FP1: Chapter 3 Coordinate Systems Dr J Frost (jfrost@tiffin.kingston.sch.uk) Last modified: 29th August 2015

Conic Sections The axis of the parabola is parallel to the side of the cone. In FP1 and FP3, we’ll be examining different types of curves. All the ones you’ll see can be obtained by taking ‘slices’ of a cone (known as a conic section). C2: Circles FP1: Rectangular Hyperbolas FP3 Ellipses FP1: Parabolas FP3: Hyperbolas

Parametric vs Cartesian Equations ! A Cartesian equation is one which says how the 𝑥, 𝑦 (𝑧 etc.) values are related for any point on the line. Examples: 𝑥+2𝑦=1 𝑥 2 −3 𝑦 2 =2 ! A parametric equation is one where each of 𝑥, 𝑦 (𝑧 etc.) are expressed in terms of an independent variable. Examples: 𝑥= tan 2 𝜃 𝑦= sin 𝜃

Example (This has been used as a Computer Science interview question at both Oxford and Cambridge) A ladder of length 2l is initially vertical up against a wall. The ladder gradually slides down to the floor. Determine the equation of the trajectory of the midpoint of the ladder. x l Cartesian Equation: (Hint: Pythagoras) (x,y) ? If we draw a line down from (x,y), we get a right-angled triangle. x2 + y2 = l2 i.e. The trajectory is a circle with radius l. l  y Parametric Equation: (Based on a parameter  we could introduce for the angular inclination of the ladder) Cartesian Variables: x, y Parametric Variables:  Constants: l ? ? ? ? By simple trigonometry: x = cos , y = sin 

Sketching Parametric Curves Sketch the curve with equation 𝑥=𝑎 𝑡 2 , 𝑦=2𝑎𝑡, 𝑡∈ℝ, where a is a positive constant. Let’s just try a few values for the parameter and see what coordinates we get... t -3 -2 -1 1 2 3 x 9a 4a a y -6a -4a -2a 2a 6a ? ? 6a 4a 2a O -2a -4a -6a 2a 4a 6a 8a

Converting Parametric  Cartesian We can use either substitution or elimination to turn parametric equations into a Cartesian one. Find the Cartesian equation for the parametric equations x = at2, y = 2at, where a is a positive constant. ? y2 = 4ax Find the Cartesian equation for the parametric equations x = ct, y = c/t, where c is a positive constant. ? y = c2/x We could either have obtained this by substituting c, or by observing that the t’s cancel when we multiply x by y.

Exercise 3A Create a table of values for the parametric equations 𝑥=2 𝑡 2 , 𝑦=4𝑡, as t varies between -4 and 4, in unit increments (include also t = 0.5 and t = -0.5). Hence draw the curve. Find the Cartesian equation of the curves given by these parametric equations. a) 𝑥=𝑡, 𝑦= 1 𝑡 → 𝒙𝒚=𝟏 d) 𝑥= 𝑡 5 , 𝑦= 3 𝑡 →𝒙𝒚= 𝟏 𝟐𝟓 1 5 ? ? ? 6 Sketch the curve with parametric equations 𝑥=3𝑡, 𝑦= 3 𝑡 by finding the Cartesian equation. ? 𝒚= 𝟗 𝒙 (Appropriate reciprocal graph sketch!) 4 Find the Cartesian equation of the curves given by these parametric equations. a) 𝑥=5 𝑡 2 , 𝑦=10𝑡 → 𝒚 𝟐 =𝟐𝟎𝒙 h) 𝑥=6𝑡, 𝑦=3 𝑡 2 → 𝒙 𝟐 =𝟏𝟐𝒚 ? ?

Parabola - Recap y = ax2 + bx + c y = ax2 y2 = ax You may already be familiar that the name of the line governed by a quadratic equation is known as a parabola. For any vertically aligned parabola: y = ax2 + bx + c If we consider just those just centred at the origin, we know its equation will be of the form: y = ax2 However, in this chapter we’ll only be considering parabolas which are horizontally aligned, by just swapping x and y. We’ll also only consider those where the constant a is positive: y2 = ax x y y x x y x

Loci – GCSE recap ! No need to write this down A locus of points is a set of points satisfying a certain condition. ? Loci involving: Thing A Thing B Interpretation Resulting Locus of points Point - A given distance from point A A Reveal Line - A given distance from line A A Reveal Perpendicular bisector Point Point Equidistant from 2 points. Reveal A B In the context of a parabola, we call this line the directrix. Line Line Equidistant from 2 lines Angle bisector A Reveal ...and this point the focus of the parabola. B Point Line Equidistant from point A and line B A Parabola Reveal B

Parabola 𝒙=𝒂 𝒕 𝟐 , 𝒚=𝟐𝒂𝒕, 𝒕∈ℝ 𝒚 𝟐 =𝟒𝒂𝒙 ! Write all this down A horizontally-aligned parabola centred at the origin has equations: (for some positive constant 𝑎) Parametric: 𝒙=𝒂 𝒕 𝟐 , 𝒚=𝟐𝒂𝒕, 𝒕∈ℝ Cartesian: 𝒚 𝟐 =𝟒𝒂𝒙 A general point 𝑃 on this curve has coordinates 𝑃 𝑥,𝑦 or 𝑃 𝑎 𝑡 2 , 2𝑎𝑡 y A parabola is a locus of points such that the distance from any point to the focus is the same as the distance to the directrix. 𝑃 𝑥,𝑦 𝑥 = −𝑎 Focus: 𝒂,𝟎 Directrix: Line 𝑥=−𝑎 Vertex: 𝟎,𝟎 x −𝑎 𝑎 FOCUS VERTEX AXIS OF SYMMETRY DIRECTRIX

Equations of Parabolas Find an equation of the parabola with focus (7,0) and directrix x + 7 = 0 a = 7, so y2 = 28x ? Q ... and with focus ( 3 4 , 0) and directrix 𝑥=− 3 4 y2 = √3 x ? Q Find the coordinates of the focus and an equation for the directrix of a parabola with equation: 𝑦 2 =24𝑥 𝑦 2 = 32 𝑥 Focus: (6, 0) Directrix: x = -6 Focus: (√2, 0) Directrix: x = - √2 ? ? Quickfire Questions: Equation: y2 = 16x y2 = 100x y2 = 24x x2 = 12y y = x2 Focus: (4, 0) (25,0) (6,0) (0,3) (0, 0.25) Directrix: x = -4 x = -25 x = -6 y = -3 y = -0.25 ? ? ? ? ? ? ? ? You wouldn’t be asked this in an exam. ? ?

Proof Can we prove that the equation of a parabola with locus (a,0) and directrix x = -a is y2 = 4ax? (Hint: express algebraically the distances PX and PS) y X P(x,y) ? PX = x + a PS = √[(x-a)2 + y2] So (x-a)2 + y2 = (x+a)2 This simplifies to y2 = 4ax x = -a S x -a a A challenge for your own time: Can you generalise this, and find the parabola for any focus (q,r) and any directrix y = ax + b?

Exercise 3B (In textbook)

Coordinate Geometry involving Parabolas A point P(8, -8) lies on the parabola C with equation y2 = 8x. The point S is the focus of the parabola. The line l passes through S and P. Find the coordinates of S. (2, 0), as we just quarter the 8. Find a equation for l, giving your answers in the form ax + by + c = 0, where a, b, c are integers. m = -8/6 = -4/3 Using point S: y – 0 = -4/3(x – 2) Rearranging: 4x + 3y – 8 = 0 The line l meets the parabola C again at the point Q. The point M is the mid-point of PQ. Find the coordinates of Q. l: 4x + 3y – 8 = 0 C: y2 = 8x Solving simultaneously gives us (1/2, 2) Find the coordinates of M. (17/4, -3) Draw a sketch showing parabola C, the line l and the points P, Q, S and M. ? ? ? ? y C: y2 = 8x Q(0.5, 2) ? S(2, 0) x M(17/4, -3) L: 4x + 3y – 8 = 0 P(8, -8)

Exercise 3C Q4 onwards.

Hyperbolas A hyperbola is a different kind of curve which we’ll more fully explore in FP3. Source: Wikipedia FP3 preview: We know that the equation of a circle with unit radius is: x2 + y2 = 1 The corresponding hyperbola would be: x2 – y2 = 1 and would look like the red curves on the left. A hyperbola has TWO focal points (F1 and F2) rather than one, and two directrices D1 and D2. Unlike parabolas, where the distance to the directrix and focus was equal, there’s now a (constant) factor difference, denoted by e (known as the ‘eccentricity’).

Rectangular Hyperbolas ! A rectangular hyperbola is a hyperbola whose asymptotes are perpendicular. 𝑥 2 4 − 𝑦 2 9 =1 Example: x2 – y2 = 1 The asymptotes of a hyperbola are not necessarily perpendicular y = (3/2)x y = -(3/2)x

Rectangular Hyperbolas In FP1, you only need to know about rectangular hyperbola whose asymptotes are vertical and horizontal. You previously identified this type of equation at GCSE as a: Recriprocal equation! ? Source: WolframAlpha.com “Reciprocal graph” is not a formal mathematical name for the line represented by equation y = k/x. We call the line a (rectangular) hyperbola, as the online calculator WolframAlpha.com identifies it:

Rectangular Hyperbola ! Write all this down A rectangular hyperbola with asymptotes x = 0 and y = 0, has the equations: Parametric: 𝒙=𝒄𝒕, 𝒚= 𝒄 𝒕 , 𝒕∈ℝ, 𝒕≠0 Cartesian: 𝒚= 𝒄 𝟐 𝒙 or 𝒙𝒚= 𝒄 𝟐 where c is a positive constant. A general point P on this curve has coordinates 𝑃 𝑥,𝑦 or 𝑃 𝑐𝑡, 𝑐 𝑡 y (Note: You do not need to know how to find the vertices, foci or directrices) x

Equations of Tangents and Normals The point P, where x = 2, lies on the rectangular hyperbola H with equation xy = 8. Find: The equation of the tangent T. The equation of the normal N to H at the point P, giving your answer in the form 𝑎𝑥+𝑏𝑦+𝑐=0. Bro Tip: This requires nothing more than C1 knowledge! ? y = 8x-1 dy/dx = -8x-2 = -8/x2 When x = 2, y = 4, mT = -2, mN = 1/2, So equation of tangent: y – 4 = -2(x – 2) This becomes 2x + y – 8 = 0 Equation of normal: x – 2y + 6 = 0

Equations of Tangents and Normals The distinct points A and B, where x = 3 lie on the parabola C with equation 𝑦 2 =27𝑥. The line l1 is the tangent to C at A and the line l2 is the tangent to C at B. Given that at A, y > 0, Find the coordinates of A and B. Draw a sketch showing the parabola C. Indicate A, B, l1 and l2. Find equations for l1 and l2, giving your answer in the form 𝑎𝑥+𝑏𝑦+𝑐=0. y2 = 81, so y = 9 ? c) 𝑦=± 3 𝑥 1 2 𝑑𝑦 𝑑𝑥 =± 3 2 𝑥 𝑚 𝑙 1 = 3 2 … 𝑙 1 :3𝑥−2𝑦+9=0 𝑙 2 :3𝑥+2𝑦+9=0 ? ? l1 B(3,9) B(3,-9) l2

Exercise 3D Odd numbered questions.

Coordinate Geometry using Parametric Equations The point P(at2, 2at) lies on the parabola C with equation y2 = 4ax, where a is a positive constant. Show that an equation of the normal to C at P is y + tx = 2at + at3 ? y = 2√a x1/2 dy/dx = √a / √x At P, x = at2, so dy/dx = 1/t mT = 1/t, so mN = -t y – 2at = -t(x – at2) ...some rearrangement y + tx = 2at + at3 Bro Tip: To find tangents/normals when you have a point expressed parametrically… ! STEP 1: Find 𝑑𝑦 𝑑𝑥 using the Cartesian equation. STEP 2: Sub in 𝑥 (e.g. 𝑥=𝑎 𝑡 2 ) to get gradient just in terms of 𝑡 and constants.Find − 1 𝑚 if necessary. STEP 3: Use 𝑦− 𝑦 1 =𝑚 𝑥− 𝑥 1

Another Example Q The point 𝑃 𝑐𝑡, 𝑐 𝑡 lies on the rectangular hyperbola H with equation xy = c2 where c is a positive constant. Show that an equation of the tangent to H at P is 𝑥+ 𝑡 2 𝑦=2𝑐𝑡 A rectangular hyperbola G has equation 𝑥𝑦=9. The tangent to G at the point A and the tangent to G at the point B meet at the point (-1, 7). Find A and B. ? 𝑦= 𝑐 2 𝑥 −1 ∴ 𝑑𝑦 𝑑𝑥 =− 𝑐 2 / 𝑥 2 At P, x = ct and hence we find 𝑚 𝑇 =− 1 𝑡 2 𝑦− 𝑐 𝑡 =− 1 𝑡 2 𝑥−𝑐𝑡 ... 𝑥+ 𝑡 2 𝑦=2𝑐𝑡 For G, c = 3. Using general equation for G, we find t = -1/7, 1 P has coordinates (ct, c/t) = (3t, 3/t) For to two values of t, this gives us coordinates (-3/7, -21) and (3, 3).

Exercise 3E Odd numbered questions.

Summary A parabola is: a locus of points such that the distance from any point to the focus is the same as the distance to the directrix. PARABOLA y ? X 𝑃(𝑥,𝑦) 𝑥=−𝑎 For a parabola: Cartesian equation: 𝑦 2 =4𝑎𝑥 Parametric equations: (based on parameter t) x = at2 y = 2at ? S 𝑎 ? If 𝒚 𝟐 =𝟒𝟎𝒙 then: Directrix: x = -10 Focus: (10, 0) Vertex: (0, 0) ? ? ? A rectangular hyperbola is a hyperbola with perpendicular asymptotes, and if these asymptotes are vertical and horizontal: Cartesian equation: 𝑦= 𝑐 2 𝑥 Parametric equation: 𝑥=𝑐𝑡, 𝑦= 𝑐 𝑡 ? ?

Parabolas in real life If rays are fired at the parabola parallel to its axis of symmetry, then the reflected rays will all pass the focus*. This is known as a parabolic reflector, and has obvious applications to satellite dishes, where a receiver is placed at the focus to receive the waves. * The proof is based on the fact that the distance of points on the parabola to the focus and directrix are the same. The trajectory of a projectile can be described using a quadratic equation, and hence the shape is parabolic. Zero gravity is achieved by a certain parabolic trajectory. For this reason, rollercoaster humps often have this shape.

Parabolas in real life caternary When a cable is hung between two points and hangs under only its own weight (such that the force at any point on the cable acts in the direction of the cable from tension), the shape is not a parabola, but a different type of curve known as a caternary.

Parabolas in real life parabola However, if the cable is connected to the deck of the bridge, there are additional forces on the cable – the tension from holding up the bridge. If the weight of the deck is evenly distributed across the curve, the shape of the curve becomes parabolic.