2}}, \]

while its distance to the directrix is $ y + p $ . (It lies y units above the x-axis, which itself lies p units above the directrix, as in the figure above.)

The two distances we’ve just computed are, by the parabola’s definition, equal. That is,

\[ \sqrt{x^{2}+(y-p)^{2}}=y+p. \]

This equation is satisfied by our variable point (and thus by every point) on the parabola, so it is the parabola’s equation. To polish it, we square both sides of the equation and then simplify. The resulting polished equation of the parabola is, as you should verify, $ y = ( ) x^{2} $ .

The equation of the parabola with vertex $ (0,0) $ and focus $ (0,p) $ is

\[ \boldsymbol{y}=\frac{1}{4\boldsymbol{p}}\boldsymbol{x}^{2} \]

We’ll use this fact to establish two little preliminary results that we’ll then parlay into a big theorem. Here’s the first little result.

Claim 1. The graph of $ y = x^{2} $ is not merely U-shaped, but parabolic.

Proof. The equation $ y = x^{2} $ has the form in the box above, with p = 1/4. Hence, its graph is a parabola with vertex $ (0, 0) $ and focus $ (0, 1/4) $ . (And its directrix is y = -1/4.)

The second of our two preliminary results concerns stretched parabolas. Suppose we stretch a parabola. The result will certainly be U-shaped, but will it still be parabolic? Remarkably, it turns out that if we stretch any parabola by any factor in any direction (not just vertically or horizontally), the result will still be a parabola! Although this full statement is too difficult for us to prove here, we can easily prove one very special – and very useful – case.

Claim 2. If we stretch the graph of $ y = x^{2} $ vertically, the result is still a parabola.

Proof. If we stretch the graph of $ y = x^{2} $ vertically by a factor of k, its new equation will be $ y = kx^{2} $ . This matches our boxed equation for a parabola with $ k = 1/(4p) $ ; or equivalently, with p = 1/4k. Hence, the graph of $ y = kx^{2} $ is indeed a parabola with vertex $ (0, 0) $ and focus $ (0, 1/(4k)) $ .

With our two preliminary results squared away, let’s turn to our big theorem.

Theorem. The graphs of all quadratic functions are parabolas.

Proof. In an earlier section (“Graphs of Quadratic Functions”), we proved that any quadratic’s graph can be obtained from the graph of $ y = x^{2} $ by following a certain sequence of transformations: first a vertical stretch (sometimes accompanied by a reflection), and then some shifts.

Since then, you’ve learned that the graph of $ y = x^{2} $ is not just a U, but a parabola (Claim 1), and that a vertical stretch will preserve its parabolic nature (Claim 2). What about reflections and shifts? Might they disrupt a stretched parabola’s parabolic nature, turning it into a mere U-shape? No, this won’t happen. After all, reflections and shifts don’t change a graph’s shape at all (they only change the graph’s location), so they must obviously preserve a parabola’s parabolic nature, too.

To sum up, we’ve now seen that the graph of each and every quadratic function can be obtained from one specific parabola $ (y = x^{2}) $ by some sequence of “parabola-preserving” transformations. It follows that the graph of every quadratic function must be a parabola, as claimed.

Combining this last theorem with our earlier work on graphing quadratics, we may conclude that

The graph of every quadratic function (i.e. function of the form $ y = ax^{2} + bx + c $ ) is a parabola.

• The parabola opens up if the leading coefficient is positive, and down if it is negative.

• To graph a quadratic, we complete the square and apply the transformations revealed thereby.

Exercises

43. Pick a random number. Call it n. Square it. The point $ (n, n^{2}) $ lies on the graph of $ y = x^{2} $ . Compute its distances to the parabola’s focus and directrix, and verify that these are indeed equal.

44. Is $ y = 2x^{2} $ the graph of a parabola? If so, find its vertex, focus, and directrix.

45. Is $ y = ()x^{2} $ the graph of a parabola? If so, find its vertex, focus, and directrix.

46. Find the equation of the parabola with vertex $ (0,0) $ and focus $ (0,5) $ .

47. Find the equation of the parabola with vertex $ (0,0) $ and focus $ (0,1/5) $ .

48. Prove that if we stretch the graph of $ y = x^{2} $ horizontally, the result will still be a parabola.

49. Find the focus and directrix of the parabola whose equation is $ y = 3x^{2} - 12x + 13 $ .

[Hint: Complete the square, think about how each geometric transformation affects the focus & directrix.]

50. Find the focus and directrix of the parabola whose equation is $ y = -3x^{2} - 6x + 1 $ .

The Reflection Property of Parabolas

At any point P on a parabola, draw the parabola’s tangent there. Next, draw the line segment joining P to the focus. Finally, draw the ray from P that is parallel to the parabola’s axis of symmetry. We can prove that the line segment and ray make equal angles with the tangent. This so-called “reflection property” of parabolas does not hold for other U-shapes: only parabolas. Esoteric though it may seem, this parabolic reflection property has remarkable physical consequences, which you’ll be able to appreciate after reading the following paragraphs about optics.

Light rays bounce off of flat surfaces in a simple manner: When a ray strikes a flat surface, it reflects off at the same angle at which it struck. The figure at right depicts a light ray striking (and then departing) a reflective surface at an angle of $ 24^{} $ .

If a light ray strikes a curved surface, the same rule holds, but now the angles are measured between the light ray and the tangent to the surface where the ray hits it. You can see why this is so if you imagine zooming in on the point of tangency so closely that the curve and the tangent line become indistinguishable. This virtual identity of a curve and its tangent line – when viewed at a microscopic scale – is, incidentally, a major theme of calculus.

Now let us return to the reflection property of parabolas. If light leaves a parabola’s focus and hits the parabola at point P, in what direction will it be reflected? If you’ve understood the three preceding paragraphs, you should be able to convince yourself of the following: Because of the reflection property, the ray that strikes the parabola at P will “bounce off” the parabola parallel to the parabola’s axis of symmetry. This being so, suppose that we illuminate a bulb at the parabola’s focus, so that light streams out of it in all directions. Amazingly, all the light that hits the parabola will bounce back in the same direction: parallel to the parabola’s axis of symmetry! It is for this reason that parabolic mirrors are used in flashlights, headlights, and so forth.

Parabolic mirrors can also be used in reverse to concentrate parallel rays into a point. For example, solar rays that reach us on Earth are effectively parallel, since the Earth and Sun are so tiny compared to the vast distance between them. If we capture solar rays in a parabolic mirror, we can concentrate them into one very hot point at the mirror’s focus. (Hence the name focus, which means “hearth” in Latin.) You can find images of such “burning mirrors” or “solar furnaces” online, including the world’s largest, located in a French village in the Pyrenees. Another example: Satellite receiving dishes are parabolic in shape to concentrate the satellite’s signals into the dish’s focus, where the dish’s transmitter is located.

Proof of the Parabola’s Reflection Property

Our proof will make use of the following fairly obvious geometric fact.

The perpendicular bisector of a line segment AB divides the plane into three sets of points: those on A’s side of the bisector (which are closer to A than B); those on B’s side of the bisector (which are closer to B than A); and those on the bisector itself (which are equidistant from A and B);

For example, in the figure at right, we must have PA = PB, but QB < QA.

Claim. Light emitted from a parabola’s focus reflects off the parabola parallel to its axis of symmetry.

Proof. First we’ll set the stage. Let P be any point on a parabola.

Now draw segment FP joining it to the focus, and ray $ PD’ $ parallel to the axis of symmetry. Among all the straight lines through P that do not enter angle $ FD’ $ , its obvious that only one makes equal angles with FP and $ PD’ $ . I’ve drawn this line $ (PS’) $ in gray. Next, we extend ray $ PD’ $ backwards until it hits the directrix at D. Since $ PD’ $ is parallel to the axis of symmetry, the extension must be perpendicular to the directrix. Finally, let $ FD’ $ intersection with the gray line be called S. The stage is now set.

In the proof’s first act (of three), we will prove that the gray line is $ FD^{} $ ‘s perpendicular bisector. The key is to show that $ FPS DPS $ , which we can do as follows: First, FP = PD by the parabola’s defining property. Next, $ FS = DS $ (since both angles are equal to $ D’S’ $ ; one by the gray line’s definition, the other by vertical angles). Finally, SP is common to both triangles. Therefore, by SAS-congruence, $ FPS DPS $ as claimed. It follows that FS = DS. That is, the gray line bisects FD. Moreover, $ FP = DP $ , and since they form a straight line, these equal angles must be right angles. Thus the gray line is indeed $ FD’ $ ’s perpendicular bisector.

In act two, we’ll prove the gray line is tangent to the parabola by showing that all the parabola’s points (besides P, of course) lie on one side of the gray line. To this end, let Q be any other point on the parabola, and drop perpendicular QR to the directrix. By the parabola’s definition, QF = QR, which is less than QD (in any right triangle, leg < hypotenuse), so QF < QD. Thus, Q is on F’s side of FD’s perpendicular bisector, the gray line. Since Q was an arbitrary point, all the parabola’s points (besides P) lie on F’s side of the gray line, which is thus tangent to the parabola at P, as claimed.

The third and final act: Imagine that a ray of light emitted from the parabola’s focus F strikes the parabola at P, as in the figure above. As explained earlier, the angle at which the ray is reflected equals the angle at which it strikes the parabola’s tangent at P. But this tangent is the gray line. Since the ray strikes this line at angle $ FS $ , it reflects off at an angle of the same magnitude. But by definition of the gray line, that angle is $ D’S’ $ . That is, the light reflects off the parabola along ray $ PD’ $ , which, by definition, is parallel to the parabola’s axis of symmetry.