Supporting Australian Mathematics Project. A guide for teachers Years 11 and 12. Functions: Module 4. Quadratics

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1 1 Supporting Australian Mathematics Project A guide for teachers Years 11 and 12 Functions: Module 4 Quadratics

2 Quadratics A guide for teachers (Years 11 12) Principal author: Peter Brown, Universit of NSW Dr Michael Evans, AMSI Associate Professor David Hunt, Universit of NSW Dr Daniel Mathews, Monash Universit Editor: Dr Jane Pitkethl, La Trobe Universit Illustrations and web design: Catherine Tan, Michael Shaw Full bibliographic details are available from Education Services Australia. Published b Education Services Australia PO Bo 177 Carlton South Vic 353 Australia Tel: (3) Fa: (3) info@esa.edu.au Website: Education Services Australia Ltd, ecept where indicated otherwise. You ma cop, distribute and adapt this material free of charge for non-commercial educational purposes, provided ou retain all copright notices and acknowledgements. This publication is funded b the Australian Government Department of Education, Emploment and Workplace Relations. Supporting Australian Mathematics Project Australian Mathematical Sciences Institute Building 161 The Universit of Melbourne VIC 31 enquiries@amsi.org.au Website:

3 Assumed knowledge Motivation Content The basic parabola Transformations of the parabola Intercepts The ais of smmetr The quadratic formula and the discriminant Applications to maima and minima Sum and product of the roots Constructing quadratics Intersection of lines and parabolas Focus-directri definition of the parabola Parametric equations of a parabola Chords, tangents and normals The reflection propert of the parabola Links forward Conic sections Histor and applications Answers to eercises

4 Quadratics Assumed knowledge The content of the module Quadratic equations (Years 9 1). The content of the module Algebra review, which includes factoring quadratic epressions, completing the square and solving quadratic equations using various techniques including the quadratic formula. The content of the module Coordinate geometr. Motivation Quadratic equations arose in antiquit in the contet of problem solving. These problems were generall epressed verball, but their solution involved some method for solving a quadratic equation often also epressed verball. The Greeks studied the parabola and its geometr. This curve occurs naturall as the edge of the cross-section of a cone sliced at an angle to its main ais. The parabola can also be thought of as the graph of the quadratic function. It models the path taken b a ball thrown under gravit at an angle to the horizontal, assuming we ignore air resistance. Since the Greeks did not have the notion of a coordinate sstem, their studies were heavil geometric and often hard to understand. With the advent of coordinate geometr, the stud of the parabola became much easier and a full and complete analsis became possible. In this module, we will summarise all of the main features and properties of quadratic equations and functions.

5 A guide for teachers Years 11 and 12 {5} We will also look at a geometric definition of the parabola and its representation using parametric equations. One of the man interesting characteristics of the parabola is its reflective propert, used in collecting light from a distant source. It is not intended that all the material in this module be taught in one topic in the classroom. Our aim here is to present the teacher with a comprehensive overview of quadratic functions and the geometr of the parabola. Content The basic parabola The graph of the quadratic function = 2 is called a parabola. The basic shape can be seen b plotting a few points with = 3, 2, 1,, 1, 2 and 3. = 2 We will refer to the above graph as the basic parabola. As we will see, all parabolas can be obtained from this one b translations, rotations, reflections and stretching. The basic parabola has the following properties: It is smmetric about the -ais, which is an ais of smmetr. The minimum value of occurs at the origin, which is a minimum turning point. It is also known as the verte of the parabola. The arms of the parabola continue indefinitel.

6 {6} Quadratics Transformations of the parabola Translations We can translate the parabola verticall to produce a new parabola that is similar to the basic parabola. The function = 2 + b has a graph which simpl looks like the standard parabola with the verte shifted b units along the -ais. Thus the verte is located at (,b). If b is positive, then the parabola moves upwards and, if b is negative, it moves downwards. Similarl, we can translate the parabola horizontall. The function = ( a) 2 has a graph which looks like the standard parabola with the verte shifted a units along the -ais. The verte is then located at (a,). Notice that, if a is positive, we shift to the right and, if a is negative, we shift to the left. = = 2 2 = ( + 2)2 = ( 1) These two transformations can be combined to produce a parabola which is congruent to the basic parabola, but with verte at (a,b).

7 A guide for teachers Years 11 and 12 {7} For eample, the parabola = ( 3) has its verte at (3,4) and its ais of smmetr has the equation = 3. In the module Algebra review, we revised the ver important technique of completing the square. This method can now be applied to quadratics of the form = 2 + q + r, which are congruent to the basic parabola, in order to find their verte and sketch them quickl. Eample Find the verte of = and sketch its graph. Solution Completing the square, we have = = ( 2) Hence the verte is (2,4) and the ais of smmetr has equation = 2. The graph is shown below. Ais of smmetr = = We can, of course, find the verte of a parabola using calculus, since the derivative will be zero at the -coordinate of the verte. The -coordinate must still be found, and so completing the square is generall much quicker.

8 {8} Quadratics Reflections Parabolas can also be reflected in the -ais. Thus the parabola = 2 is a reflection of the basic parabola in the -ais. = 2 (a,a2) = 2 a (a, a2) a Eample Find the verte of = and sketch its graph. Solution Completing the square, we have = ( ) = ( ( 3) 2 1 ) = 1 ( 3) 2. Hence the verte is (3,1) and the parabola is upside down. The equation of the ais of smmetr is = 3. The graph is shown below. 1 = = 1 ( 3)

9 A guide for teachers Years 11 and 12 {9} Eercise 1 Find the verte of each parabola and sketch it. a = b = Stretching Not all parabolas are congruent to the basic parabola. For eample, the arms of the parabola = 3 2 are steeper than those of the basic parabola. The -value of each point on this parabola is three times the -value of the point on the basic parabola with the same -value, as ou can see in the following diagram. Hence the graph has been stretched in the -direction b a factor of 3. = 32 (a,3a2) = 2 (a,a2) In fact, there is a similarit transformation that takes the graph of = 2 to the graph of = 3 2. (Map the point (, ) to the point ( 1 3, 1 3 ).) Thus, the parabola = 32 is similar to the basic parabola. In general, the parabola = a 2 is obtained from the basic parabola = 2 b stretching it in the -direction, awa from the -ais, b a factor of a. Eercise 2 Sketch the graphs of = and = 2 on the same diagram and describe the relationship between them.

10 {1} Quadratics This transformation of stretching can now be combined with the other transformations discussed above. Once again, the basic algebraic technique is completing the square. Eample Find the verte of the parabola = and sketch its graph. Solution B completing the square, we obtain = ( = ) 2 ( = 2 ( + 1) ) 2 = 2( + 1) Hence the verte is at ( 1,7). The basic parabola is stretched in the -direction b a factor of 2 (and hence made steeper) and translated Rotations The basic parabola can also be rotated. For eample, we can rotate the basic parabola clockwise about the origin through 45 or through 9, as shown in the following two diagrams.

11 A guide for teachers Years 11 and 12 {11} 45º = 2 Algebraicall, the equation of the first parabola is complicated and generall not studied in secondar school mathematics. The second parabola can be obtained from = 2 b interchanging and. It can also be thought of as a reflection of the basic parabola in the line =. Its equation is = 2 and it is an eample of a relation, rather than a function (however, it can be thought of as a function of ). Summar All parabolas can be obtained from the basic parabola b a combination of: translation reflection stretching rotation. Thus, all parabolas are similar.

12 {12} Quadratics This is an important point, since in the module Polnomials, it is seen that the graphs of higher degree equations (such as cubics and quartics) are not, in general, obtainable from the basic forms of these graphs b simple transformations. The parabola, and also the straight line, are special in this regard. For the remainder of the Content section of this module, we will restrict our attention to parabolas whose ais is parallel to the -ais. It will be to these tpes that we refer when we use the term parabola. Intercepts Thus far we have concentrated on the transformational aspects of the graphs of quadratics and the coordinates of the verte. Of course one ma also be interested in the points at which the graph crosses the - and -aes. These are called the intercepts. It is eas to find the -intercept of a parabola it alwas has one b putting =. Not all parabolas, however, have -intercepts. When -intercepts eist, the are obtained b putting = and solving the resulting quadratic equation. As we know, there are man was to solve a quadratic equation, and there are obvious choices of strateg depending on the given problem. If we have alread completed the square to find the verte, then we can use that to quickl find the -intercepts. Eample Find the verte and intercepts of = and sketch the parabola. Solution Completing the square gives = ( + 1) 2 7, so the verte is at ( 1, 7). Putting =, we find that the -intercept is 6. Putting = to find the -intercepts, we have ( + 1) 2 7 = and so ( + 1) 2 = 7, which gives = 1 + 7, =

13 A guide for teachers Years 11 and 12 {13} Completing the square will also tell us when there are no -intercepts. For eample, given the quadratic = , completing the square ields = ( + 3) We can see that is never zero, since the right-hand side is at least 4 for an value of. In some instances especiall when completing the square is harder we ma prefer to use the factor method instead. For eample, the quadratic = factors to = ( 3)( 2), and so we can easil see that the -intercepts are at = 2 and = 3. There are a number of strategies to find the verte. You will have seen in the eamples that, since the graph of the quadratic function is smmetric about the verte, the -intercepts are smmetric about the verte. Hence the average of the -intercepts gives the -coordinate of the verte. This is ver useful in those situations in which we have alread found the -intercepts. For eample, the quadratic function = has its -intercepts at = 2 and = 3. Hence the -coordinate of the verte is the average of these numbers, which is From this we can find the -coordinate of the verte, which is 1 4. Of course, not all parabolas have -intercepts. In the net section, we give a formula for the -coordinate of the verte of = a 2 + b + c. The ais of smmetr All parabolas have eactl one ais of smmetr (unlike a circle, which has infinitel man aes of smmetr). If the verte of a parabola is (k,l), then its ais of smmetr has equation = k. = k (k,l)

14 {14} Quadratics We can find a simple formula for the value of k in terms of the coefficients of the quadratic. As usual, we complete the square: = a 2 + b + c [ = a 2 + b a + c ] a [( = a + b ) 2 c ( b ) 2 ] + 2a a. 2a We can now see that the -coordinate of the verte is b. Thus the equation of the ais 2a of smmetr is = b 2a. We could also find a formula for the -coordinate of the verte, but it is easier simpl to substitute the -coordinate of the verte into the original equation = a 2 + b + c. Eample Sketch the parabola = b finding the verte and the -intercept. Also state the equation of the ais of smmetr. Does the parabola have an -intercepts? Solution Here a = 2, b = 8 and c = 19. So the ais of smmetr has equation = b 2a = 8 4 = 2. We substitute = 2 into the equation to find = 2 ( 2) ( 2) + 19 = 11, and so the verte is at ( 2,11). Finall, putting = we see that the -intercept is 19. There are no -intercepts

15 A guide for teachers Years 11 and 12 {15} Eercise 3 A parabola has verte at (1,3) and passes through the point (3,11). Find its equation. The quadratic formula and the discriminant The quadratic formula was covered in the module Algebra review. This formula gives solutions to the general quadratic equation a 2 + b + c =, when the eist, in terms of the coefficients a, b, c. The solutions are = b + b 2 4ac 2a provided that b 2 4ac., = b b 2 4ac, 2a The quantit b 2 4ac is called the discriminant of the quadratic, often denoted b, and should be found first whenever the formula is being applied. It discriminates between the tpes of solutions of the equation: > tells us the equation has two distinct real roots = tells us the equation has one (repeated) real root < tells us the equation has no real roots. Eercise 4 For what values of k does the equation (4k + 1) 2 6k + 4 = have one real solution? A quadratic epression which alwas takes positive values is called positive definite, while one which alwas takes negative values is called negative definite. Quadratics of either tpe never take the value, and so their discriminant is negative. Furthermore, such a quadratic is positive definite if a >, and negative definite if a <. Δ <, a > Δ <, a <

16 {16} Quadratics Eample Show that the quadratic epression alwas takes positive values for an value of. Solution In this case, a = 4, b = 8 and c = 7. So = ( 8) = 48 < and a = 4 >. Hence the quadratic is positive definite. Eercise 5 For what values of k does the equation (4k + 1) 2 2(k + 1) + (1 2k) = have one real solution? For what values of k, if an, is the quadratic negative definite? Applications to maima and minima Since ever parabola has a verte, the -coordinate of the verte gives the maimum or minimum value of for all possible values of. For eample, the verte of the parabola = 4 2 is (2,4) and the greatest -value is 4. 4 (2,4) 2 4 There are a number of geometric problems and word problems in which the variables are connected via a quadratic function. It is often of interest to maimise or minimise the dependent variable.

17 A guide for teachers Years 11 and 12 {17} Eample A farmer wishes to build a large rectangular enclosure using an eisting wall. She has 4 m of fencing wire and wishes to maimise the area of the enclosure. What dimensions should the enclosure have? Solution 4 2 Eisting wall With all dimensions in metres, let the width of the enclosure be. Then the three sides to be made from the fencing wire have lengths, and 4 2, as shown in the diagram. Thus the area A is given b A = (4 2) = This gives a quadratic relationship between the area and the width, which can be plotted. A Clearl the area will be a maimum at the verte of the parabola. Using = b, we see 2a that the -value of the verte is = 1. Hence the dimensions of the rectangle with maimum area are 1 m b 2 m, and the maimum area of the enclosure is 2 m 2. Note. Differential calculus can be used to solve such problems. While this is quite acceptable, it is easier to use the procedure outlined above. You ma use calculus, however, to solve the following eercise.

18 {18} Quadratics Eercise 6 A piece of wire with length 2 cm is divided into two parts. One part is bent to form a circle while the other is bent to form a square. How should the wire be divided in order that the sum of the areas of the circle and square is minimal? Sum and product of the roots We have seen that, in the case when a parabola crosses the -ais, the -coordinate of the verte lies at the average of the intercepts. Thus, if a quadratic has two real roots α,β, then the -coordinate of the verte is 1 2 (α + β). Now we also know that this quantit is equal to b 2a. Thus we can epress the sum of the roots in terms of the coefficients a,b,c of the quadratic as α + β = b a. In the case when the quadratic does not cross the -ais, the corresponding quadratic equation a 2 + b + c = has no real roots, but it will have comple roots (involving the square root of negative numbers). The formula above, and other similar formulas shown below, still work in this case. We can find simple formulas for the sum and product of the roots simpl b epanding out. Thus, if α,β are the roots of a 2 + b + c =, then dividing b a we have 2 + b a + c a = ( α)( β) = 2 (α + β) + αβ. Comparing the first and last epressions we conclude that α + β = b a and αβ = c a. From these formulas, we can also find the value of the sum of the squares of the roots of a quadratic without actuall solving the quadratic. Eample Suppose α,β are the roots of =. Find the value of α 2 + β 2 and eplain wh the roots of the quadratic cannot be real. Solution Using the formulas above, we have α + β = 2 and αβ = 7 2. Now (α + β)2 = α 2 + β 2 + 2αβ, so α 2 +β 2 = = 3. If the roots were real, then the sum of their squares would be positive. Since the sum of their squares is 3, the roots cannot be real.

19 A guide for teachers Years 11 and 12 {19} Eercise 7 Find the monic quadratic with roots 2 3 5, Eercise 8 Suppose α,β are the roots of =. Find the value of a 1 α + 1 β b 1 α β 2. Note that, in the previous eercise, the desired epressions are smmetric. That is, interchanging α and β does not change the value of the epression. Such epressions are called smmetric functions of the roots. Eercise 9 Suppose α,β are the roots of a 2 + b + c =. Find a formula for (α β) 2 in terms of a, b and c. You ma recall that the arithmetic mean of two positive numbers α and β is 1 2 (α + β), while their geometric mean is αβ. Thus, if a quadratic has two positive real solutions, we can epress their arithmetic and geometric mean using the above formulas. Eercise 1 Without solving the equation, find the arithmetic and geometric mean of the roots of the equation =. Constructing quadratics It is a fundamental fact that two points uniquel determine a line. That is, given two distinct points, there is one and onl one line that passes through these points. How man (distinct) non-collinear points in the plane are required to determine a parabola? The answer is three. This idea ma be epressed b the following theorem: Theorem If two quadratic functions f () = a 2 + b + c and g () = A 2 + B + C take the same value for three different values of, then f () = g () for all values of.

20 {2} Quadratics Proof We start b observing that a quadratic equation can have at most two solutions. Hence, if the quadratic equation d 2 + e + f = has more than two solutions, then d = e = f =. Now, if f () and g () agree for = α, = β and = γ, then f () g () = (a A) 2 + (b B) + (c C ) = when = α, = β and = γ. In that case, the quadratic equation above has three different solutions and so the coefficients are. Thus, a = A, b = B and c = C, and so f () = g () as claimed. Eample Find the equation of the quadratic function whose graph passes through ( 1, 2), (, 3) and (1,6). Solution Suppose the equation is = a 2 +b+c. Substituting the coordinates of the three points, we have 2 = a b + c, 3 = c, 6 = a + b + c. Thus, c = 3 and so a b = 1 and a + b = 3. Therefore a = 1 and b = 2. Hence the function is = Eercise 11 Epress 2 in the form a( 1) 2 + b( 2) 2 + c( 3) 2. Intersection of lines and parabolas In this section we investigate under what conditions a line and a parabola might meet in the plane. It is eas to see that there are three main cases to consider. A line ma: intersect a parabola at two points touch a parabola at one point (in this case we sa the line is tangent to the parabola) not intersect the parabola at all.

21 A guide for teachers Years 11 and 12 {21} The three cases are shown in the diagrams below. There is also a special case when the line is parallel to the -ais. In this case the line will cut the parabola once. Having noted this case, we will not need to consider it further, since the intersection point is easil found. Suppose the line has the form = m +d and the parabola is = a 2 +b +c. Equating, we have a 2 + b + c = m + d = a 2 + (b m) + (c d) =. This gives a quadratic equation. Hence we can conclude: A line can meet a parabola in at most two points. If the discriminant of the resulting quadratic is negative, the line does not meet the parabola at all. If the discriminant of the resulting quadratic is positive, the line meets the parabola in two distinct points. If the discriminant of the resulting quadratic is zero, then the line is tangent to the parabola.

22 {22} Quadratics Eample Discuss the intersection of the line = and the parabola = Solution Equating the two equations, we have = = =. Now the discriminant of this quadratic is b 2 4ac = = and so the equation has onl one solution. Hence the line is tangent to the parabola. Eercise 12 Use the ideas above to show that the line = is tangent to the circle = 5. Focus-directri definition of the parabola The Greeks defined the parabola using the notion of a locus. A locus is a set of points satisfing a given condition. These points will generall lie on some curve. For eample, the circle with centre O and radius r is the locus of a point P moving so that its distance from the point O is alwas equal to r. P r O The locus definition of the parabola is onl slightl more complicated. It is ver important and gives us a new wa of viewing the parabola. One of the man applications of this is the refection principle, which we will look at in a later section. We fi a point in the plane, which we will call the focus, and we fi a line (not through the focus), which we will call the directri. It is easiest to take the directri parallel to the -ais and choose the origin so that it is equidistant from the focus and directri. Thus, we will take the focus at S(, a) and the directri with equation = a, where a >. This is as shown in the following diagram.

23 A guide for teachers Years 11 and 12 {23} S(,a) Focus Directri = a We now look at the locus of a point moving so that its distance from the focus is equal to its perpendicular distance to the directri. Let P(, ) be a point on the locus. Our aim is to find the equation governing the coordinates and of P. P(,) S(,a) Focus PT = PS Directri T a = a The distance PS is equal to 2 + ( a) 2, and we can see from the diagram that the perpendicular distance PT of P to the directri is simpl + a. Since PS = PT, we can square each length and equate, giving 2 + ( a) 2 = ( + a) 2 = 2 = 4a. The last equation ma also be written as = 2 4a and so we see that it is the equation of a parabola with verte at the origin. The positive number a is called the focal length of the parabola. An parabola of the form = A 2 + B +C can be put into the standard form ( p) 2 = ±4a( q), with a >, where (p, q) is the verte and a is the focal length. When a parabola is in this standard form, we can easil read off its verte, focus and directri.

24 {24} Quadratics Eample Find the verte, focal length, focus and directri for the parabola = Solution We complete the square and rearrange as follows: = = = ( 2) 2 1 = ( 2) 2 = 4 1 ( + 1). 4 Hence the verte is (2, 1) and the focal length is 1 4. The focus is then (2, 3 4 ) and the equation of the directri is = 5 4. = S(2, 4 ) (2, 1) 5 = 4 A parabola ma also have its directri parallel to the -ais and then the standard form is ( p) 2 = ±4a( q), with a >. = q a (q,p) S (q + a,p)

25 A guide for teachers Years 11 and 12 {25} Eercise 13 Put the parabola 2 2 = 4 5 into the above standard form (for a directri parallel to the -ais) and thus find its verte, focus and directri, and sketch its graph. Parametric equations of a parabola Since ever parabola is congruent to 2 = 4a, for some choice of a >, we can stud the geometr of ever parabola b confining ourselves to this one! This shows the power of both coordinate geometr and transformations. Imagine a particle moving in the plane along a curve C as shown. P(,) = (f(t),g(t)) The - and -coordinates of the particle can be thought of as functions of a new variable t, and so we can write = f (t), = g (t), where f, g are functions of t. In some phsical problems t is thought of as time. The equations = f (t) = g (t) are called the parametric equations of the point P(, ) and the variable t is called a parameter. It is often ver useful to take a cartesian equation = F () and introduce a parameter t so that the - and -coordinates of an point on the curve can be epressed in terms of this parameter.

26 {26} Quadratics Here is a ver simple eample of a parametrisation. Suppose we begin with the line whose equation is = 3 4. We can introduce a new variable t and write = t. Then we have = 3t 4. Thus we can rewrite the equation = 3 4 in parametric form as = t = 3t 4. Eercise 14 Eliminate the parameter t from the equations = 2t 3 = 5 3t and describe the resulting function. Another common parametrisation is that used for a circle. The cartesian equation of a circle of radius r centred at the origin is = r 2. We can put = r cosθ, = r sinθ, and we see that = r 2 cos 2 θ + r 2 sin 2 θ = r 2 (cos 2 θ + sin 2 θ) = r 2, so these satisf the original equation. The angle θ here is the parameter and measures the angle between the positive -ais and a radial line to the circle. P(r cos θ, r sin θ) r θ Thus we can describe the circle parametricall b = r cosθ = r sinθ, where θ < 2π.

27 A guide for teachers Years 11 and 12 {27} The parabola 2 = 4a, where a >, ma also be parametrised and there are man was to do this. We will choose the parametrisation = 2at = at 2. The parameter in this case is t. This is the standard parametrisation for the parabola 2 = 4a and the reason for this parametrisation will become apparent later in this module. You should check that this parametrisation actuall satisfies the original equation. Eample Epress the parabola 2 = 12 parametricall. Solution In this case we see that a = 3 and so the general point on the parabola can be epressed parametricall as = 6t = 3t 2. Chords, tangents and normals Chords Again, we consider the parabola 2 = 4a, for some a >, and the parametrisation = 2at = at 2. When we take two distinct points P,Q on the parabola, we can join them to form a line segment called a chord. Sometimes we will refer to the line through P and Q as the chord. Let the value of the parameter t at the points P and Q be p and q respectivel. Thus P has coordinates (2ap, ap 2 ) and Q has coordinates (2aq, aq 2 ).

28 {28} Quadratics 2 = 4a Q(2aq aq2) P(2ap ap2) A chord of the parabola. The gradient of the line PQ is given b ap 2 aq 2 2ap 2aq = a(p2 q 2 ) 2a(p q) = p + q 2. So the gradient of the chord is the average of the values of the parameter. (One of the man reasons this parametrisation was chosen.) Hence the equation of the chord is ap 2 = p + q ( 2ap), 2 which rearranges to = p + q 2 apq. It is best to reconstruct this equation each time rather than memorise it. A chord which passes through the focus of a parabola is called a focal chord. A given chord will be a focal chord if the point (, a) lies on it. Substituting these coordinates into the equation of the chord above we have a = p + q 2 apq, which is equivalent to pq = 1. This gives us a simple condition to tell when the chord joining two points is a focal chord. Thus PQ is a focal chord if and onl if pq = 1. The basic fact should be committed to memor.

29 A guide for teachers Years 11 and 12 {29} Eample Suppose P(2ap, ap 2 ) and Q(2aq, aq 2 ) are two points ling on the parabola 2 = 4a. Prove that, if OP and OQ are perpendicular, then pq = 4. Solution The gradient of OP is ap2 2ap = p 2 and similarl the gradient of OQ is q. Since these lines 2 are perpendicular, the product of their gradients is 1 and so p 2 q = 1, which implies 2 that pq = 4. Note that, in this case, PQ can never be a focal chord. Eercise 15 Suppose P(2ap, ap 2 ) and Q(2aq, aq 2 ) are two points ling on the parabola 2 = 4a. Prove that, if PQ is a focal chord, then the length of the chord PQ is given b a ( p + 1 p ) 2. Tangents We have seen that the gradient of the chord joining two distinct points P(2ap, ap 2 ) and Q(2aq, aq 2 ) on the parabola 2 = 4a is 1 2 (p + q). If we now imagine the point Q moving along the parabola to the point P, then the line through P and Q will become a tangent to the parabola at the point P. Putting q = p, we see that the gradient of this tangent is 1 2 (p + p) = p. 2 = 4a P(2ap ap2) A tangent to the parabola.

30 {3} Quadratics We could also find the gradient of the tangent b differentiating = 2 4a and substituting = 2ap. In either case, the gradient of the tangent to 2 = 4a at the point P(2ap, ap 2 ) is p. (This is the main reason for this choice of parametrisation.) Hence the equation of the tangent to the parabola at P is given b ap 2 = p( 2ap) = = p ap 2. This formula is best derived each time it is needed rather than memorised. Normals The line through a point P perpendicular to the tangent to the curve at P is called the normal to the curve at P. Since the normal and tangent are perpendicular, the product of their gradients is 1 and so the gradient of the normal at P(2ap, ap 2 ) is 1 p. Hence the equation of the normal at P is ap 2 = 1 p ( 2ap) = + p = 2ap + ap3. Once again, this is best derived each time rather than memorised. 2 = 4a P(2ap ap2) normal Eample Find the point of intersection of the tangents to the parabola 2 = 4a at the points P(2ap, ap 2 ) and Q(2aq, aq 2 ). What can be said about this intersection point if the chord PQ is a focal chord?

31 A guide for teachers Years 11 and 12 {31} Solution The equations of the tangents at P and Q are respectivel = p ap 2 and = q aq 2. Equating these we find p ap 2 = q aq 2 = = a(p + q). Substituting back we find that the -coordinate is = apq and so the point of intersection is ( a(p + q), apq ). Now, if PQ is a focal chord, then pq = 1 and so the point of intersection becomes ( a(p 1 p ), a). While the -coordinate of this point can var, the -coordinate is alwas a and so the point of intersection alwas lies on the directri. Eercise 16 Find the point of intersection R of the normals to the parabola 2 = 4a at the points P(2ap, ap 2 ) and Q(2aq, aq 2 ). Show that, if PQ is a focal chord, then the point R alwas lies on a certain parabola, and find the equation of this parabola. The reflection propert of the parabola While the parabola has man beautiful geometric properties as we have seen, it also has a remarkable propert known as the reflection propert, which is used in such diverse places as car headlights and satellite dishes. A car headlight emits light from one source, but is able to focus that light into a beam so that the light does not move off in all directions as lamplight does but is focused directl ahead. Satellite dishes do the opposite. The collect electromagnetic ras as the arrive and concentrate them at one point, thereb obtaining a strong signal. Headlight reflectors and satellite dishes are designed so their cross-sections are parabolic in shape and the point of collection or emission is the focus of the parabola. To show how this works we need a basic fact from phsics: when light (or an electromagnetic radiation) is reflected off a surface, the angle of incidence equals the angle of reflection. That is, when a light ra bounces off the surface of a reflector, then the angle between the light ra and the normal to the reflector at the point of contact equals the angle between the normal and the reflected ra. This is shown in the following diagram.

32 {32} Quadratics reflected ra normal light ra Note that this implies that the angle between the ra and the tangent is also preserved after reflection, which is a more convenient idea for us here. Let P(2ap, ap 2 ) be a point on the parabola 2 = 4a with focus at S and let T be the point where the tangent at P cuts the -ais. Suppose PQ is a ra parallel to the -ais. Our aim is to show that the line PS will satisf the reflection propert, that is, QPB is equal to SPT. 2 = 4a Q B P(2ap ap2) S(,a) tangent at P (, ap2) T Notice that, since QP is parallel to ST, QPB is equal to ST P, so we will show that ST P = SPT. Now S has coordinates (, a), and T has coordinates (, ap 2 ), obtained b putting = in the equation of the tangent at P. Hence SP 2 = (2ap ) 2 + (ap 2 a) 2 = a 2 (p 2 + 1) 2

33 A guide for teachers Years 11 and 12 {33} after a little algebra. Also, ST 2 = (a + ap 2 ) 2 = a 2 (1 + p 2 ) 2 and so SP = ST. Hence ST P is isosceles and so ST P = SPT. Thus the reflection propert tells us that an ra parallel to the ais of the parabola will bounce off the parabola and pass through the focus. Conversel, an ra passing through the focus will reflect off the parabola in a line parallel to the ais of the parabola (so that light emanating from the focus will reflect in a straight line parallel to the ais). S Links forward Conic sections The parabola is one of the curves known as the conic sections, which are obtained when a plane intersects with a double cone. The non-degenerate conic sections are the parabola, ellipse, hperbola and circle. Parabola 2 = 4a Ellipse b 2 2 a + = 1 2 b 2 a a b Hperbola a a 2 2 a 2 b = 1 2 Circle a = a2 a a a

34 {34} Quadratics The degenerate case produces also a line, a pair of lines and a point. Locus definitions The ellipse and hperbola can be defined as a locus in various was. Suppose we fi two distinct points F 1 and F 2 in the plane. Then the ellipse can be defined as the locus of points P such that the sum of the distances PF 1 + PF 2 is a fied constant 2a. (Imagine pinning down the ends of a piece of string at the points F 1 and F 2, then using a pencil to pull the string tight and allowing the pencil to move around the page. The sum of the distances is then the length of the string, which remains constant.) P F1 F2 PF1 + PF2 = 2a Locus definition of the ellipse. Thus, if we set the points to be F 1 ( c,) and F 2 (c,), with < c < a, then the sum of the distances is ( c) ( + c) = 2a. Taking the second square root onto the other side, squaring, simplifing and squaring again, we obtain the equation 2 a a 2 c 2 = 1. If we write b 2 = a 2 c 2, then the equation becomes 2 a b 2 = 1. This is the cartesian equation of the ellipse.

35 A guide for teachers Years 11 and 12 {35} Eercise 17 Perform the algebra suggested above to derive the equation of the ellipse. If, instead, we require the difference of the distances PF 1 PF 2 to be a fied constant 2a, then we obtain part of a hperbola. The equation has the form 2 a 2 2 b 2 = 1. P 2 2 a = 1 2 b 2 F1 PF1 PF2 = 2a F2 Locus definition of the hperbola. Locus definition of the hperbola. Focus-directri definitions Recall that the parabola can be defined as the locus of the point P moving so that the distance PS, from P to the focus S, is equal to the perpendicular distance of P to the P D directri l. Thus, if D is the nearest point on l to P, we have PS = PD, which we can write as PS PD = 1. The ellipse and the hperbola also have focus-directri definitions. We choose a fied point S, which will be the focus, and a line l (not containing S). Take the locus of all points P such that, if D is the nearest point on l to P, then the ratio of the distances PS PD is equal to a fied positive number e, called the eccentricit. PS PD = e S l If e is equal to 1, then we have, of course, the parabola. The name parabola comes from the Greek meaning fall beside since, in the case of the parabola, the distance PS is equal to (i.e., falls beside) the distance PD. If e is less than 1, then we obtain an ellipse (from the Greek meaning fall short of the original sense was that PS falls short of PD). If e is greater than 1, then we have a hperbola (from the Greek meaning eceed the original sense was that PS eceeds PD).

36 {36} Quadratics D1 l1 P1 P2 l2 D2 S1 S2 Focus-directri definition of the ellipse. As we saw above, the cartesian equation of the ellipse is 2 a b 2 = 1. It can be shown that the eccentricit e of the ellipse is connected to the quantities a and b b the equation b 2 = a 2 (1 e 2 ). The foci can be written as (±ae,), while the equations of the directrices are = ± a e. Similarl, for the hperbola 2 a 2 2 b 2 = 1, the eccentricit is connected to the quantities a and b b the equation b 2 = a 2 (e 2 1). The ellipse also has a reflective propert in that an light beam emanating from one of the two foci will reflect off the curve and pass through the other focus. This idea has been widel eploited in acoustics and optics. Second-degree equations Algebraicall, ever second-degree equation of the form A 2 + B 2 +C + D + E = F can be shown, b means of linear changes of variables, to be similar to one of the conic sections (including the degenerate cases) we have listed above. For eample, the equation = can be written as ( 2) 2 + ( + 3) 2 = 13, which is a circle. Similarl, the equation 2 2 = can be written as ( )( + ) = and so as = or =, which is a pair of lines. An equation such as = 1 in fact represents a hperbola, but one which is rotated about the origin. The theor of matrices is generall used to analse such equations, but this is not part of secondar school mathematics.

37 A guide for teachers Years 11 and 12 {37} Three-dimensional analogues Each of these curves has a 3-dimensional analogue. For eample, the parabolic reflector is an eample of a paraboloid. The basic paraboloid has equation z = and its cross-section in the z plane is a parabola. Similarl, an ellipsoid has the basic shape of a football, with equation 2 a b 2 + z2 c 2 = 1. In the case when a = b = c, we have a sphere. z Paraboloid z Ellipsoid There are also 3-dimensional hperboloids of one and two sheets. z z

38 {38} Quadratics Histor and applications The histor of the algebraic aspects of quadratics was covered in the module Quadratic equations (Years 9 1). Historicall, the geometric properties of the parabola were studied b the ancient Greeks. Menaechmus (c BCE) appears to have been the first to stud the properties of the parabola along with the hperbola and ellipse. These curves arose through the stud of the cone. Apollonius of Perga ( BCE) wrote a major treatise on the conic sections, but his definition of the parabola was in terms of ratios and without an use of coordinates. The focus-directri definition used in this module goes back to Pappus (29 35 CE). It was not until the advent of coordinate geometr at the time of Descartes that further real progress could be made, although the reflective propert was known to the Greeks. Galileo ( ) realised that the motion of a projectile under gravit formed a parabolic path. Kepler ( ) was the first to realise that the planets revolve around the sun in orbits that are ver close to being elliptical. Newton proved this using his universal law of gravitation and the calculus. Path of a projectile. The reflective properties of the parabola and the ellipse have been eploited in architecture to obtain remarkable acoustic properties in large church buildings, and later in the design of powerful telescopes and reflectors.

39 A guide for teachers Years 11 and 12 {39} Answers to eercises Eercise 1 a b 3 ( 3, 4) Eercise 2 The arms of the parabola = 2 are steeper. = 2 = Eercise 3 = 2( 1) Eercise 4 The discriminant is 36k 2 16(4k + 1) and we set this equal to zero in order to obtain one real solution to the given equation. This gives 9k 2 16k 4 =, which has solutions k = 2 or k = 2 9.

40 {4} Quadratics Eercise 5 The discriminant is = 36k 2. One real solution occurs when =, which is when k =. The quadratic is never negative definite, since for all values of k. Eercise 6 Divide the wire into 4 cm for the circle and (2 4) cm for the square. (The 4 is to make the algebra easier.) The total area is then S = 42 π + (5 )2. Since the leading coefficient is positive, the parabola has a minimum. We set ds d = to find the stationar point, which occurs when = 5π 2π 4+π. Hence, we use 4+π cm for the circle and 8 4+π cm for the square. Eercise Eercise 8 a 4 7 b Eercise 9 We have (α β) 2 = (α + β) 2 4αβ ( = b ) 2 4c a a = b2 4ac a 2. (Note that, if a = 1, then (α β) 2 is the discriminant of the quadratic. Similarl, we can define the discriminant of a monic cubic with roots α,β,γ to be (α β) 2 (α γ) 2 (β γ) 2, and so on for higher degree equations.) Eercise 1 The arithmetic mean is ± 145 6!) Eercise 11 and the geometric mean is 2. (Note that the actual roots are Successivel substitute = 1, = 2, = 3 and solve the resulting equations for a,b,c to obtain 2 = 3( 1) 2 3( 2) 2 + ( 3) 2.

41 A guide for teachers Years 11 and 12 {41} Eercise 12 Substituting the line into the circle we obtain =. The discriminant of this quadratic is and so the line is tangent to the circle at (2,1). Eercise 13 The parabola can be written as ( 1) 2 = 4( 1). So the verte is (1,1), the focal length is a = 1, the focus is (2,1), and the -ais is the directri. 1 S(2,1) 1 Eercise 14 The line = 1. Eercise 15 If PQ is a focal chord, then pq = 1. Hence we can write the coordinates of Q as ( 2a p, a p 2 ). Thus ( PQ 2 = 2ap + 2a ) 2 + (ap 2 a ) 2 p p 2 = 4a 2 p 2 + 8a 2 + 4a2 p 2 = a 2( p 4 + 4p p p 4 ) = a 2( p + 1 p ) 4, + a2 p 4 2a 2 + a2 p 4 where we recall the numbers 1,4,6,4,1 from Pascal s triangle and the binomial theorem. ( So PQ = a p + 1 ) 2. p

42 {42} Quadratics Eercise 16 The equations of the normals at P and Q are respectivel + p = 2ap + ap 3 and + q = 2aq + aq 3. Subtracting we have (p q) = a(p 3 q 3 ) + 2a(p q) = a(p q)(p 2 + q 2 + pq + 2) so = a(p 2 + q 2 + pq + 2). Substituting back we have = apq(p + q) after simplifing. Hence R is the point ( apq(p + q), a(p 2 + q 2 + pq + 2) ). Now, if PQ is a focal chord, then pq = 1. Thus the - and -coordinates of R become = a(p +q) and = a(p 2 +q 2 +1) = a ( (p +q) 2 +3 ), since pq = 1. We can then eliminate p and q to obtain ( ( ) ) 2 = a + 3 = 2 a a + 3a, which is a parabola. Eercise 17 Squaring ( c) = 2a ( + c) 2 + 2, we have a ( + c) = a 2 + c. Squaring again and rearranging, we arrive at 2 (a 2 c 2 ) + a 2 2 = a 2 (a 2 c 2 ). Dividing b a 2 (a 2 c 2 ) gives the desired equation.

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