Locus and the Parabola

Transcription

1 Locus and the Parabola TERMINOLOGY Ais: A line around which a curve is reflected eg the ais of symmetry of a parabola Cartesian equation: An equation involving two variables and y Chord: An interval joining any two points on a curve In this chapter, any two points on a parabola Circle: The locus of a point moving so that it is equidistant from a fied point on a plane surface Directri: A fied line from which all points equidistant from this line and a fied point called the focus form a parabola Focal chord: A chord that passes through the focus Focal length: The distance between the focus and the verte of a parabola or the shortest distance between the verte and the directri Focus: A fied point from which all points equidistant from this point and the directri form a parabola Latus rectum: A focal chord that is perpendicular to the ais of the parabola Locus: The path traced out by a point that moves according to a particular pattern or rule Locus can be described algebraically or geometrically Parabola: The locus of a point moving so that it is equidistant from a fied point called the focus and a fied line called the directri Parametric equations: A set of equations where variables and y are both written in terms of a third variable, called a parameter, usually p or t Tangent: A straight line that touches a curve at a single point only Verte: The turning point (maimum or minimum point) of a parabola It is the point where the parabola meets the ais of symmetry

2 Chapter Locus and the Parabola 579 INTRODUCTION THIS CHAPTER EXPANDS THE work on functions that you have already learned It shows a method of finding the equation of a locus In particular, you will study the circle and the parabola, defined as a locus A parabola can also be defined as a set of parametric equations, and you will study these in this chapter DID YOU KNOW? Locus problems have been studied since very early times Apollonius of Perga (6 90 BC), a contemporary (and rival) of Archimedes, studied the locus of various figures In his Plane Loci, he described the locus points whose ratio from two fied points is constant This locus is called the Circle of Apollonius Apollonius also used the equation y = l for the parabola René Descartes ( ) was another mathematician who tried to solve locus problems His study of these led him to develop analytical (coordinate) geometry Locus A relation can be described in two different ways It can be a set of points that obey certain conditions, or a single point that moves along a path according to certain conditions A locus is the term used to describe the path of a single moving point that obeys certain conditions

3 580 Maths In Focus Mathematics Etension Preliminary Course EXAMPLES Describe the locus of the following A pencil on the end of compasses The path of the pencil is a circle with centre at the point of the compasses A person going up an escalator (standing still on one step) What would the locus be if the person walks up the escalator? The body travels along a straight line parallel to the escalator 3 A doorknob on a closing door

4 Chapter Locus and the Parabola 58 If the door could swing right around it would follow a circle So a door closing swings through an arc of a circle 4 A point on the number line that is 3 units from 0 The locus is! 3 5 A point in the number plane that moves so that it is always 3 units from the y -ais The locus is vertical lines with equations =! 3 Class Discussion Describe the path of a person abseiling down a cliff Eercises Describe the locus of the following: a racing car driving around a track a person climbing a ladder 4 a ball s flight when thrown 5 a person driving up to the 5th floor of a car park 3 a child on a swing

5 58 Maths In Focus Mathematics Etension Preliminary Course 6 a point that moves along the number line such that it is always less than units from 0 7 a point on the number plane that moves so that it is always units from the origin 8 a point that moves so that it is always unit from the -ais 9 a point that moves so that it is always 5 units from the y -ais 0 a point that moves so that it is always units above the -ais a point that moves so that it is always unit from the origin a point that moves so that it is always 4 units from the point ^, -h 3 a point that is always 5 units below the -ais 4 a point that is always 3 units away from the point (, ) 5 a point that is always 7 units to the left of the y -ais 6 a point that is always 3 units to the right of the y -ais 7 a point that is always 8 units from the -ais 8 a point that is always 4 units from the y -ais 9 a point that is always 6 units from the point (-, 4) 0 a point that is always unit from the point (- 4, 5) A locus describes a single point Py ^, h that moves along a certain path The equation of a locus can often be found by using Py ^, h together with the information given about the locus EXAMPLES Find the equation of the locus of a point Py ^, h that moves so that it is always 3 units from the origin You studied this formula in Chapter 7 It is easier to use d than d to find the equation of the locus You may recognise this locus as a circle, centre ^0, 0h radius 3 units Its equation is given by + y = 9 Alternatively, use the distance formula d = _ - i + _ y - y i i or d = _ - i + _ y - y

6 Chapter Locus and the Parabola 583 Let Py ^, h be a point of the locus We want PO = 3 ie PO = 9 ^ - 0h + ^y - 0h = 9 + y = 9 Place P anywhere on the number plane Find the equation of the locus of point P ^, yh that moves so that distance PA to distance PB is in the ratio : where A = ^- 3, h and B = ^, -h Let Py ^, h be a point of the locus PA : PB = : PA ie = PB PA = PB ` PA = ] PBg = 4PB ie [ -^- 3 h] + ^y - h = 4$ ^ - h + [ y -^- h] ^ + 3 h + ^y - h = 4[ ^ - h + ^y + h ] y - y + = 4^ y h = y + 6y = y + 8y + or y + 8y + = 0 Use the distance formula as in Eample This is the equation of a circle 3 Find the equation of the locus of a point P ^, yh that moves so that the line PA is perpendicular to line PB, where A = ^, h and B = ^-3, -h CONTINUED

7 584 Maths In Focus Mathematics Etension Preliminary Course These results come from Chapter 7 The locus is a circle with diameter AB Let P ^, yh be a point of the locus For perpendicular lines, mm =- y - y Using m = - y - PA: m = - y - ]-g PB: m = - ]-3g y + = + 3 For PA perpendicular to PB ie y - y + # = y - y - = y - y - = - ^ + - 3h + + y - y - 5 = 0 = Find the equation of the locus of point Py ^, h that is equidistant from fied point A ^, -h and fied line with equation y = 5 This is the equation of a parabola Can you see where the parabola lies? Let Py ^, h be a point of the locus B has coordinates ^, 5h We want PA = PB ie PA = PB ^ - h + [ y -^- h] = ^ - h + ^y - 5h ^ - h + ^y + h = ^y - 5h y + 4y + 4 = y - 0y y - 0 = 0

8 Chapter Locus and the Parabola 585 Eercises Find the equation of the locus of point Py ^, h that moves so that it is always unit from the origin Find the equation of the locus of point Py ^, h that moves so that it is always 9 units from the point ^-, -h 3 Find the equation of the locus of a point that moves so that it is always units from the point ^5, -h 4 Find the equation of the locus of point Py ^, h that moves so that it is equidistant from the points ^3, h and ^-, 5h 5 Find the equation of the locus of a point that moves so that it is equidistant from the points ^-4, 6h and ^, -7h 6 Find the equation of the locus of point Py ^, h that moves so that it is equidistant from the -ais and the y -ais 7 Find the equation of the locus of a point P that moves so that PA is twice the distance of PB where A = ^0, 3h and B = ^4, 7h 8 Find the equation of the locus of point Py ^, h that moves so that the ratio of PA to PB is 3 : where A = ^-6, 5h and B = ^3, -h 9 Find the equation of the locus of a point that moves so that it is equidistant from the point ^, -3h and the line y = 7 0 Find the equation of the locus of a point that moves so that it is equidistant from the point ^0, 5h and the line y =- 5 Find the equation of the locus of a point that moves so that it is equidistant from the point ^, 0h and the line = 6 Find the equation of the locus of a point that moves so that it is equidistant from the point ^, -h and the line y = 3 3 Find the equation of the locus of a point that moves so that it is equidistant from the point ^0, -3h and the line y = 3 4 Find the equation of the locus of a point Py ^, h that moves so that the line PA is perpendicular to line PB where A = ^, -3h and B = ^4, 5h 5 Find the equation of the locus of a point Py ^, h that moves so that the line PA is perpendicular to line PB, where A = ^-4, 0h and B = ^, h 6 Find the equation of the locus of a point Py ^, h that moves so that the line PA is perpendicular to line PB where A = ^, 5h and B = ^-, -3h 7 Point P moves so that PA + PB = 4 where A = ^3, -h and B = ^-5, 4h Find the equation of the locus of P 8 Point P moves so that PA + PB = where A = ^-, - 5h and B = ^, 3h Find the equation of the locus of P 9 Find the equation of the locus of a point that moves so that its distance from the line 3 + 4y + 5 = 0 is always 4 units

9 586 Maths In Focus Mathematics Etension Preliminary Course 0 Find the equation of the locus of a point that moves so that its distance from the line - 5y - = 0 is always unit Find the equation, in eact form, of the locus of a point that moves so that its distance from the line - y - 3 = 0 is always 5 units Find the equation of the locus of a point that moves so that it is equidistant from the line 4-3y + = 0 and the line 3 + 4y - 7 = 0 3 Find the equation of the locus of a point that moves so that it is equidistant from the line 3 + 4y - 5 = 0 and the line 5 + y - = 0 4 Given two points A ^3, -h and B ^-, 7h, find the equation of the locus of Py ^, h if the gradient of PA is twice the gradient of PB 5 If R is the fied point ^3, h and P is a movable point ^, yh, find the equation of the locus of P if the distance PR is twice the distance from P to the line y =- PROBLEM Can you see mistakes in the solution to this question? Find the locus of point Py ^, h that moves so that its perpendicular distance from the line + 5y - = 0 is always 3 units Let Py ^, h be a point of the locus a + by + c d = a + b 5 + y - 3 = y - = y - = y - = 3 ` 39 = 5 + y - 0 = 5 + y - 40 Can you find the correct locus?

10 Chapter Locus and the Parabola 587 Circle as a Locus The locus of point P (, y) that is always a constant distance from a fied point is a circle The circle, centre ^0, 0h and radius r, has the equation + y = r Proof Find the equation of the locus of point Py ^, h that is always r units from the origin Let P ^, yh be a point of the locus OP = r ie OP = r ^ - 0h + ^y - 0h = r + y = r So + y = r is the equation of the locus It describes a circle with radius r and centre ^0, 0h The circle, centre ^a, bh and radius r, has the equation ^ - ah + ^y - bh = r Proof Find the equation of the locus of point Py ^, h that is always r units from point Aab ^, h

11 588 Maths In Focus Mathematics Etension Preliminary Course Let Py ^, h be a point of the locus AP = r ie AP = r ^ - ah + ^y - bh = r So ] - ag + ^y - bh = r is the equation of the locus It describes a circle with radius r and centre ^a, bh EXAMPLES Find the equation of the locus of a point that is always units from the point ^-, 0h You could find this equation by using P (, y) and treating the question as a locus problem This is a circle with radius and centre ^-, 0h Its equation is in the form ^ - ah + ^y - bh = r ie [ -^- h] + ^y - 0h = ^ + h + y = y = y - 3 = 0 Find the radius and the coordinates of the centre of the circle + + y - 6y - 5 = 0 You learned how to complete the square in Chapter 3 We put the equation into the form ^ - ah + ^ y - bh = r To do this we complete the square In general, to complete the square on b + b, add c m to give: b b + b + c m = c + m First we move any constants to the other side of the equation, then complete the square To complete the square on +, we add c m =

12 Chapter Locus and the Parabola 589 To complete the square on y - 6y, we add 6 c m = y - 6y - 5 = y - 6y = y - 6y + 9 = ^ + h + ^y - 3h = 5 ^ -]- gh + ^y - 3h = 5 The equation is in the form ^ - ah + ^ y - bh = r This is a circle, centre ^-, 3h and radius 5 3 Eercises Find the length of the radius and the coordinates of the centre of each circle (a) + y = 00 (b) + y = 5 (c) ^ - 4h + ^y - 5h = 6 (d) ^ - 5h + ^y + 6h = 49 (e) + ^y - 3h = 8 Find the equation of each circle in epanded form (without grouping symbols) (a) Centre (0, 0) and radius 4 (b) Centre (3, ) and radius 5 (c) Centre ^-, 5h and radius 3 (d) Centre (, 3) and radius 6 (e) Centre ^-4, h and radius 5 (f) Centre ^0, -h and radius (g) Centre (4, ) and radius 7 (h) Centre ^-3, -4h and radius 9 (i) Centre ^-, 0h and radius 5 (j) Centre ^-4, -7h and radius 3 3 Find the equation of the locus of a point moving so that it is unit from the point ^9, -4h 4 Find the equation of the locus of a point moving so that it is 4 units from the point ^-, -h 5 Find the equation of the locus of a point moving so that it is 7 units from the point ^, 0h 6 Find the equation of the locus of a point moving so that it is units from the point ^-3, 8h 7 Find the equation of the locus of a point moving so that it is units from the point ^5, -h 8 Find the equation of a circle with centre ^0, 0h and radius 3 units 9 Find the equation of a circle with centre ^, 5h and radius unit 0 Find the equation of a circle with centre ^-6, h and radius 6 units Find the equation of a circle with centre ^4, 3h and radius 3 units Find the equation of a circle with centre ^0, -3h and radius units 3 Find the coordinates of the centre and the length of the radius of each circle (a) y - y - 4 = 0 (b) y - 4y - 5 = 0 (c) + y - y = 0

13 590 Maths In Focus Mathematics Etension Preliminary Course Concentric circles have the same centre (d) y + 6y - = 0 (e) + + y - y + = 0 (f) - + y = 0 (g) y - 8y = 0 (h) y - 4y + 40 = 0 (i) y + y + 5 = 0 (j) + + y + 4y - 5 = 0 4 Find the centre and radius of the circle with equation given by y + y - 6 = 0 5 Find the centre and radius of the circle with equation given by y - 0y + 4 = 0 6 Find the centre and radius of the circle with equation given by + + y + y - = 0 7 Find the centre and radius of the circle with equation given by y - 4y + = 0 8 Find the centre and radius of the circle with equation given by y - y - 3 = 0 9 Sketch the circle whose equation is given by y - y + = 0 0 Prove that the line 3 + 4y + = 0 is a tangent to the circle y + 4y - 5 = 0 (a) Show that - + y + 4y + = 0 and - + y + 4y - 4 = 0 are concentric (b) Find the difference between their radii Given two points A ^, -5h and B ^-4, 3h, find the equation of the circle with diameter AB 3 Find the eact length of the tangent from ^4, -5h to the circle y - y - = 0 4 Find the eact length of AB where A and B are the centres of the circles y = 0 and y + 6y - 3 = 0 respectively 5 (a) Find the length of XY where X and Y are the centres of the circles y - y + = 0 and y - y + = 0 respectively (b) Find the radius of each circle (c) What conclusion can you draw from the results for (a) and (b)? 6 Show that the circles + y = 4 and + + y - 4y - 4 = 0 both have 3 + 4y + 0 = 0 as a tangent 7 A circle has centre C ^-, 3h and radius 5 units (a) Find the equation of the circle (b) The line 3 - y + = 0 meets the circle at two points Find their coordinates (c) Let the coordinates be X and Y, where Y is the coordinate directly below the centre C Find the coordinates of point Z, where YZ is a diameter of the circle (d) Hence show + ZXY = 90c 8 (a) Find the perpendicular distance from P ^, -5h to the line 5 + y - = 0 (b) Hence find the equation of the circle with centre P and tangent 5 + y - = 0

14 Chapter Locus and the Parabola 59 Parabola as a Locus The locus of a point that is equidistant from a fied point and a fied line is always a parabola The fied point is called the focus and the fied line is called the directri Work on the parabola as a locus is very important, as the properties of the parabola are useful to us The parabola is used in lenses of glasses and cameras, in car headlights, and for bridges and radio telescope dishes DID YOU KNOW? Any rope or chain supporting a load (eg a suspension bridge) is in the shape of a parabola Find some eamples of suspension bridges that have a parabola shaped chain Other bridges have ropes or chains hanging freely These are not in the shape of a parabola, but are in a shape called a catenary Can you find some bridges with this shape? More recent bridges are cable-stayed, where ropes or chains are attached to towers, or pylons, and fan out along the sides of the bridge An eample is the Anzac Bridge in Sydney There are many different bridge designs One famous bridge in Australia is the Sydney Harbour Bridge Research different bridge designs and see if you can find some with parabolic shapes Parabola with verte at the origin Just as the circle has a special equation when its centre is at the origin, the parabola has a special equation when its verte is at the origin Both also have a more general formula

15 59 Maths In Focus Mathematics Etension Preliminary Course The locus of a point that is equidistant from a fied point and a fied line is always in the shape of a parabola If the fied point is (0, a ) and the fied line is y =- a (where a 0), then one of the equidistant points is the origin (0, 0) The distance between the points (0, 0) and (0, a ) is a units The point on y =- a directly below the origin is ^0, -ah and the distance from (0, 0) to ^0, -ah is also a units y (0, a) a a (0, -a) y =- a To find the equation of the parabola, we use the general process to find the equation of any locus The features of the parabola have special names A parabola is equidistant from a fied point and a fied line The fied point is called the focus The fied line is called the directri The turning point of the parabola is called the verte The ais of symmetry of the parabola is called its ais The distance between the verte and the focus is called the focal length An interval joining any two points on the parabola is called a chord A chord that passes through the focus is called a focal chord The focal chord that is perpendicular to the ais is called the latus rectum A tangent is a straight line that touches the parabola at a single point

16 Chapter Locus and the Parabola 593 PARABOLA = 4 ay The locus of point Py ^, h moving so that it is equidistant from the point ^0, ah and the line y =- a is a parabola with equation = 4ay Proof Let Py ^, h be a point of the locus Taking the perpendicular distance from P to the line y =- a, point B = ^, -ah PA = PB ` PA = PB ^ - 0 h + ^y - ah = ^ - h + [ y -^-ah] + ^y - ah = ^y + ah + y - ay + a = y + ay + a = 4ay The parabola = 4ay has focus at ^0, ah directri with equation y =- a verte at ^0, 0h ais with equation = 0 focal length the distance from the verte to the focus with length a latus rectum that is a horizontal focal chord with length 4 a Since the focal length is a, a is always a positive number Class Investigation Find the equation of the locus if point Py ^, h is equidistant from ^0, -ah and y = a

17 594 Maths In Focus Mathematics Etension Preliminary Course EXAMPLES Find the equation of the parabola whose focus has coordinates ^0, h and whose directri has equation y =- The focus has coordinates in the form ^0, ah and the directri has equation in the form y =- a, where a = ` the parabola is in the form = 4ay where a = ie = 4 ( ) y = 8y (a) Find the coordinates of the focus and the equation of the directri of the parabola = 0y (b) Find the points on the parabola at the endpoints of the latus rectum and find its length (a) The parabola = 0y is in the form = 4ay 4a = 0 ` a = 5 The focal length is 5 units We can find the coordinates of the focus and the equation of the directri in two ways Method : Draw the graph = 0y and count 5 units up and down from the origin as shown y = 0y 5 5 (0, 5) (0, -5) y = -5 The focus is (0, 5) and the directri has equation y = -5

18 Chapter Locus and the Parabola 595 Method : The focus is in the form (0, a ) where a = 5 So the focus is (0, 5) The directri is in the form y =- a where a = 5 So the directri is y =- 5 (b) The latus rectum is a focal chord that is perpendicular to the ais of the parabola as shown y = 0 y (0, 5) The endpoints of the latus rectum will be where the line y = 5 and the parabola intersect Substitute y = 5 into the parabola = 0y = 0] 5g = 00 =! 00 =! 0 So the endpoints are (- 0, 5) and (0, 5) y = 0 y (-0, 5) (0, 5) (0, 5) From the graph, the length of the latus rectum is 0 units The latus rectum is 4 a units long which gives 0 units CONTINUED

19 596 Maths In Focus Mathematics Etension Preliminary Course 3 Find the equation of the focal chord to the parabola = 4y that passes through (- 4, 4) The parabola = 4y is in the form = 4ay 4a = 4 ` a = The focal length is unit The focus is unit up from the origin at (0, ) and the focal chord also passes through (- 4, 4) y (-4, 4) = 4y (0, ) You used these formulae in Chapter 7 We can find the equation of the line between (0, ) and (- 4, 4) by using either formula y - y y - y y - y = m or _ - i - = - y - y y - y - = - y = y - 3 = - 4-4^y - h = 3-4y + 4 = 3 0 = 3 + 4y - 4 As you saw in the previous chapter, a parabola can be concave downwards Can you guess what the equation of this parabola might be? PARABOLA =- 4 ay The locus of a point P (, y ) moving so that it is equidistant from the point ^0, -ah and the line y = a is a parabola with equation =- 4ay

20 Chapter Locus and the Parabola 597 Proof y B(, a) y = a P(, y) A(0, -a) Let P (, y ) be a point of the locus Taking the perpendicular distance from P to the line y = a, point B = ^, ah PA = PB ` PA = PB ^ - 0h + 7y -^- aha = ^ - h + ^y - ah + ^y + ah = ^y - ah + y + ay + a = y - ay + a =-4ay The parabola =- 4ay has focus at ^0, -ah directri with equation y = a verte at (0, 0) ais with equation = 0 focal length a latus rectum a horizontal focal chord with length 4 a

21 598 Maths In Focus Mathematics Etension Preliminary Course EXAMPLES Find the equation of the parabola with focus ^0, -4h and directri y = 4 If we draw this information, the focus is below the directri as shown So the parabola will be concave downwards (the parabola always turns away from the directri) y y = (0, -4) The focal length is 4 so a = 4 The parabola is in the form =- 4ay where a = 4 =-4ay =-4] 4g y =-6y Find the coordinates of the verte, the coordinates of the focus and the equation of the directri of the parabola =- y The parabola =- y is in the form =- 4ay 4a = ` a = 3 The focal length is 3 units The verte is (0, 0) We can find the coordinates of the focus and the equation of the directri in two ways Method : Draw the graph =- y and count 3 units up and down from the origin as shown (The parabola is concave downward)

22 Chapter Locus and the Parabola 599 y y = (0, -3) =-y Counting down 3 units, the focus is ^0, -3h Counting up 3 units, the directri has equation y = 3 Method : The focus is in the form ^0, -ah where a = 3 So the focus is ^0, -3h The directri is in the form y = a where a = 3 So the directri is y = 3 3 Find the equation of the parabola with focal length 5 and whose verte is ^0, 0h and equation of the ais is = 0 Verte ^0, 0h and ais given by = 0 give a parabola in the form =! 4ay, since there is not enough information to tell whether it is concave upwards or downwards This gives two possible parabolas CONTINUED

23 600 Maths In Focus Mathematics Etension Preliminary Course Focal length of 5 means a = 5 The equation is =! 45 ( ) y ie =! 0y 4 Eercises Find the equation of each parabola (a) focus (0, 5), directri y =- 5 (b) focus (0, 9), directri y =- 9 (c) focus (0, ), directri y =- (d) focus (0, 4), directri y =- 4 (e) focus (0, 0), directri y =- 0 (f) focus (0, 3), directri y =- 3 (g) focus (0, 6), directri y =- 6 (h) focus (0, ), directri y =- (i) focus (0, ), directri y =- (j) focus (0, ), directri y =- Find the equation of each parabola (a) focus (0, - ), directri y = (b) focus (0, - 3), directri y = 3 (c) focus (0, - 4), directri y = 4 (d) focus (0, - 7), directri y = 7 (e) focus (0, - 6), directri y = 6 (f) focus (0, - 9), directri y = 9 (g) focus (0, - 8), directri y = 8 (h) focus (0, - ), directri y = (i) focus (0, - 5), directri y = 5 (j) focus (0, - 3), directri y = 3 3 Find (i) the coordinates of the focus and (ii) the equation of the directri of (a) = 4y (b) = 8y (c) = 6y (d) = 36y (e) = 40y (f) = 44y (g) = y (h) = 6y (i) = 0y (j) = 5y 4 Find (i) the coordinates of the focus and (ii) the equation of the directri of (a) =- 4y (b) =- 4y (c) =- 8y (d) =- 48y (e) =- 0y (f) =- 6y (g) =- 3y (h) =- 40y (i) =- y (j) y 5 Find the equation of the parabola with (a) coordinates of the focus ^0, 7h and equation of the directri y =- 7 (b) coordinates of the focus ^0, h and equation of the directri y =- (c) coordinates of the focus ^0, -6h and equation of the directri y = 6 (d) coordinates of the focus ^0, h and coordinates of the verte ^0, 0h

24 Chapter Locus and the Parabola 60 (e) coordinates of the verte ^0, 0h, equation of the ais = 0 and focal length 3 (f) coordinates of the verte ^0, 0h, equation of the ais = 0 and focal length 8 (g) coordinates of the verte ^0, 0h and equation of the ais = 0, and passing through the point ^-8, h (h) coordinates of the verte ^0, 0h and equation of the ais = 0, and passing through the point ^-, 7h 6 Find the coordinates of the focus, the equation of the directri and the focal length of the parabola (a) = 8y (b) = 4y (c) =- y (d) = y (e) =- 7y (f) = y 7 Find the equation of the focal chord that cuts the curve = 8y at ^-4, h 8 The tangent with equation - y - 4 = 0 touches the parabola = 4y at A Find the coordinates of A 9 The focal chord that cuts the parabola =- 6y at ^6, -6h cuts the parabola again at X Find the coordinates of X 0 Find the coordinates of the endpoints of the latus rectum of the parabola =- 8 y What is the length of the latus rectum? The equation of the latus rectum of a parabola is given by y =- 3 The ais of the parabola is = 0, and its verte is ^0, 0h (a) Find the equation of the parabola (b) Find the equation of the directri (c) Find the length of the focal chord that meets the parabola at c, - m 3 (a) Show that the point ^-3, 3h lies on the parabola with equation = 3 y (b) Find the equation of the line passing through P and the focus F of the parabola (c) Find the coordinates of the point R where the line PF meets the directri 3 (a) Find the equation of chord PQ where P c-, m and Q ^, h 4 lie on the parabola = 4 y (b) Show that PQ is not a focal chord (c) Find the equation of the circle with centre Q and radius units (d) Show that this circle passes through the focus of the parabola 4 (a) Show that Q_ aq, aq i lies on the parabola = 4 ay (b) Find the equation of the focal chord through Q (c) Prove that the length of the latus rectum is 4 a

25 60 Maths In Focus Mathematics Etension Preliminary Course Investigation Sketch the parabola = y You may like to complete the table below to help you with its sketch y Is this parabola a function? What is its ais of symmetry? The parabola that has y rather than in its equation is a sideways parabola It still has the same properties, but generally the and y values are swapped around PARABOLA y = 4 a The locus of point Py ^, h moving so that it is equidistant from the point ^a, 0h and the line =- a is a parabola with equation y = 4a Proof Find the equation of the locus of point Py ^, h, which moves so that it is equidistant from the point ^a, 0h and the line =- a Coordinates of B are ^-a, yh We want PA = PB ie PA = PB ^ - ah + ^y - 0h = [ -^- ah ] + ^y - yh ^ - ah + y = ^ + ah - a + a + y = + a + a y = 4a

26 Chapter Locus and the Parabola 603 The parabola y = 4a has focus at ^a, 0h equation of directri =- a verte at ^0, 0h ais with equation y = 0 focal length the distance from the verte to the focus with length a latus rectum that is a vertical focal chord with length 4 a EXAMPLES Find the equation of the parabola with focus (7, 0) and directri =- 7 If we draw this information, the focus is to the right of the directri as shown (the parabola always turns away from the directri) So the parabola turns to the right y = (7, 0) CONTINUED

27 604 Maths In Focus Mathematics Etension Preliminary Course The focal length is 7 so a = 7 The parabola is in the form y = 4a where a = 7 y = 4a = 4^7h = 8 Find the coordinates of the focus and the equation of the directri of the parabola y = 3 The parabola y = 3 is in the form y = 4a 4a = 3 ` a = 8 The focal length is 8 units Method : Draw the graph y = 3 and count 8 units to the left and right from the origin as shown (The parabola turns to the right) y = (8, 0) y = 3 Counting 8 units to the right, the focus is (8, 0) Counting 8 units to the left, the directri has equation =- 8 Method : The focus is in the form ( a, 0) where a = 8 So the focus is (8, 0) The directri is in the form =- a where a = 8 So the directri is =- 8 A parabola can also turn to the left

28 Chapter Locus and the Parabola 605 PARABOLA y =- 4 a The locus of a point P (, y ) moving so that it is equidistant from the point ^-a, 0h and the line = a is a parabola with equation y =- 4a Proof y P(, y) B(a, y) A( - a, 0) = a Let P (, y ) be a point of the locus Taking the perpendicular distance from P to the line = a, point B = ^a, yh PA = PB ` PA = PB 7 -^ - aha + ^ y - 0h = ^ - ah + ^ y - yh ^ + ah + y = ^ - ah + a + a + y = - a + a y =-4a

29 606 Maths In Focus Mathematics Etension Preliminary Course The parabola y =- 4a has focus at (- a, 0) directri with equation = a verte at (0, 0) ais with equation y = 0 focal length a latus rectum a vertical focal chord with length 4 a EXAMPLES Find the equation of the parabola with focus (- 4, 0) and directri = 4 Drawing this information shows that the parabola turns to the left y ( - 4, 0) 4 4 = 4 The focal length is 4 so a = 4 The parabola is in the form y =- 4a where a = 4 y =-4a =-4^4h =-6 Find the coordinates of the focus and the equation of the directri of the parabola y =- The parabola y =- is in the form y =- 4a 4a = ` a = The focal length is unit

30 Chapter Locus and the Parabola 607 Method : Draw the graph y =- and count unit to the left and right from the origin as shown (The parabola turns to the left) y ( -, 0 ) = Counting units to the left, the focus is c-, 0m Counting units to the right, the directri has equation = Method : The focus is in the form (- a, 0) where a = So the focus is c-, 0m The directri is in the form = a where a = So the directri is = 5 Eercises Find the equation of each parabola (a) focus (, 0), directri =- (b) focus (5, 0), directri =- 5 (c) focus (4, 0), directri =- 4 (d) focus (9, 0), directri =- 9 (e) focus (8, 0), directri =- 8 (f) focus (6, 0), directri =- 6 (g) focus (7, 0), directri =- 7 (h) focus (3, 0), directri =- 3 (i) focus (4, 0), directri =- 4 (j) focus (, 0), directri =- Find the equation of each parabola (a) focus (- 9, 0), directri = 9 (b) focus (- 4, 0), directri = 4 (c) focus (- 0, 0), directri = 0 (d) focus (- 6, 0), directri = 6 (e) focus (-, 0), directri = (f) focus (-, 0), directri = (g) focus (-, 0), directri = (h) focus (- 5, 0), directri = 5 (i) focus (- 3, 0), directri = 3 (j) focus (- 7, 0), directri = 7

31 608 Maths In Focus Mathematics Etension Preliminary Course 3 Find (i) the coordinates of the focus and (ii) the equation of the directri of (a) y = 8 (b) y = (c) y = 6 (d) y = 4 (e) y = 8 (f) y = 3 (g) y = 4 (h) y = 36 (i) y = (j) y = 8 4 Find (i) the coordinates of the focus and (ii) the equation of the directri of (a) y =- 8 (b) y =- (c) y =- 8 (d) y =- 4 (e) y =- 4 (f) y =- 5 (g) y =- 60 (h) y =- (i) y =- 6 (j) y =- 5 5 Find the equation of the parabola with (a) coordinates of the focus ^5, 0h and equation of the directri =- 5 (b) coordinates of the focus ^, 0h and equation of the directri =- (c) coordinates of the focus ^-4, 0h and equation of the directri = 4 (d) coordinates of the focus ^3, 0h and coordinates of the verte ^0, 0h (e) coordinates of the verte ^0, 0h equation of the ais y = 0 and focal length 9 (f) coordinates of the verte ^0, 0h, equation of the ais y = 0 and focal length (g) coordinates of the verte ^0, 0h and equation of the ais y = 0 and passing through the point ^3, 6h (h) coordinates of the verte ^0, 0h and equation of the ais y = 0 and passing through the point ^, h 6 Find the coordinates of the focus, the equation of the directri and the focal length of the parabola (a) y = 8 (b) y = 4 (c) y =- (d) y = 6 (e) y =- 5 (f) 3y = 7 Find the equation of the focal chord that cuts the curve y = 6 at ^4, 8h 8 Find the length of the latus rectum of the parabola y = What are the coordinates of its endpoints? 9 The line with equation - 3y - 7 = 0 meets the parabola y = 4 at two points Find their coordinates 0 Let R c, -m be a point on the 5 parabola y = 0 (a) Find the equation of the focal chord passing through R (b) Find the coordinates of the point Q where this chord cuts the directri (c) Find the area of D OFQ where O is the origin and F is the focus (d) Find the perpendicular distance from the chord to the point P ^-, -7h (e) Hence find the area of D PQR

32 Chapter Locus and the Parabola 609 Application A parabolic satellite dish receives its signals through the focus If the dish has height m and a span of 0 m, find where the focus should be placed, to the nearest mm SOLUTION The parabola is of the form = 4ay and passes through (0, ) and (- 0, ) Substituting (0, ) gives 0 4 a () = 00 = 48a 083 = a So the focus should be placed 083 m from the verte This is 083 mm to the nearest millimetre Here is a summary of the 4 different types of parabola with the verte at the origin = 4ay y Focus (0, a) = 4ay Directri y =-a

33 60 Maths In Focus Mathematics Etension Preliminary Course =- 4ay y Directri y = a Focus (0, -a) =-4ay 3 y = 4a y Directri =-a Focus (a, 0) y = 4a 4 y =- 4a y Directri = a Focus (-a, 0) y =-4a General Parabola When the parabola does not have its verte at the origin, there is a more general formula Since we use a to mean the focal length, we cannot use ( a, b ) as the verte We use ( h, k ) instead

34 Chapter Locus and the Parabola 6 PARABOLA ( - h) = 4a(y - k) The concave upwards parabola with verte ( h, k ) and focal length a has equation ^ - hh = 4a^y -kh Proof Find the equation of the parabola with verte ^h, kh and focal length a Counting up a units from verte V gives the focus F = ^h, k + ah Counting down a units from V gives the point on the directri D = ^h, k - ah So the equation of the directri is given by y = k - a We find the equation of the locus of P ^, yh that is equidistant from point F ^h, k + ah and line y = k - a B has coordinates ^, k - ah We want PF = PB ie PF = PB ^ - hh + [ y - ^k + ah] = ^ - h + [ y - ^k + ah] ^ - hh + ^y - k - ah = ^y - k + ah ^ - hh = ^y - k + ah -^ y - k - ah = [ ^ y - k + ah + ^ y - k - ah] # [ ^ y - k + ah -^ y - k - ah] ^ difference of two squaresh = ^y - kh^ah = 4ay - 4ak = 4a^ y - kh

35 6 Maths In Focus Mathematics Etension Preliminary Course The parabola ^ - hh = 4a ^y - kh has ais parallel to the y -ais verte at ^h, kh focus at ^h, k + ah directri with equation y = k - a EXAMPLES Find the equation of the parabola with focus ^, 3h and directri with equation y =- 7 Draw a diagram to find the verte and to find a Coordinates of B are ^, -7h The verte is the midpoint of ^, 3h and ^, -7h ` verte = ^, -h Focal length is the distance from the focus to the verte ` a = 5 From the diagram the parabola is concave upwards The equation is in the form ^ - hh = 4a^y - kh ie ^ - h = 4^5h[ y -^-h] = 0^y + h = 0y y - 36 = 0 Find the coordinates of the verte and the focus, and the equation of the directri, of the parabola with equation + 6 -y - 3 = 0

36 Chapter Locus and the Parabola 63 Complete the square on + 6 -y - 3= = y = y ^ + 3h = y + = ( y + ) So the parabola has equation ^ + 3h = ^ y + h Its verte has coordinates ^-3, -h 4a = ` a = 3 The parabola is concave upwards as it is in the form ^ - hh = 4a^y - kh Count up 3 units to the focus ` focus = ^-3, h Count down 3 units to the directri ` directri has equation y =- 4 It is easy to find the focus and the directri by counting along the y -ais PARABOLA ( - h) =- 4a(y - k) The concave downwards parabola with verte ( h, k ) and focal length a has equation ^ - hh = -4a^y - kh Proof Find the equation of the concave downwards parabola with verte ( h, k ) and focal length a

37 64 Maths In Focus Mathematics Etension Preliminary Course Counting down a units from the verte V gives the focus F = ^h, k - ah Counting up a units from the verte V gives the point on the directri D = ^h, k + ah So the equation of the directri is given by y = k + a We find the equation of the locus of P (, y ) that is equidistant from point Fhk ^, - ah and line y = k + a y B y = k + a P(, y) F (h, k- a) B has coordinates ^, k + ah We want PF = PB PF = PB ^ - hh + 7y -^k - aha = ^ - h + 7y - ^k + aha ^ - hh + ^y - k + ah = ^y - k - ah ^ - hh = ^y - k - ah -^y - k + ah = 7^y - k - ah+ ^y - k + aha7^y - k - ah-^y - k + aha ( difference of two squares) = ^y - kh^-ah =- 4ay + 4ak =-4a^y -kh

38 Chapter Locus and the Parabola 65 The parabola ^ - hh = -4a^y - kh has ais parallel to the y -ais verte at ( h, k ) focus at ^h, k - ah directri with equation y = k + a EXAMPLES Find the equation of the parabola with focus (-, ) and directri y = 3 y B (-, ) y = 3 Coordinates of B are (-, 3) The verte is the midpoint of (-, ) and (-, 3) ` verte = (-, ) Focal length a = From the diagram the curve is concave downwards The equation is in the form ie ^ - hh = -4a^ y - kh 7 -^ - ha = -4] g^ y - h ^ + h = -4^ y - h = - 4y y - 4 = 0 Find the coordinates of the verte and focus, and the equation of the directri of the parabola y - 6 = 0 CONTINUED

39 66 Maths In Focus Mathematics Etension Preliminary Course Complete the square on y - 6 = 0-8 = - 8y = - 8y ^ - 4h = - 8y + 3 =-8^y- 4h So the parabola has equation ^ - 4h = -8^ y - 4h Its verte has coordinates (4, 4) 4a = 8 ` a = The parabola is concave downwards as it is in the form ^ - hh = -4a^ y - kh y y = (4, 4) (4, ) 3 4 Count down units to the focus ` focus = ^4, h Count up units to the directri ` directri has equation y = 6 PARABOLA ( y - k) = 4a( - h) The parabola with verte ( h, k ) and focal length a that turns to the right has equation ^y - kh = 4a^ - hh Proof Find the equation of the parabola that turns to the right with verte ( h, k ) and focal length a

40 Chapter Locus and the Parabola 67 Counting a units to the right from the verte V gives the focus F = ^h + a, kh Counting a units to the left from the verte V gives the point on the directri D = ^h - a, kh So the equation of the directri is given by = h - a We find the equation of the locus of P (, y ) that is equidistant from point Fh ^ + ak, h and line = h - a = h - a y B P(, y) F (h+a, k) B has coordinates ^h - a, yh We want PF = PB PF = PB 7 - ^h + aha + ^y - kh = 7 -^h - aha + ^y - yh ^ - h - ah + ^y - kh = ^ - h - ah ^y - kh = ^ - h + ah -] - h - ag = 7^ - h + ah+ ^ - h - aha7^ - h + ah-^ - h - aha ( difference of two squares) = ^ - hh^ah = 4a - 4ah = 4a ^ - hh

41 68 Maths In Focus Mathematics Etension Preliminary Course The parabola ^y - kh = 4a^ - hh has ais parallel to the -ais verte at ^h, kh focus at ^h + a, kh directri with equation = h - a EXAMPLES Find the equation of the parabola with focus (, - ) and directri =- 5 =-5 y B (, -) Coordinates of B are (- 5, - ) The verte is the midpoint of (- 5, - ) and (, - ) ` verte = ^-, -h Focal length a = 3 From the diagram the parabola curves to the right The equation is in the form ^ y - kh = 4a] - hg ie 7y -^- ha = 43 ] g7 -^-ha ^ y + h = ] + g y + y + = + 4 y + y = 0

42 Chapter Locus and the Parabola 69 Find the coordinates of the verte and focus, and the equation of the directri of the parabola y + y = 0 Complete the square on y y + y = 0 y + y = y + y + 36 = ^ y + 6h = = 4^ + h So the parabola has equation ^y + 6h = 4^ + h or 7y -]- 6gA = 46 -]-g@ Its verte has coordinates (-, - 6) 4a = 4 ` a = The parabola turns to the right as it is in the form ^ y - kh = 4a^ - hh y =- (-, -6) (-0, -6) Count unit to the right for the focus ` focus = ^-0, -6h Count unit to the left for the directri ` directri has equation =- PARABOLA (y k) = 4a( h) The parabola with verte ( h, k ) and focal length a that turns to the left has equation ^y - kh = -4a^ - hh

43 60 Maths In Focus Mathematics Etension Preliminary Course Proof Find the equation of the parabola that turns to the left with verte ( h, k ) and focal length a Counting a units to the left from the verte V gives the focus F = ^h - a, kh Counting a units to the right from the verte V gives the point on the directri D = ^h + a, kh So the equation of the directri is given by = h + a We find the equation of the locus of P (, y ) that is equidistant from point Fh ^ - ak, h and line = h + a y = h + a P (, y) B F (h -a, k) B has coordinates ^h + a, yh We want PF = PB PF = PB 7 - ^h - aha + ^y - kh = 7 - ^h + aha + ^y - yh ^ - h + ah + ^y - kh = ^ - h - ah ^y - kh = ^ - h - ah -^ - h + ah = 7^ - h - ah+ ^ - h + aha7^ - h - ah-^ - h + aha (difference of two squares) = ^ - hh^-ah =- 4a + 4ah =-4a ^ -hh =-4a^y -kh

44 Chapter Locus and the Parabola 6 The parabola ^ y - kh = -4a] - hg has ais parallel to the -ais verte at ( h, k ) focus at ^h - a, kh directri with equation = h + a EXAMPLES Find the equation of the parabola with focus (, ) and directri = 3 y = 3 (, ) B (, ) Coordinates of B are (3, ) The verte is the midpoint of (3, ) and (, ) ` verte = c, m Focal length a = From the diagram the parabola curves to the left The equation is in the form ^y - kh = -4a^ - hh ie ^y - h = -4c m c - m ^y - h = -c - m y - y + = y - y = 0 Find the coordinates of the verte and focus, and the equation of the directri of the parabola y + 4y = 0 CONTINUED

45 6 Maths In Focus Mathematics Etension Preliminary Course Complete the square on y y + 4y = 0 y + 4y = y + 4y + 4 = ^ y + h = =-8] -g So the parabola has equation ^ y + h = -8] -g or 7y -]- ga = -8] -g Its verte has coordinates ^, -h 4a = 8 ` a = The parabola turns to the left as it is in the form ^y - kh = -4a^ - hh y = (-, -) 3 - (, -) - Count units to the left for the focus ` focus = ^-, -h Count units to the right for the directri ` directri has equation = 3 6 Eercises Complete the square on to write each equation in the form ] - hg =! 4a^ y - kh (a) -6-8y - 5= 0 (b) -0-4y + = 0 (c) - -4y - = 0 (d) y - 0 = 0 (e) - -8y - 0 = 0 (f) y + = 0 (g) y - 6= 0 (h) y + 9 = 0 (i) + -8y - 7= 0 (j) y + = 0

46 Chapter Locus and the Parabola 63 Complete the square on y to write each equation in the form ^ y - kh =! 4a] - hg (a) y - 8y - 4 = 0 (b) y - y - 8-5= 0 (c) y + 4y = 0 (d) y - 0y = 0 (e) y + 6y = 0 (f) y -y = 0 (g) y + 0y = 0 (h) y + 4y - 4 = 0 (i) y - 4y = 0 (j) y + 8y + 8 = 0 3 Find the equation of each parabola (a) focus ^-, 3h, directri y =- (b) focus ^-4, h, directri y =- (c) focus (, 0), directri y =- 4 (d) focus (3, 6), directri y = (e) focus ^-, 5 h, directri y =- 3 (f) focus ^-, -4h, directri y = 4 (g) focus ( 4, - 3), directri y = 7 (h) focus ^-5, h, directri y = 5 (i) focus ^-3, -6h, directri y = 0 (j) focus ^0, -7h, directri y =- 5 (k) focus (, 3), directri =- 4 (l) focus ^-, 4h, directri =- 3 (m) focus (6, 0), directri = (n) focus ( 3, - ), directri =- 5 (o) focus ^, -h, directri =- 3 (p) focus ^-, -4h, directri = 4 (q) focus (, ), directri = 4 (r) focus ^-5, 3 h, directri = 3 (s) focus ^-, h, directri = 0 (t) focus (3, ), directri = 4 4 Find (i) the coordinates of the focus and (ii) the equation of the directri of (a) - 6-4y - 3 = 0 (b) - - 8y - 7 = 0 (c) + 4-4y = 0 (d) -8 - y + 4 = 0 (e) + 0-8y + = 0 (f) y + = 0 (g) + + 8y - 5= 0 (h) y = 0 (i) y + 4 = 0 (j) y - = 0 5 Find (i) the coordinates of the focus and (ii) the equation of the directri of (a) y + y = 0 (b) y -8y = 0 (c) y - 6y = 0 (d) y + 4y -6 - = 0 (e) y - y = 0 (f) y + 0y = 0 (g) y + 4y = 0 (h) y - y = 0 (i) y - 4y = 0 (j) y + 6y = 0 6 Find the equation of the parabola with verte ^0, 3h if it is concave upwards and a = 3 7 Find the equation of the parabola with verte ^-, -h, focal length, and ais parallel to the y -ais 8 A parabola has its verte at ^, -5h and its focal length as If the parabola is concave upwards, find its equation 9 A parabola has its ais parallel to the -ais If its verte has coordinates ^, 6h and a = 3, find its equation if it turns to the left 0 Find the equation of the parabola with verte at ^, 0h and focus at ^, 4h Find the equation of the parabola that has verte ^, h and focus ^, 8h A parabola has its verte at ^, -h and focus at ^-4, -h Find its equation

47 64 Maths In Focus Mathematics Etension Preliminary Course 3 Find the equation of the parabola with verte ^0, 3h and focus ^8, 3h 4 Find the equation of the parabola with verte ^3, 3h and equation of directri y = 5 The latus rectum of a parabola has endpoints ^-, 3h and ^6, 3h Find two possible equations for the parabola 5 Find the equation of the parabola with verte ^3, -h and directri =- 6 A parabola has directri y = 5 and focus ^-3, 3h Find its equation 7 Find the equation of the locus of a point moving so that it is equidistant from the point ^, h and the line y =- 4 8 Find the equation of the parabola with focus ^, -h and directri = 0 9 Find the coordinates of the verte and focus and the equation of the directri for the parabola (a) + 4-8y + = 0 (b) -6 - y + 33= 0 (c) - + 4y + 5 = 0 (d) y -8y = 0 (e) y + 4y = 0 (f) y = 0 0 For the parabola + + 8y - = 0, find the coordinates of its verte and focus, and the equations of its directri and ais What is its maimum value? (a) Find the equation of the arch above (b) Find the coordinates of its focus and the equation of its directri 3 (a) Sketch y = + - 8, showing intercepts and the minimum point (b) Find the coordinates of the focus and the equation of the directri of the parabola 4 Find the equation of the parabola with verte ^-, 3h that also passes through ^, h and is concave downwards 5 A parabolic satellite dish has a diameter of 4 m at a depth of 04 m Find the depth at which its diameter is 35 m, correct to decimal place DID YOU KNOW? The word directri is due to the Dutch mathematician Jan De Witt (69 7) He published a work called Elementa curvarum, in which he defined the properties of the parabola, ellipse, circle and hyperbola These curves are all called conic sections

48 Chapter Locus and the Parabola 65 De Witt was well known as the Grand Pensionary of Holland He took part in the politics and wars of his time, opposing Louis XIV When the French invaded Holland in 67, De Witt was seized and killed Tangents and Normals Remember that the gradient of the tangent to a curve is given by the derivative The normal to the curve is perpendicular to its tangent at that point That is, m m =- for perpendicular lines EXAMPLES Find the gradient of the tangent to the parabola = 8y at the point ^4, h ` = 8y y = 8 dy = d 8 = 4 CONTINUED

49 66 Maths In Focus Mathematics Etension Preliminary Course dy 4 At ^4, h, = d 4 = So the gradient of the tangent at ^4, h is Find the equation of the normal to the parabola = 4y at the point ^-8, 6h So At (-86, ): y = 4 dy = d 4 = dy -8 = d =-4 So the gradient of the tangent m =-4 = 4y The normal is perpendicular to the tangent ` mm =- So ]- 4g m = - ` m = 4 The equation of the normal is given by y - y = m( - ) ie y - 6 = [ -(-8)] 4 = ( + 8) 4 4y - 64 = = - 4y Eercises Find the gradient of the tangent to the parabola = y at the point where = Find the gradient of the tangent to the parabola =- 3y at the point ^6, -h 3 Find the gradient of the normal to the parabola = 4y at the point where = 4 Find the gradient of the tangent to the parabola = 6y at the point ^4, h

50 Chapter Locus and the Parabola 67 5 Show that the gradient of the tangent to the curve = y at any point is its -coordinate 6 Find the equation of the tangent to the curve = 8y at the point ^4, h 7 Find the equation of the normal to the curve = 4y at the point where =- 4 8 Find the equations of the tangent and normal to the parabola =- 4y at the point ^, -6h 9 Find the equations of the tangent and normal to the parabola = 6y at the point where = 4 0 Find the equation of the tangent to the curve =- y at the point ^4, -8h This tangent meets the directri at point M Find the coordinates of M Find the equation of the normal to the curve = y at the point ^6, 3h This normal meets the parabola again at point P Find the coordinates of P The normal of the parabola = 8y at ^-6, h cuts the parabola again at Q Find the coordinates of Q 3 Find the equations of the normals to the curve =- 8y at the points ^-6, -3h and c-, - m Find their point of intersection and show that this point lies on the parabola 4 Find the equation of the tangent at ^8, 4h on the parabola = 6 y This tangent meets the tangent at the verte of the parabola at point R Find the coordinates of R 5 (a) Show that the point P_ p, p i lies on the parabola = 4 y (b) Find the equation of the normal to the parabola at P (c) Show that p + = 0 if the normal passes through the focus of the parabola ^p! 0h Parametric Equations of the Parabola An equation involving and y, for eample = 4 ay, is called a Cartesian equation Equations can also be written in parametric form In this form, and y are both written in terms of a third variable called a parameter An eample of a Cartesian equation is y = - An eample of parametric equations is = t + 3, y = t - Any Cartesian equation can be written in parametric form The word Cartesian comes from the name of the mathematician Descartes