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1 Review Eercise ( i j+ k) ( i+ j k) i j k = = i j+ k (( ) ( ) ) (( ) ( ) ) (( ) ( ) ) = i j+ k = ( ) i ( ( )) j+ ( ) k = j k Hence ( ) ( i j+ k) ( i+ j k) = ( ) + ( ) = 8 = Formulae for finding the vector product are given in the Edecel formulae booklet which is provided for the eamination. One form it gives is i j k a b = a a a and that has been used here. b b b You use the formula for the magnitude of a vector i+ yj+ zk = + + ( y z ) a We do this with the determinant method: i j k 0 = 0+ k = k k 0 + k + k b The magnitude of the vector product is given by + ( k) + ( + k) Then, in order to minimise it, we need to minimise 9+ 9k k+ k = 0k + 4k+ b In quadratic epressions of the form ak + bk+ c the minimum is realised by k = a, so in this case the least possible magnitude is obtained when k = 5 In particular, the value of this magnitude is = = Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.

2 AB= b a, AC = c a AB AC = ( b a) ( c a) = b c b a a c+ a a As a b= b a, c a= a c and a a = 0 AB AC = b c+ a b+ c a = a b+ b c+ c a You multiply out the brackets using the usual rules of algebra. You must take care with the order in which the vectors are multiplied as the vector product is not commutative. For a vector product y= y. The area of the triangle,, say, is given by = bcsina = AC ABsinA = AB AC = a b+ b c+ c a, as required. 4 a AB = (5i+ 8j+ k) (i+ 7 j k) = i+ j+ k AC = (6i+ 7j+ 4 k) (i+ 7 j k) = 4i+ 5k AB AC = (i+ j+ k) (4i+ 5 k) i j k = = i j+ k = 5i j 4k b AD= (i + j 9 k) (i+ 7 j k) = 0i 6j 8k AD ( AB AC) = (0i 6j 8 k) (5i j 4 k) = ( 6) ( ) + ( 8) ( 4) = = 00 The magnitude of the vector product a b is a b sin θ, where θ is the angle between the vectors. The vector product can be interpreted as a vector with magnitude twice the area of the triangle which has the vectors as two of its sides. AB = OB OA It is important to get the vectors the right way round. It is a common error to use AB = OA OB and obtain the negative of the correct answer. 0i 6j 9k = (5i j 4 k) so AD and AB AC are parallel. As the vector product is perpendicular to AB and AC, it follows that the line AD is perpendicular to the plane of the triangle ABC. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.

3 4 c The volume of the prism, P say, is given by P = AD ( AB AC) = 00= 50 5 a b 6 AC = 4 = AD= 6 = ( 5) 5 AC AD = 5 = ( 0) ( 5) 5 5 ( 0) 0 = 5 45 r = + λ c For B to lie on the line there must be a value of λ for which 5 = + λ Equating the components of the vectors 5= λ λ= Checking this value of λ for the other components y component: + λ ( ) = + ( ) ( ) =, as required. z component: + λ = + ( ) =, as required. Hence, B lies on the line. In this case, the volume of the prism is the area of the triangle ABC, which is half the magnitude of AB AC, multiplied by the distance AD. (Even if the line AD is not perpendicular to the plane of the triangle ABC, the triple scalar product is still twice the volume of the prism.) For writing vectors, you can use either the form with is, js and ks, or column vectors, which are used in this solution. Sometimes it may even be appropriate to use a miture of the two. The form using i, j and k usually gives a more compact solution but many find column vectors quicker to write. The choice is entirely up to you and you may choose to vary it from question to question. 0 The vector 5 is perpendicular to both AC and AD 45 This vector or any multiple of it may be used for the equation of the line. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.

4 5 d 5 AB= = 0 AB ( AC AD) = 5 = ( 0) + ( 5) + ( ) = = 0 The volume of the tetrahedron, V say, is given by V = AB ( AC AD) = 0 = 0= The volume of the tetrahedron is one sith of the triple scalar product. 6 a b AB = = 0 0 AC = = AB AC = 4 = A vector equation of Π is r = = = Once you have a vector n perpendicular to Let the plane, you can find a vector equation r = y of the plane using rn. = an., where a is the z position vector of any point on the plane. 6 6 Here the position vector of A has been used but the position vectors of B and C r = y = 6+ y 4z = 4 would do just as well. As the scalar 4 z 4 product is quite quickly worked out, it is a useful check to recalculate, using one of A Cartesian equation of Π is the other points. All should give the same answer, here 4 6+ y 4z= 4 Divide throughout by y+ z = 7, as required. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 4

5 6 c A vector equation of the line l is 5 r = 5 + t Parametric equations of l are = 5+ t, y= 5 t, z = t Substituting the parametric equations into y+ z = 7 (5+ t) (5 t) + ( t) = t 5+ t+ 6 4t = 7 The coordinates of T are t = 9 t = (5+ t, 5 t, t) = (5 6, 5+, + 6) = (, 8, 9 ) The two vector forms of a straight line ( r a) b = 0 and r= a+ tb are equivalent and you can always interchange one with the other. Here 5 a = 5 and b= d BT = OT OB = 8 = From part a AB = When A, B and T lie on the same straight line, AB and BT are in the same direction. So you show that the vectors AB and BT are parallel. Hence AB = BT and AB is parallel to BT. Hence A, B and T lie in the same straight line. Points which lie on the same straight line are called collinear points. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 5

6 7 a Equating the components t = u () Equating the y components + t = + u t = u () Substituting () into () u = u u = 4 As t = u, t = 4 Checking the z components For l: t = = 4 4 For m: + u = + = 4 4, so the lines do not intersect. 4 4 To show that two lines do not intersect, you find the values of the two parameters, here t and u, which make two of the components equal and then you show that, with these values, the third components are not equal. b 0 OA= + t OB = + u 0 AB = OB OA= + u t t u = t + u 4+ t + u = ( t u ) i+ ( t + u ) j+ ( 4 + t + u ) k Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 6

7 7 c If AB is perpendicular to l t u t + u = 0 4 t u + + 4t 4u t + u + 4 t u = 0 6t + 4u = 6 () If AB is perpendicular to m t u t + u = 0 4 t u t + 4u t + u 4+ t + u = 0 (4) () 0u = 6 u = 5 Substituting u = into (4) t + = 6 t = = AB AB 4t + 6u = 6 (4) t u 5 = t + u = + = t u = + = = The length of AB is given by 49 7 AB = =, as required. 5 5 As AB is perpendicular to l, the scalar product of AB with the direction of l, which is, is zero. This gives one equation in t and u Carrying out the same process with the direction of m, gives you a second equation in t andu You solve these simultaneous equations for t andu and use these values to find AB. The magnitude of this vector is the length you have been asked to find. This length is the shortest distance between the two skew lines. This question illustrates one of the methods by which this shortest distance can be found. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 7

8 8 a Since the direction vector of the line is given by 5 =, whose magnitude is 7, 0 the direction cosines l, m and n are easily found as follows: l = m= n= y 5 z b In order to find a Cartesian equation, we use the standard formula: it is = = 9 a AB= i+ j (i j+ 4 k) = 4i+ j 4k AC = 5i j+ 7 k (i j+ 4 k) = i j+ k AB AC = ( 4i+ j 4 k) (i j+ k) i j k = = i j+ k = i+ 4j+ k AB = OB OA It is important to get the vectors the right way round. It is a common error to use AB = OA OB and obtain the negative of the correct answer. b An equation of Π is r ( i+ 4j+ k) = (i j+ 4 k) ( i+ 4j+ k) = + ( ) 4+ 4 = 4+ 8= 7 So r ( i+ 4j+ k ) = 7 Once you have a vector n perpendicular to the plane, you can find a vector equation of the plane using r n = a n, where a is the position vector of any point on the plane. Here the position vector of A has been used but the position vectors of B and C would do just as well. As the scalar product is quite quickly worked out, it is a useful check to recalculate, using one of the other points. All should give the same answer, here 7 c AD = 5i+ j+ k (i j+ 4 k) = i+ j k AD ( AB AC) = (i+ j k) ( i+ 4j+ k ) = ( ) = + = The volume, V say, of the tetrahedron is given by V = AD ( AB AC) = = 6 6 Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 8

9 i j k 0 a n = = i j+ k 0 0 = i+ 8j 4k b n = i+ j k If the equation of a plane is given to you in the form r = a+ ub+ vc, then you can find a normal to the plane by finding b c. The Cartesian equation + y z = can be written in the vector form r ( i+ j k) =, where r = i+ yj+ zk Comparison with the standard form, r n= p, gives you that i+ j k is perpendicular to i j k c n n = = i j+ k = 4i j 5k = (4i + j+ 5 k) The scalar product n nis perpendicular to both n and n So to show that a vector, r say, is perpendicular to two other vectors, you can show that r is parallel to the vector product of the two other vectors. An alternative method is to show that the scalar product of r with each of the other two vectors is zero. n nis perpendicular to the plane containing n and n, and 4 i+ j+ 5 k is a multiple of n n Hence 4i+ j+ 5k is perpendicular to both n and n d r= i+ j+ k+ t(4i+ j+ 5 k ) This diagram illustrates that the line of intersection of the planes and lies in the direction of n n In this case, 4i+ j+ 5k = n nand can be used as the direction of the line, as can any other multiple of this vector. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 9

10 a a(4i+ j+ k) ( i 5j+ k) = a(4 + ( 5) + ) = a(4 5+ 6) = 5a Hence, A lies in the plane Π, as required. b BA= a(4i+ j+ k) a(i+ j 4 k) = a(i 0j+ 6 k) BA= a( i 5j+ k) For A to lie on the plane with equation r.n = 5a, when r is replaced by the position vector of A, r.n must give 5a. When a plane has an equation of the form r.n = p, the vector n is perpendicular to the plane. BA is parallel to the vector i 5j+ k, which is perpendicular to the plane Π Hence BA is perpendicular to the plane Π, as required. c The angle OBA is the angle between BO and BA Both these line segments have a definite sense and so you must use the scalar product BO BA to find θ If you used OB BA, you would obtain the supplementary angle (80 θ), which is not the correct answer. Let OAB = θ BO = a (( ) + ( ) + 4 ) = a (4) BA = a ( + ( 0) + 6 ) = a (40) BO BA= a( i j+ 4 k) a(i 0j+ 6 k) BO BA θ = a + + ( ) cos ( ) ( ) ( 0) 4 6 (4) (40) cos ( 4 0 4) a a θ = a cosθ = = (4) (40) θ=. (tothenearestonetenthof adegree) Finding the angle between two vectors using the scalar product is part of the C4 specification. Knowledge of the C4 specification is a pre-requisite of the FP specification. a A vector n perpendicular to l and l is given by n= (i+ k) ( i j+ k) i j k = = i j+ k = 6i+ j 4k The vector i+ k is parallel to l and i j+ k is parallel to l The vector product of two vectors is perpendicular to both vectors and so is perpendicular to the plane containing both lines. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 0

11 b An equation for Π has the form r (6i+ j 4 k) = p p= ( i+ 6 j k) (6i+ j 4 k ) = = 6 A vector equation of Π is r (6i+ j 4 k ) = 6 A Cartesian equation of Π is given by ( i+ yj+ zk) (6i+ j 4 k) = 6 6+ y 4z= 6, as required. To obtain a Cartesian equation of a plane when you have a vector equation in the form r.n = p, replace r by i+ yj+ zk and work out the scalar product. c The point with coordinates (, p, 0) lies on l and, hence, must lie on Π Substituting (, p, 0) into the result of part b 8+ p = 6 p = d The line of intersection lies in the direction given by i j k ( i+ j+ k) (6i+ j 4 k) = 6 4 = i j+ k = 9i+ 0j k To find one point that lies on both Π and Π Π : 6+ y 4z = 6 Π : + y+ z = () () () + 4 () gives 0+ 9y= 4 Choose =, y = 6 Substitute into () + + z = z = 7 One point on the line is (, 6, 7) An equation of the line is ( ) r ( i+ 6j 7 k) ( 9i+ 0j k ) = 0 You need to find just one point that is on both planes and there are infinitely many possibilities. Here you can choose any pair of values of and y which fit this equation. A choice here has been made which gives whole numbers but you may find, for eample, y = 0, =.4 easier to see. The form ( r a) b = 0, for the equation of a straight line, represents a line that passes through the point with position vector a and is parallel to the vector b. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.

12 a b 4 RP = 0 = c c 4 RQ = 0 = c c 4 RP RQ = c c ( c) ( + c) 5 4c = c 4( + c) = 6 5c 4 5 4c 6 5c = d Equating the components 5 4c= 4c= 8 c= Equating the y components d = 6 5c= 6 5 ( ) = 6+ 0 = 4, as required. In this solution, the vector product has be found a b ab ab using the formula a b = ab ab a b ab ab This formula can be found in the Edecel formulae booklet which is provided for the eamination. c When c = 5 4c 5+ 8 RP RQ= 6 5c = 6+ 0 = 4 An equation of Π is r 4 = 4 = + r 4 = 7 Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.

13 d In this diagram, the point N is the intersection of SS and the plane. As S is the reflection of S in Π, SS is perpendicular to Π and N is the midpoint of SS. Hence the translation (or displacement) from S to N is the same as the translation (or displacement) from N to S. The method used in this solution is to find the position vector of N and, then, to find the translation which maps S to N. This translation can then be used to find the position vector of S from the position vector of N. A vector equation of SS is r = 5 + t 4 0 Parametric equations for SS are = + t, y= 5+ 4 t, z = 0+ t () A Cartesian equation of Π is + 4y+ z = 7 () To find the position vector of N, the point of intersection of SS and Π, substitute () into () (+ t) + 4(5+ 4 t) + 0+ t = 7 + 9t+ 0+ 6t+ 0+ t = 7 6t = 6 t = + t The position vector of N is 5+ 4t = 5 4 = 0+ t 0 9 The translation which maps S to N is represented by the vector SN = 5 = The position vector of S is given by = 9 8 The translation which maps S to N will also map N to S. The position vector of S is the position vector of N added to the vector representing the translation. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.

14 4 a b a= i+ j 4 k ( i+ j k) = i k c a= 5i j k ( i+ j k) = 4i 5j k i j k ( b a) ( c a) = = i j+ k = 5i 0j 0k The vector ( b a) ( c a ) is perpendicular to both AB and AC and, so, is perpendicular to the plane Π You can use this vector, or any parallel vector, as the n in the equation r.n = p in part b. Here each coefficient has been divided by 5 This eases later working and avoids negative signs. b A vector perpendicular to Π is i+ j+ k A vector equation of Π is r (i+ j+ k) = ( i+ j k) (i+ j+ k) = + 6 = 7 c The line l is parallel to the vector i j k ( i+ k) (i+ j+ k) = 0 = i+ j+ k To find one point on both Π and Π For Π + z = Let z= 0, then = The form ( r p) q = 0 is that of a line passing through a point with position vector p, parallel to the vector q. So you need to find one point on the line; that is any point which is on both Π and Π As there are infinitely many such points, there are many possible answers to this question. The choice of z = 0 here is an arbitrary one. Substituting z = 0, = into a Cartesian equation of Π + y+ z = 7 9+ y+ 0 = 7 y = One point on Π and Π and, hence on l is (,, 0) Hence, a vector equation of l is ( ) r ( i j) ( i+ j+ k ) = 0 d A vector equation of l is r = ( i j) + t( i+ j+ k) = ( t) i+ ( + t) j+ tk At P, r is perpendicular to l ( t t t ) ( ) i+ ( + ) j+ k ( i+ j+ k) = t + t+ 4t = 0 9t = 7 t = 9 The coordinates of P are 4 ( t, + t, t) =,, At the point P which is nearest to the origin O, the position vector of P, r, is perpendicular to the direction of the line, q. Forming the scalar product r.q and equating to zero gives you an equation in t. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 4

15 5 a a b= ( i k) (4i+ j+ k) i j k = 0 = i 6j+ 6k 4 b Substituting (0, 0, 0) into y + z = 0 So the plane with equation y + z = 0 contains O. Similarly as 0+ ( ) = = 0 and 4 + = 4 6+ = 0, the plane with equation y + z = 0 contains A(, 0, ) and B(4,, ) Verify means check that the equation is satisfied by the data of this particular question. To do this you can just show that the coordinates of O, A and B satisfy y + z = 0 You do not have to show any general methods. c For B to lie on the plane with equation r. n= d (4i+ j+ k) ( i+ j k ) = d d = ( ) = + = 4 d The line of intersection L lies in the direction given by i j k ( i j+ k) ( i+ j k) = = 0i+ 7j+ 7k A vector parallel to 7j + 7k is j + k and this is parallel to the line L. The point B, which has position vector 4i + j + k, lies on both Π and Π and, hence, on L. A vector equations of L is r = 4i + j + k + t(j + k) e Rearranging the answer to part d r = 4i + ( + t)j + ( + t)k At the point X on L where OX is perpendicular to L ( ) r.( j+ k) = 0 4 i+ ( + t) j+ ( + t) k ( j+ k ) = + t+ + t = 0 The positive vector of X is 4 i+ ( ) j+ ( ) k = 4i+ j k t = 4 t = The vector n = i j + k is perpendicular to Π and the vector n = i + j k is perpendicular to Π This diagram illustrates the line of intersection of the planes is parallel to n n This gives you the direction of L. To find the equation of L, you also need one point on L. In this case, the information given in the question shows you that you already have such a point, B. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 5

16 6 a Let a= j+ k, b= i+ j+ k, and c= i+ j+ k b a= i+ j+ k ( j+ k) = i+ j k c a= i+ j+ k ( j+ k) = i+ k A vector which is perpendicular to Π is i j k ( b a) ( c a) = 0 = i j k The vector product ( b c) ( c a ) is, by definition, perpendicular to both b a and c a and, so, it is perpendicular to both AB and AC. It will also be perpendicular to the plane containing AB and AC. b ABC = ( b a ) ( c a ) = i j k ( ( ) ( ) ) = + + = 7 The vector product can be interpreted as a vector with magnitude twice the area of the triangle which has the vectors as two of its sides. c A vector equation of Π is r (i j k) = ( j+ k) (i j k) = 0 4 = 7 d A Cartesian equation of Π is y z = 7 a The vector equation r b = p and the c Cartesian equation a+ by+ cz = p are equivalents. e The distance from a point ( α, β, γ ) to a plane n+ n y+ n z+ d = 0 is nα+ n β+ nγ+ d ( n + n + n ) Hence the distance from (0, 0, 0) to y z = 7 This formula is given in the Edecel formulae booklet. If you use a formula from the booklet, it is sensible to quote it in your solution. The distance of a point from a plane is defined to be the shortest distance from the point to the plane; that is the perpendicular distance from the point to the plane. is 7 7 ( + ( ) + ( ) ) = 7 Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 6

17 6 f Let the angle between AD and Π be θ AD = ( 0) + (4 ) + ( ) = = 9 AD = 9 ( ) 7 sin = = θ 9 θ =. (d.p.) 7 a b 4 5 AB= = 0 AC = = 0 5 AB AC = = 5 0 An equation of the line, l say, which passes through D and is perpendicular to Π is r = + t 5 AD = = AD ( AB AC) = 5 = = The volume of the tetrahedron, V say, is given by V = AD ( AB AC) = = The vector product AB AC is, by definition, perpendicular to both AB and AC. So it will also be perpendicular to the plane containing AB and AC. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 7

18 7 c An equation for Π is r 5 = 5 = 5+ r 5 = d A Cartesian equation for Π is + 5y+ z = a The vector equation r b = p and the Cartesian c equation a+ by+ cz = p are equivalents and one can always be replaced by the other. Parametric equation corresponding to the equation of l found in part a are = t, y= + 5 t, z = + t Substituting these parametric equations into the Cartesian equation for Π ( t) + 5(+ 5 t) + + t = + 9t+ 0+ 5t + + t = 5t = t = 5 The coordinates of E are given by ( t, + 5 t, + t) = +, 5, =,, Use your calculator to help you work out these awkward fractions. Of course, 5 = 5 and this 5 7 is acceptable as part of the answer. However, the subsequent working is easier if all the coordinates have the same denominator. e DE = = = = = The distance d between points with coordinates (,, ) and ( y, y, y ) is given by d = ( y ) + ( y ) + ( y ) Hence 5 DE = = =, as required Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 8

19 7 f The translation mapping D to E is represented by the vector DE = = The position vector of D is given by OD' = OE+ DE = + = The coordinates of D are,, a Equating the components s= t () Equating the y components + s= + t () () + () s= s= Substitute s= into () 4= + t t = 5 Checking the z components For l : 4+ s = 4+ 6= For l : 7 t = 7 5= These are the same, so l and l intersect. The lines l and l are parallel to i+ j+ k and i+ j k respectively. ( i+ j+ k) ( i+ j k ) = + = 0 Hence l is perpendicular to l As D' is the reflection of D in Π, E is the midpoint of DD and the translation which maps D to E also maps E to D. So you can find the position vector of D by adding to the position vector of E. 5 To show that two lines intersect, you find the values of the two parameters, here s and t, which make two of the components equal and then you show that these values make the third components equal. As the scalar product a b= a b cos θ, where θ is the angle between the vectors, if, for non-zero vectors, the scalar produce is zero then cosθ = 0and θ = 90 b Substituting s = into the equation for l, the common point has position vector i+ j 4k+ ( i+ j+ k) = 5i+ 4j+ k Using r = a+ µ ( b a), an equation of l is r = 5i+ 4j+ k+ µ (4i+ λj k ( 5i+ 4j+ k)) = 5i+ 4j+ k+ µ (9 i+ ( λ 4) j 5 k) r = a+ u( b a ) is the appropriate form of the equation of a straight line going through two points with position vectors a and b a= 5i+ 4j+ k Here b= 4i+ λj k Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 9

20 8 c A vector n perpendicular to the plane, Π say, containing l and l is n= ( i+ j k) ( i+ j+ k) i j k = = 4i+ 5j+ k Let the angle between l and Π be θ n = = 4 9 i+ ( λ 4) j 5k = 9 + ( λ 4) + ( 5) = + λ λ+ + = λ λ n (9 i+ ( λ 4) j 5 k) = n (9 i+ ( λ 4) j 5 k) cos(90 θ) (4i+ 5 j+ k) (9 i+ ( λ 4) j 5 k) = + 4 ( λ 8λ ) sin ( λ 4) + ( 5) sinθ = = 4 ( λ 8λ+ ) 5λ+ 4 ( λ 8 λ+) θ The cosine of the angle between n and l can be found using the scalar product of n and a vector parallel to l This cosine is sinθ d If l, l and l are coplanar then θ = 0 and sinθ = 0 Hence 5λ+ = 0 λ= 5 Looking at the diagram in part b above, if l lies in the plan Π, then θ = 0 Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 0

21 9 a The Cartesian equations of the planes are P : y+ z = 9 P : 4+ y z = 8 () + () 0+ 5y = 5 + y = 5 () () Let = t, then y = 5 = 5 t From () z= 4+ y 8 = 4t+ (5 t) 8= 7 t The general point on the line of intersection of the planes has coordinates ( t, 5 t, 7 t) The distance, y say, from O to this general point is given by y = t + (5 t) + (7 t) = t + 5 0t+ 4t t + 4t = 9t 48t + 74 () Differentiating both sides of () with respect to t dy y 8t 48 dt = At a minimum distance d y 0 dt = t 48= 0 t = = 8 Substituting into () Points on the line of intersection of the two planes can be found by solving simultaneous the Cartesian equations of the two planes. As there are equations in unknowns, there are infinitely many solutions. A free choice can be made for one variable, here is given the parameter t, and the other variables cab then be found in terms of t. This is the equivalent of the parametric equations of the common line = t, y = 5 t, z= 7 t The equivalent vector equation of this line is r = 5j+ 7 k+ t( i j k) A calculus method of finding the minimum distance is shown here. You could instead use the property that, at the shortest distance, the position vector of the point is perpendicular to the common line. This method is illustrated in Question. 8 8 y = = = 0 The shortest distance from O to the line of intersection of the planes is 0 b The line of intersection of P and P has vector equation r = 5j+ 7 k+ t( i j k ) Hence the vector i j k is perpendicular to Π An equation of P is r ( i j k) = ( j+ k) ( i j k) = 4 = 6 The common on line of P and P is a normal to the plane P which is perpendicular to P and P Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.

22 9 c Substituting ( t, 5 t, 7 t) into y z = 6 t (5 t) (7 t) = 6 t 0+ 4t 4+ 4t = 6 9t = 8 t = The position vector of the common point is ti+ (5 t) j+ (7 t) k = i+ j+ k 0 Equating the equation of the hyperbola with that of the straight line we find that the y-coordinates of y 4y 6 = 64 A and B are given by the solutions of ( ) This equation can be rewritten as 4y 6y = 64 y 4y 6= 0 which has solutions ± 5 Using = and rationalising we find that the -coordinate of the midpoint is 64 y The point that lies on the three planes is given by substituting the general point on the line of intersection of P and P into the Cartesian equation of P = = 8 The y-coordinate is then obtained from the equation of the straight line at the point = 8, so that the coordinates of M are ( 8,) a You have to recognise that = t, y= 6t is a parabola and draw it passing through the origin with the correct orientation. b For the intersections, substitute into y = 7 6t = t 7 t 6t 7 = 0 ( ) t t 4 = 0 ( t 6)( t+ 4) = 0 t = 6, 4 For A, say, t = 6 = t = = = 08, y 6t 6 For B, say, t = 4 = t = = = = = t, y= 6t ( 4) 48, y 6t 4 Using d = ( ) + ( y y ) AB AB = (08 48) + (6 ( 4)) = = 60 = = ( 60 ) 60 ( t, 6 t) is a general point on the parabola. The points A and B must be of this form and, if they also lie on the line with equation y = 7, the points on the parabola must also satisfy the equation of the line. The question does not tell you which point is A and which point is B but, as you are only asked to find the distance between them, it does not matter which is which and you can make your own choice. You are asked to give your answer as a surd in its simplest form is not acceptable as it is not a surd and 700is not the simplest form. A surd in its simplest form contains the square root of the smallest possible single number. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.

23 Since P and Q lie on the parabola, their coordinates are (, ) and (9, 6) respectively. Therefore, the line through these points has equation y= +, and the midpoint of PQ is R=(5, 4) Then we can easily find that the equation of the perpendicular bisector to PQ is y 4= ( 5) y 4= + 0 y= + 4 Now if we intersect this with the equation of C we find that the -coordinates of M and N are the solutions of ( + 4) = 4 We solve this: = = = 0 5± 9, = In particular, they can be written in the form 5 5 ± 9, so λ =, µ= You should mark on your diagram any points given in the question. Here mark S, P and Q. Diagrams often help you check your working. Here, for eample, it is obvious from the diagram that Q must have a negative y-coordinate. If you get y = 4 (a mistake it is easy to make), you would know you were wrong and look for an error in your working. a S(4, 0) b Using y y = y y with (, y ) = (4, 0) and (, y ) = (6,6), an equation of SP is y 0 4 = y 4 = 6 y = 4 6 ( 4) y = y 6 = 0 The focus of the parabola with equation y = 4a has coordinates (a, 0). Here a = 4. The question asks you to write down the answer, so you do not have to show working. Methods for finding the equation of a straight line are given in Chapter 5 of Edecel AS and A Level Modular Mathematics AS and A-Level, Core Mathematics You can use any correct method for finding the line. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.

24 c From (b) y+ 6 = 4 Substitute for in y y = y = 6 4 y+ 6 6 = y y 64= 0. ( y 6)( y+ 4) = 0 y = 6 corresponds to the point P. For Q, y = = = = 4 4 The coordinates of Q are (, 4). To find Q you solve the simultaneous equations 4 y 6= 0and y = 6 The method of using substitution, when one equation is linear and the other is quadratic, is given in Chapter of Edecel AS and A Level Modular Mathematics AS and A- Level, Core Mathematics 4 First of all, we need to find the coordinates of the focus S. Obviously, the y-coordinate is 0; as for the -coordinate, if we consider the equation of C to be y = then we know that the coordinate must be 0 4 = 5 4 Now, this allows us to find the equation of the straight line l: it is y = ( 5) This intersects C in two points, whose y-coordinates satisfy We solve this: 4 y y = 5 0 y y = 0 5 5y = y 00 y 5y 00= y y = 5 0 This has only the one positive solution = 0 Then the point P has coordinates (0, 0) The area of the region R is given by the area enclosed by C between O and P minus the area of the triangle PQS, where Q is the projection of P on the -ais. As in the first Cartesian quadrant the parabola is the graph of the function y= 5, the area can be written as follows: d = ( ) = = = 5(40 5) 50= = = Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 4

25 5 The Cartesian equation of the hyperbola is y = 6 Therefore, the -coordinates of P and Q are the solutions of the equation (+ 4) = 6 We solve this: + 4= 6 + 8= 0 = ±, Then P = (, 8) and Q= ( 4, 4) 6 6 Substitute = 8 t, y = into y = + 4 t 4 6 = 8 t+ 4 t 4 6 = t+ 4 t Multiplying by t and rearranging t + 4t 6 = 0 ( ) t + t 8 = ( t+ 4)( t ) = 0 t =, 4 For A, say, t = = 8t = 8 = and y = 8 t = = The coordinates of A are (6, 8) For B, say, t = 4 = 8t = 8 4= 6 6 and y = 4 t = 4 = The coordinates of B are (, 4) The -coordinate of the midpoint of AB is given by M 6 = = 8 The y-coordinate of the midpoint of AB is given by y M 8 4 = = The coordinates of M are ( 8, ) 6 8 t, is a general point on the t rectangular hyperbola. The points A and B must be of this form and, if they also lie on the line with equation y = + 4, the points on the parabola 4 must also satisfy the equation of the line. The question does not tell you which point is A and which point is B but, as the midpoint is not affected by the choice, it does not matter which is which and you can make your own choice. The coordinates of the midpoint M of A(, y ) and B(, y) are given by + y + y ( M, ym), = Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 5

26 7 a The y-coordinates of P and Q are the solutions of the equation ( y) y= 0 We solve this: ( y) y = 0 y y = 0 y 6y+ 5= 0 ( y 5)( y ) = 0 We substitute y=5 and y= into the equation of the hyperbola to find the coordinates of P and Q: P=(, 5) and Q=(0, ) b The line intersects the coordinate ais in the points (0, 6) and (, 0) The area of the triangle these points form with the origin is 6 The area enclosed by the hyperbola between the points P and Q is 0 0 d = 0 = 0 d= = 0(ln0 ln ) = = 0ln5 Therefore, the area of the triangle minus the area enclosed by the hyperbola is 6 0ln 5 However, we must also remove the parts that go from the intersection with the aes to P and Q: the trapezoid that is determined by P, the point (0, 6) and their projections on the -ais has area, while the triangle determined by Q, its projection and (0, ) has area Then the area of the shaded region is 6 0ln5 = 4 0ln5 Hence a = 4, b = 0 and c = 5 8 a (4 t, 48 t ) must satisfy the equation y = 44 t 4t 48t = = = t = For P, = 4t = 4 = 6 y = 48t = 48 = 4 The coordinates of P are (6, 4) The point with coordinates ( at, at) always lies on the parabola with equation y = 4 a, in this case a = 4, so P is on the parabola for all t. There will however only be one value of t for which P also lies on the rectangular hyperbola and you find it by substituting (4 t, 48 t ) into y = 44 Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 6

27 8 b y = 96 y = (96) = = 6 6= 4 6.So dy 6 = = = = d (4 6) 4 6 dy 6 At = 6, = = d 6 Using y y = m( ), an equation of the tangent to the parabola at P is y 4= ( 6) = y= + 44 y = = 44 dy 44 = 44 = d dy 44 At = 6, = = 4 d 6 c Using y y = m( ), an equation of the tangent to the hyperbola at P is y 4= 4( 6) = 4+ 4 y= Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 7

28 9 The t > 0 in the question implies that you only need to consider the part of the parabola where y > 0, that is the part above the -ais. To find an equation of the tangent PT y = 4a y = a ( y > 0) y = a a dy = a = d Using d n d ( ) = n, = = d d n At = at, = a t dy a = = d t a t Using y y = m( ) with (, y) = ( at, at), an equation of the tangent to the parabola at P is y at= at t ty at = at ( ) ty= + at () To find -coordinate of T, substitute y = 0 into () 0= + at = at Using d = ( ) + ( y + y ) with (, y ) = ( at, at) and (, y ) = ( at, 0) PT = ( at ( at )) + (at 0) = ( at ) + 4a t = 4a t + 4a t = 4 a t ( t + ) () To find an equation of the normal PN Using mm =, m = m = t t The tangent crosses the -ais where y = 0 The normal is perpendicular to the tangent. From working earlier in the question, you know that the gradient of the tangent is t Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 8

29 9 cont. Using y y = m ( ) with (, y) = ( at, at), an equation of the normal to the parabola at P is y at = t( at ) = t+ at + = at+ at () y t To find the -coordinate of N, substitute y = 0 into () t = at+ at = a+ at The normal crosses the -ais where y = 0 Using d = ( ) + ( y y ) with (, y ) = ( at, at) and (, y ) = ( a+ at, 0) PN = ( at ( a+ at )) + (at 0) = ( a) + ( at) = 4a + 4a t a t From () and (4) PT PN Hence = 4 ( + ) (4) 4 a t ( t + ) = = t 4 a ( t + ) PT t, as required. PN = Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 9

30 0 a 9 y = = 9 dy 9 = 9 = d dy 9 When = t, = = d ( ) t t Using y y = m( ) with (, y) = t,, t the tangent to H y = ( t ) t t t ( t ) t y t = + + t y = t 6, as required. d ( n ) n, Using = n d d 9 (9 ) = 9 = 9 = d You can use (, y) = t, in the t formula y y = m( ) in eactly the same way as you use coordinates with numerical values like, say, (6, 4) b For A, substitute, y = 0 into = 6t OA= 6t For, B, substitute, = 0 into Area 6 6 t y = 6t y = OB = t t + t y= 6t + t y= 6t 6 OAB = OA OB = 6t = 8 t This area, 8, is a constant independent of t. This result means that no matter which point you take on this rectangular hyperbola the area of the triangle OAB is always the same, 8 Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 0

31 a c y = = c dy c = c = d At P, = ct y c = = d c t t d For the gradient of the normal, using mm' =, m' = m' = t t Using y y = m( ) with (, y) ct, c = t, an equation of the normal to the hyperbola at P is c y = t ( ct) t c y = t ct t 4 ( t) yt c= t ct + = 0 4 t ty ct c 4 t ty c( t ) = 0, as required b For G, substitute y = into the result in part (a) 4 t t c( t ) = 0 4 ( t t) = c( t ) 4 c( t ) c( t ) ( t + ) ct + c c = = = = ct+ t t t( t ) t t c c The coordinates of G are ct+, ct+ t t PG = ( ) + ( y y ) c c c = ct+ ct + ct+ t t t c = + c t = c t +, asrequired. t t The normal to H at P is perpendicular to the tangent at P. To work out perpendicular gradients you will need the formula mm' = So you have to find the gradient of tangent before you can find the gradient of the normal. You find the gradient of the tangent using differentiation. When you are asked to show that an equation is true, you must use algebra to transform your equation to the equation eactly as it is printed in the question. You could not cancel the t = +, as then ( t ) terms if ( t ) would be 0, but these cases are eplicitly ruled out in the question. Using d = ( ) + ( y y ) with c c (, y) = ct+, ct+ and t t c (, y) = ct, t Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.

32 a (8, 0) b = 8 c Using y y = y y with (, y ) = (,8) and an equation of PQ is y 8 = 8 y 8 = 40 0 y 4 = 4+ 8 y+ 4 = Substitute y = = = 8 (, y ) = (, ), The coordinates of S (8, 0) satisfy the equation of PQ. Hence S lies on the joining P and Q. If y = 4 a, the focus has coordinates (a, 0) and the directri has equation = a Comparison of y = 4awith y =, shows that, in this case, a= 8 Methods for finding the equation of a straight line are given in Chapter 5 of Edecel AS and A level Modular Mathematics AS and A-Level, Core Mathematics You can use any correct method for finding the line. = (6 ) = 6 = 4 On the upper half of the parabola, in the first quadrant, the y-coordinates of P are positive. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.

33 d y = y =± 4 P is on the upper half of the parabola where y=± 4 dy = 4 = d dy At =, = = d Usingy y = m( ), the tangent to C at P is y 8= ( ) = 4 y= () Q is on the lower half of the parabola where y= 4 dy = 4 = d dy At =, = = = d 4 Using y y = m( ), the tangent to C at Q is On the lower half of the parabola, in the fourth quadrant, the y-coordinates of P are negative. y+ = ( ) = + 6 y = 6... () To find the -coordinate of the intersection of the tangents, from () and () + 4= 6 5 = 0 = 0 = 8 5 The equation of the directri is = 8and, hence, the intersection of the tangents lies on the directri. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.

34 a y = 4a y = a y = a a dy = a = d The normal is perpendicular to the tangent, so you must first find the gradient of the tangent. Then you use mm' = to find the gradient of the normal. At = at, = a t dy a = = d t a t Using mm ' = m' = m' = t t Using y y = m'( ) with (, y) = ( at, at), an equation of the normal to the parabola at P is When you are asked to show that an equation y at = t( at ) is true, you must use algebra to transform = t+ at your equation to the equation eactly as it is printed in the question. y+ t = at+ at, as required. You substitute = aq and y= aqinto the b Let the coordinates of Q be ( aq, aq ) The point Q lies on the normal at P, so aq+ taq = at+ at + = aq at atq at 0 + = a( q t) at( q t ) 0 a( q t) + at( q t)( q+ t) = 0 a( q t)( + t( q+ t)) = = tq t 0 t q = + t answer to part (a). There are two possibilities here: q t = 0and + t( q+ t) = 0 As P and Q are different points, q t, so you need only consider the second possibility. You use this to find q in terms of t. t + Replace the q in ( aq, aq) by t You need not attempt to simplify this further. The coordinates of Q are t + t + a, a t t Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 4

35 4 a Substitute (,8) into y = 4a 64 64= 4a = 8a a = = 8 8 b y = a a dy = a = d Using d n d ( ) = n, = = d d n When a= 8 and = dy 8 d = = = Using y y = m( ) the tangent to C at P is y 8= ( ) = 4 y= + 4 c At X, y= 0 0= + 4 = SO OX = At Y, = 0 y= 0+ 4 y = 4 SO OY = 4 The method for obtaining the equation of a straight line when you know its gradient and one point which it passes through is given in Chapter 5 of Edecel AS and A Level Modular Mathematics AS and A-Level, Core Mathematics Area OXY = OX OY = 4 = 4 Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 5

36 5 a a c y = = c dy c d = = At P, = ct c The normal to H at P is perpendicular to the tangent at P. To work out perpendicular gradients you will need the formula mm = So you have to find the gradient of the tangent before you can find the gradient of the normal. You find the gradient of the tangent by differentiating. dy c = = d c t t For the gradient of the normal, using mm ' =, m' = m' = t t Using y y = m( ) with c (, y ) = ct,, t equation of the normal to the hyperbola at P is c y = t ( ct) t = t ct c y = t + ct t, as required...() an When you are asked to show that an equation is true, you must use algebra to transform your equation to the equation eactly as it is printed in the question. In this case, the form of the printed equation suggests a method for the net part of the question. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 6

37 5 b c c = =...() y c y For Q, from () and () c c t + ct = t t and collect terms as a quadratic in 4 t c ct c t + ( ) = 0 ( ct)( t + c) = 0 = ctcorresponds to P c For Q, = t Substitute the -coordinate into () c c y = = = ct c t c The coordinates of Q are t t 4 ( ) c ( ) ( ) c t 4 X t c( t ) t = 4 = c( t ) 4 Y t c t t, ct+ 4 4 t ct c c( t ) X = = = t t ct c 4 4 t + ct ct ct c( t ) Y = = = t t =, as required. t ( ) Writing the equation of the rectangular hyperbola, in the form y =..., enables you to eliminate y quickly between () and (). The -coordinate of P ct, must be a solution of the equation. So ( ct) must be a factor of the quadratic and, so, you can write down the second factor using t = t and ct c = c t Multiplying all terms on the top and bottom of the fraction by t The coordinates of the midpoint of A(, y) and B(, y) are given by + y + y ( X, Y) =, Multiplying all terms on the top and bottom of the fraction by t. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 7

38 6 a c y = = c dy c d = = At = cp c dy c = = d c p p Using y y = m( ) with, y cp, c =, an p equation of the tangent to the hyperbola is c y = ( cp ) p p c c y = + p p p c y = + p p ( p ) p y = + cp,as required...() b The tangent at Q is q y cq = +...() To find the y-coordinate of N subtract () from () ( p q ) y = c( p q) c( p q) c( p q) c y = =, as required. p q ( p q) ( p+ q ) = p+ q The equation of the tangent at Q is the same as the equation of the tangent at P with the ps replaced by qs. You do not have to work out the equation twice. To find y, you eliminate from equations () and (). These equations are a pair of simultaneous linear equations and the method of solving them is essentially the same as you learnt for GCSE. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 8

39 6 c To find the -coordinate of N substitute the result of part (b) into () cp = + cp p+ q cp cp( p+ q) cp cpq = cp = = p+ q p+ q p+ q The gradient of PQ, m say, is given by c ( q p) pq c c p p m= = cp cq c ( p q) = pq The gradient of ON, m' say, is given by c p+ q cpq p+ q m' = = pq The gradient m is found using y y m= (, y) cp, c = and (, ), c y = cq p q with Given that ON is perpendicular to PQ mm ' = = p q = pq pq Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 9

40 7 a c y = = c dy c = c = d At = cp dy c = = d c p p c Using y y = m( ) with (, y) = cp,, p an equation of the tangent to the hyperbola is c y = ( cp ) p p c c y = + p p p c y+ = p p ( p ) p y+ = cp,asrequired...() b ( cp,0) The tangent crosses the -ais at y = 0 c Using ( ) ( y y) d = + with You can put y = 0 into () in your head and just write down the coordinates of S. No working is needed. (, y) cp, c = and (, y) = ( cp,0) p c c = ( ) + 0 = c p + PS cp cp 4 p 4 p + = c p + c = p p c PS = ( + p ) p p There are many possible forms for this answer. Any equivalent form would gain full marks. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 40

41 7 d To find the equation of the normal at P The working in part (a) shows the gradient of the tangent is p Let the gradient of the normal be m ' Using mm ' =, m' = m' = p p Using y y m'( ) = with ( y ) c y = p ( cp) p = p cp c p = y + cp p, cp, c = p an equation of the normal to the hyperbola at P is To find the -coordinate of R, substitute y = 0 c c p = + cp = cp p p 4 c c p + RS = cp cp cp c = + = p p p Area RPS = RS height 4 p + c 4c = c p p c 4 = ( p + ) 4 p p = p + p = p= 8 c c The coordinates of P are cp, =, c p To find an epression for the area of the triangle you can obtain the length of the side RS and use that as the base of the triangle in the formula for the area of the triangle. First you need to obtain an equation of the normal and use it to find the coordinates of R. If RS is taken as the base of the triangle, the height of the triangle is the y-coordinate of P. As p > 0 is given in the question, you need not consider the alternative solution p = 8 The coordinates of P are ct, c ; therefore, the coordinates of the midpoint between P and the origin t ct c are, t ct c c The product yields : in other words, the locus of the midpoint is described by the equation t 4 c y = 4 This is the equation of a hyperbola. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 4

42 9 A diagram helps you see and understand what is going on. You should label important points. If letters are not given in the question, you can make your own up. Putting them on a diagram makes your method clear to the eaminer. By the definition of a parabola SP = PN SP = PN S = (5,0), P= (, y) = ( ) + ( y y) SP = ( 5) + y PN = + 5 SP = PN ( 5) + y = ( + 5) y = y = 0 Using the formula d = ( ) + ( y y ), it is often easier to work with the distances squared rather than with distances. Along PN, the distance from P to the y-ais is, and it is a further distance 5 from the y-ais to N. Multiply out the brackets using ( a+ b) = a + ab+ b Then cancel the equal terms on both sides of the equation. Comparing with y = 4 a, this is the required form with a= 5 40 a When you draw a sketch, you should show the important features of the curve. When drawing an ellipse, you should show that it is a simple closed curve and indicate the coordinates of the points where the curve intersects the aes. b b = a ( e ) 9 = 6( e ) = 6 6e e = = 6 6 e= 7 4 The formula you need for calculating the eccentricity and the coordinates of the foci are given in the Edecel formula booklet you are allowed to use in the eamination. You should be familiar with the formulae in that booklet. You should quote any formulae you use in your solution. c The coordinates of the foci are given by ( ) 7 ( ± ae,0) = ± 4,0 = ± 7,0 4 Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 4

43 4 a b = a ( e ) 4 = 6( e ) = 6e e = = = e = The formula for calculating the eccentricity is b = a ( e ) It is important not to confuse this with the formula for calculating the eccentricity of an ellipse b = a ( e ) b The coordinates of the foci are given by 5 ( ± ae,0) = ± 4,0 =( ± 5,0) The formulae for the foci of an ellipse and a hyperbola are the same ( ± ae,0) Therefore the distance between the foci is 5+ 5 = 4 5 c In this sketch, you should show where the curves cross the aes. Label which curve is H and which is E. These two curves touch each other on the -ais. 4 a b = a ( e ) ( e ) 4 = 9 = 9 9e e = = e= 9 9 As the coordinates of the foci of an ellipse are (±ae,0), you first need to find the eccentricity of the ellipse using b a ( e ) and b = = with a = The coordinates of the foci are given by 5 ( ± ae,0) = ±,0 =( ± 5,0) Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 4

44 4 b In this question, you are not asked to draw a diagram but with questions on coordinate geometry it is usually a good idea to sketch a diagram so you can see what is going on. a The equations of the directrices are = ± e 9 =± =± 5 5 Let the line through P parallel to the -ais intersect the directrices at N and N', as shown in the diagram 9 8 N' N = = 5 5 The focus directri property of the ellipse gives that SP = epn and S'P = epn' SP + S'P = epn'+ epn' = e (PN + PN') = en'n 5 8 = = 6, as required. 5 If you introduce points, like N and N' here, you should define them in your solution and mark them on your diagram. This helps the eaminer follow your solution. 4 a + 4y = y + = 4 b = a e ( ) ( e ) = 4 = 4 4e 4 e = = e= 4 4 You divide this equation by Comparing the result with y + =, a = 4 a b and b = and you use b = a ( e ) to calculate e. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 44

45 4 b + 4y = Differentiate implicitly with respect to dy 6+ 8y = 0 d d y 6 = = d 8y 4y Differentiating implicitly using the chain rule, d dy d dy ( 4y ) = ( 4y ) = 8y d d dy d At, dy = = d 4 Using y y =m( ), an equation of the tangent is y = ( ) = + y = + c Sketching a diagram makes it clear that the area of the triangle is to be found using the standard epression base height base S'S and the height OG. with the The coordinates of S are,0 =,0 = (,0) ( ae ) By symmetry, the coordinates of S' are (,0) The y-coordinate of G is given by y = 0+ = SS' G = base height = ' S S OG = = You find the y-coordinate of G by substituting = 0 into the answer to part a. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 45

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