# FP1: Chapter 3 Coordinate Systems Dr J Frost Last modified: 27 th January 2014.

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FP1: Chapter 3 Coordinate Systems Dr J Frost (jfrost@tiffin.kingston.sch.uk) Last modified: 27 th January 2014

Conic Sections 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:Parabolas FP1:Rectangular Hyperbolas FP3:Hyperbolas FP3Ellipses The axis of the parabola is parallel to the side of the cone.

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 – 3y 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  y = sin 

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

Sketching Parametric Curves t-3-20123 x9a4aa0a 9a y-6a-4a-2a02a4a6a Let’s just try a few values for the parameter and see what coordinates we get... ? 2a 4a 6a 8a 6a 4a 2a O -2a -4a -6a ?

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 = at 2, y = 2at, where a is a positive constant. y 2 = 4ax Find the Cartesian equation for the parametric equations x = ct, y = c/t, where c is a positive constant. y = c 2 /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 1 4 5 6 ? ? ? ? ? ?

Parabola - Recap 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 = ax 2 + bx + c If we consider just those just centred at the origin, we know its equation will be of the form: y = ax 2 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: y 2 = ax x y x y x x y

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

Parabola Parametric: Cartesian: x y FOCUS DIRECTRIX VERTEX 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.  Write all this down AXIS OF SYMMETRY

Equations of Parabolas Find an equation of the parabola with focus (7,0) and directrix x + 7 = 0 a = 7, so y 2 = 28x Q Q y 2 = √ 3 x Q Focus: (6, 0) Directrix: x = -6 Focus: ( √ 2, 0) Directrix: x = - √ 2 ? ? ? ? Quickfire Questions: Equation: y 2 = 16x y 2 = 100x y 2 = 24x x 2 = 12y y = x 2 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 y 2 = 4ax? Can we prove that the equation of a parabola with locus (a,0) and directrix x = -a is y 2 = 4ax? x = -a -a y P(x,y) a (Hint: express algebraically the distances PX and PS) X S PX = x + a PS = √ [(x-a) 2 + y 2 ] So (x-a) 2 + y 2 = (x+a) 2 This simplifies to y 2 = 4ax ? 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? x

Exercise 3B To Do

Coordinate Geometry involving Parabolas A point P(8, -8) lies on the parabola C with equation y 2 = 8x. The point S is the focus of the parabola. The line l passes through S and P. a)Find the coordinates of S. (2, 0), as we just quarter the 8. b)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 c)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: y 2 = 8x Solving simultaneously gives us (1/2, 2) d)Find the coordinates of M. (17/4, -3) e)Draw a sketch showing parabola C, the line l and the points P, Q, S and M. y x C: y 2 = 8x L: 4x + 3y – 8 = 0 Q(0.5, 2) S(2, 0) M(17/4, -3) 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: x 2 + y 2 = 1 The corresponding hyperbola would be: x 2 – y 2 = 1 and would look like the red curves on the left. A hyperbola has TWO focal points (F 1 and F 2 ) rather than one, and two directrices D 1 and D 2. 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 The asymptotes of a hyperbola are not necessarily perpendicular Example: x 2 – y 2 = 1  A rectangular hyperbola is a hyperbola whose asymptotes are 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! “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: “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: Source: WolframAlpha.com ?

Rectangular Hyperbola  Write all this down A rectangular hyperbola with asymptotes x = 0 and y = 0, has the equations: Parametric: Cartesian: (Note: You do not need to know how to find the vertices, foci or directrices) where c is a positive constant. y x

Equations of Tangents and Normals Q Bro Tip: This requires nothing more than C1 knowledge! y = 8x -1 dy/dx = -8x -2 = -8/x 2 When x = 2, y = 4, m T = -2, m N = 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 Q a)y 2 = 81, so y =  9 b) l1l1 l2l2 B(3,-9) B(3,9) ? ? ?

Exercise 3D Odd numbered questions.

Coordinate Geometry using Parametric Equations The point P(at 2, 2at) lies on the parabola C with equation y 2 = 4ax, where a is a positive constant. Show that an equation of the normal to C at P is y + tx = 2at + at 3 Q y = 2 √ a x 1/2 dy/dx = √ a / √ x At P, x = at 2, so dy/dx = 1/t m T = 1/t, so m N = -t y – 2at = -t(x – at 2 )...some rearrangement y + tx = 2at + at 3 ? Q 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. For a parabola: Cartesian equation: y 2 = 4ax Parametric equations: (based on parameter t) x = at 2 y = 2at If y 2 = 40x then: Directrix:x = -10 Focus:(10, 0) Vertex:(0, 0) x = -a -a y P(x,y) a X S PARABOLA 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 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. caternary

Parabolas in real life 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. parabola