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Unit Mathematics Contents Polnomials and Quadratics 6 Quadratics 6 The Discriminant 66 Completing the Square 67 Sketching Parabolas 70 5 Determining the Equation of a Parabola 7 6 Solving Quadratic Inequalities 7 7 Intersections of Lines and Parabolas 76 8 Polnomials 77 9 Snthetic Division 78 0 Finding Unknown Coefficients 8 Finding Intersections of Curves 8 Determining the Equation of a Curve 86 Approimating Roots 88 Integration 89 Indefinite Integrals 89 Preparing to Integrate 9 Differential Equations 9 Definite Integrals 95 5 Geometric Interpretation of Integration 96 6 Areas between Curves 0 7 Integrating along the -ais 06 Trigonometr 07 Solving Trigonometric Equations 07 Trigonometr in Three Dimensions 0 Compound Angles Double-Angle Formulae 6 5 Further Trigonometric Equations 7 Circles 9 Representing a Circle 9 Testing a Point 0 The General Equation of a Circle 0 Intersection of a Line and a Circle 5 Tangents to Circles 6 Equations of Tangents to Circles 7 Intersection of Circles 6 - ii - HSN000

Unit Polnomials and Quadratics OUTCOME Polnomials and Quadratics Quadratics A quadratic has the form numbers, provided a 0. You should alread be familiar with the following. a + b + c where a, b, and c are an real The graph of a quadratic is called a parabola. There are two possible shapes: concave up (if a > 0 ) concave down (if a < 0 ) This has a minimum turning point This has a maimum turning point To find the roots (i.e. solutions) of the quadratic equation we can use: factorisation; completing the square (see Section ); the quadratic formula: EXAMPLES. Find the roots of ( + )( ) = 0. Solve = 0 a b c + + = 0, b ± b ac = (this is not given in the eam). a = 0. + = 0 or = 0 = =. ( )( ) + 8 + 6 = 0. + 8 + 6 = 0 + + = 0 + = 0 or + = 0 = =. Page 6 HSN000

Unit Polnomials and Quadratics. Find the roots of + = 0. We cannot factorise +, but we can use the quadratic formula: ± ( ) = ± 6 + = ± 0 = 5 = ± = ± 5. Note If there are two distinct solutions, the curve intersects the -ais twice. If there is one repeated solution, the turning point lies on the -ais. If b ac < 0 when using the quadratic formula, there are no points where the curve intersects the -ais. Page 65 HSN000

Unit Polnomials and Quadratics The Discriminant Given a + b + c, we call b ac the discriminant. This is the part of the quadratic formula which determines the number of real roots of the equation a + b + c = 0. If b If b ac > 0, the roots are real and unequal (distinct). ac = 0, the roots are real and equal (i.e. a repeated root). If b ac < 0, the roots are not real; the parabola does not cross the -ais. EXAMPLE. Find the nature of the roots of 9 + + 6 = 0. a = 9 b = c = 6 Since b b ac = 9 6 = 576 576 = 0 ac = 0, the roots are real and equal.. Find the values of q such that Since a = 6 b = c = q 6 q 0 6 q 0 + + = has real roots, b b ac 0 6 0 q q 0 q q q 6. + + = has real roots. ac 0 : two roots one root no real roots Page 66 HSN000

Unit Polnomials and Quadratics. Find the range of values of k for which the equation no real roots. k + 7 = 0 has For no real roots, we need a = k b = c = 7 b b ac < 0 : ac < 0 k ( 7) < 0 + 8k < 0 8k < k < 8 k <.. Show that ( k ) ( k ) ( k ) + + + + = 0 has real roots for all real values of k. a = k + b = k + c = k b ac 7 ( k ) ( k )( k ) 9k k ( k )( k 8) = + + = + + + = k + k + k + 9 8 k = + + ( k ) k 6 = + 6. Since b ac ( k ) = + 6 0, the roots are alwas real. Completing the Square The process of writing a b c called completing the square. = + + in the form ( ) = a + p + q is Once in completed square form we can determine the turning point of an parabola, including those with no real roots. The ais of smmetr is = p and the turning point is ( p, q). + p = + p + p. For eample, The process relies on the fact that ( ) we can write the epression + using the bracket ( + ) since when multiplied out this gives the terms we want with an etra constant term. Page 67 HSN000

Unit Polnomials and Quadratics This means we can rewrite the epression gives us the correct and terms, with an etra constant. We will use this to help complete the square for Step Make sure the equation is in the form = a + b + c. Step Take out the -coefficient as a factor of the and terms. Step Replace the + k epression and compensate for the etra constant. + k using ( + k ) since this = +. = +. ( ) = +. = ( ( + ) ) = ( + ). Step Collect together the constant terms. = ( + ) 5. Now that we have completed the square, we can see that the parabola with equation = + has turning point (, 5). EXAMPLES. Write = + 6 5 in the form = ( + p) + q. = + 6 5 = ( + ) 9 5 = ( + ). Note You can alwas check our answer b epanding the brackets.. Write + ( ) ( ) + in the form ( + p) + q. = + 9 = + 5. Page 68 HSN000

Unit Polnomials and Quadratics. Write = + 8 in the form = ( + a) + b and then state: (i) the ais of smmetr, and (ii) the minimum turning point of the parabola with this equation. = + 8 ( ) ( ) = + 6 = + 9. (i) The ais of smmetr is =. (ii) The minimum turning point is (, 9).. A parabola has equation = + 7. (a) Epress the equation in the form = ( + a) + b. (b) State the turning point of the parabola and its nature. (a) = + 7 ( ) = + 7 (( 9 ) ) ( ) ( ) = + 7 = 9 + 7 =. (b) The turning point is ( ), and is a minimum. Remember If the coefficient of is positive then the parabola is concave up. Page 69 HSN000

Unit Polnomials and Quadratics Sketching Parabolas The method used to sketch the curve with equation depends on how man times the curve intersects the -ais. = a + b + c We have met curve sketching before. However, when sketching parabolas, we do not need to use calculus. We know there is onl one turning point, and we have methods for finding it. Parabolas with one or two roots Find the -ais intercepts b factorising or using the quadratic formula. Find the -ais intercept (i.e. where = 0). The turning point is on the ais of smmetr: O The ais of smmetr is halfwa between two distinct roots. O A repeated root lies on the ais of smmetr. Parabolas with no real roots There are no -ais intercepts. Find the -ais intercept (i.e. where = 0). Find the turning point b completing the square. EXAMPLES. Sketch the graph of = 8 + 7. Since b ac = ( 8 ) 7 > 0, the parabola crosses the -ais twice. The -ais intercept ( = 0) : = ( 0) 8( 0) + 7 = 7 ( 0, 7 ). The -ais intercepts ( = 0) : 8 + 7 = 0 ( )( 7) = 0 = 0 or 7 = 0 = = 7 (, 0 ) ( 7, 0 ). The ais of smmetr lies halfwa between = and = 7, i.e. =, so the -coordinate of the turning point is. Page 70 HSN000

Unit Polnomials and Quadratics We can now find the -coordinate: ( ) ( ) = 8 + 7 = 6 + 7 = 9. So the turning point is (, 9).. Sketch the parabola with equation Since b ac ( 6) ( ) ( 9) 0 = 6 9. = =, there is a repeated root. The -ais intercept ( = 0) : = ( 0) 6( 0) 9 = 9 ( 0, 9 ). Since there is a repeated root, (, 0) is the turning point.. Sketch the curve with equation The -ais intercept ( = 0) : = 6 9 0 ( ) + 6 + 9 = 0 ( + )( + ) = 0 + = 0 = = 8 +. (, 0 ). Since b ac = ( 8 ) < 0, there are no real roots. The -ais intercept ( = 0) : = ( 0) 8( 0) + = ( 0, ). So the turning point is (, 5 ). Complete the square: = 8 + 7 O 9 = 6 9 ( ) = + = ( ) 8 + O = ( ) + 5. = + = 8 + 7 7 (, 9) 8 O (, 5) Page 7 HSN000

Unit Polnomials and Quadratics 5 Determining the Equation of a Parabola Given the equation of a parabola, we have seen how to sketch its graph. We will now consider the opposite problem: finding an equation for a parabola based on information about its graph. We can find the equation given: the roots and another point, or the turning point and another point. When we know the roots If a parabola has roots where k is some constant. = a and = b then its equation is of the form = k ( a)( b) If we know another point on the parabola, then we can find the value of k. EXAMPLES. A parabola passes through the points (, 0 ), ( 5, 0 ) and ( 0, ). Find the equation of the parabola. Since the parabola cuts the -ais where = and = 5, the equation is of the form: = k ( )( 5 ). To find k, we use the point ( 0, ): = k ( )( 5) = k ( 0 )( 0 5) = 5k k = 5. So the equation of the parabola is: 5 5 = ( )( 5) ( 6 5) = + 8 = +. 5 5 Page 7 HSN000

Unit Polnomials and Quadratics. Find the equation of the parabola shown below. (, 6) Since there is a repeated root, the equation is of the form: = k ( + 5)( + 5) = k ( + 5 ). 6 5 Hence = ( + ). 5 O To find k, we use (, 6 ) : = k ( + 5) 6 = k ( + 5) 6 6 k = 6 =. When we know the turning point Recall from Completing the Square that a parabola with turning point p, q has an equation of the form ( ) where a is some constant. = a( + p) + q If we know another point on the parabola, then we can find the value of a. EXAMPLE. Find the equation of the parabola shown below. O (, ) 7 Since the turning point is (, ) the equation is of the form: ( ) = a. = 5 6. Hence ( ), To find a, we use ( 0, 7) : ( ) = a ( ) 7 = a 0 6a = 5 a = 5 6. Page 7 HSN000

Unit Polnomials and Quadratics 6 Solving Quadratic Inequalities The most efficient wa of solving a quadratic inequalit is b making a rough sketch of the parabola. To do this we need to know: the shape concave up or concave down, the -ais intercepts. We can then solve the quadratic inequalit b inspection of the sketch. EXAMPLES. Solve + < 0. The parabola with equation = + is concave up. The -ais intercepts are given b: + = 0 ( + )( ) = 0 + = 0 = or = 0 =. Make a sketch: = + So + < 0 for < <.. Find the values of for which The parabola with equation The -ais intercepts are given b: 6 + 7 = 0 ( ) 7 6 = 0 ( + )( ) = 0 + = 0 or = 0 = 6 7 0 +. 6 7 =. = + is concave down. Make a sketch: = 6 + 7 So 6 7 0 + for. Page 7 HSN000

Unit Polnomials and Quadratics. Solve 5 + > 0. The parabola with equation = 5 + is concave up. The -ais intercepts are given b: 5 + = 0 Make a sketch: ( )( ) = 0 = 5 + = 0 or = 0 = =. So 5 + > 0 for < and >.. Find the range of values of for which the curve is strictl increasing. We have d 5 d = +. The curve is strictl increasing where + 5 > 0. + 5 = 0 Make a sketch: ( )( + 5) = 0 = 0 or + 5 = 0 = = 5. 5 5 = + + Remember Strictl increasing means d 0 d >. = + 5 So the curve is strictl increasing for < 5 and >. 5. Find the values of q for which ( ) For no real roots, b ac < 0 : a = b ac = q q b = q = ( q )( q ) q c = q = q 8q + 6 q + q + q = 0 has no real roots. ( ) ( )( ) q 0q 6. = + We now need to solve the inequalit q 0q + 6 < 0. The parabola with equation = 0 + 6 is concave up. q q Page 75 HSN000

Unit Polnomials and Quadratics The -ais intercepts are given b: q 0q + 6 = 0 ( q )( q ) 8 = 0 q = 0 or q 8 = 0 Therefore b q = q = 8. ac < 0 for q 8 no real roots when < q < 8. Make a sketch: = q 0q + 6 < <, and so ( ) 7 Intersections of Lines and Parabolas 8 + q + q = 0 has To determine how man times a line intersects a parabola, we substitute the equation of the line into the equation of the parabola. We can then use the discriminant, or factorisation, to find the number of intersections. q If b ac > 0, the line and curve intersect twice. If b ac = 0, the line and curve intersect once (i.e. the line is a tangent to the curve). If b ac < 0, the line and the parabola do not intersect. EXAMPLES. Show that the line = 5 is a tangent to the parabola and find the point of contact. Substitute = 5 into: = + 5 = + 0 + = + = 0 ( )( ) = 0. Since there is a repeated root, the line is a tangent at =. = + To find the -coordinate, substitute = into the equation of the line: = 5 =. So the point of contact is (, ). Page 76 HSN000

Unit Polnomials and Quadratics. Find the equation of the tangent to = + that has gradient. The equation of the tangent is of the form = m + c, with m =, i.e. = + c. Substitute this into c = + + c = + + = 0. Since the line is a tangent: b ( ) ( c ) = 0 ac = 0 9 + c = 0 c = 5 c = 5. Therefore the equation of the tangent is: = 5 5 = 0. Note You could also do this question using methods from Differentiation. 8 Polnomials Polnomials are epressions with one or more terms added together, where each term has a number (called the coefficient) followed b a variable (such as ) raised to a whole number power. For eample: 5 6 + + or 8 + 0. The degree of the polnomial is the value of its highest power, for eample: 5 6 + + has degree 5 8 + 0 has degree 8. Note that quadratics are polnomials of degree two. Also, constants are 0 polnomials of degree zero (e.g. 6 is a polnomial, since 6 = 6 ). Page 77 HSN000

Unit Polnomials and Quadratics 9 Snthetic Division Snthetic division provides a quick wa of evaluating polnomials. For eample, consider f ( ) = 9 + +. Evaluating directl, we find f ( 6) =. We can also evaluate this using snthetic division with detached coefficients. Step Detach the coefficients, and write them across the top row of the table. Note that the must be in order of decreasing degree. If there is no term of a specific degree, then zero is its coefficient. Step Write the number for which ou want to evaluate the polnomial (the input number) to the left. 9 6 9 Step Bring down the first coefficient. 6 9 Step Multipl this b the input number, writing the result underneath the net coefficient. Step 5 Add the numbers in this column. Repeat Steps and 5 until the last column has been completed. The number in the lower-right cell is the value of the polnomial for the input value, often referred to as the remainder. 6 9 + 6 9 6 9 8 0 0 = f ( 6) Page 78 HSN000

Unit Polnomials and Quadratics EXAMPLE. Given f ( ) = + 0, evaluate f ( ) using snthetic division. 0 So f ( ) = 0. using the above process 0 0 0 0 Note In this eample, the remainder is zero, so f ( ) = 0. This means + 0 = 0 when =, which means that = is a root of the equation. So + must be a factor of the cubic. We can use this to help with factorisation: ( ) f ( ) = ( + ) q( ) where q ( ) is a quadratic Is it possible to find the quadratic q ( ) using the table? Tring the numbers from the bottom row as coefficients, we find: ( + )( 0) = + = 0 = f ( ). 0 0 So using the numbers from the bottom row as coefficients has given the correct quadratic. In fact, this method alwas gives the correct quadratic, making snthetic division a useful tool for factorising polnomials. EXAMPLES. Show that is a factor of is a factor = is a root. 9 5 8 0 9 5 +. 0 0 0 0 Since the remainder is zero, = is a root, so is a factor. Page 79 HSN000

Unit Polnomials and Quadratics. Given f ( ) = 7 + 8, show that = 7 is a root of f ( ) = 0, and hence full factorise f ( ). 7 0 7 8 7 9 8 7 0 Since the remainder is zero, = 7 is a root. Hence we have f ( ) = 7 + 8. Show that = 5 is a root of factorise the cubic. 5 7 9 0 0 5 0 6 0 ( 7)( 7 ) = + + Using snthetic division to factorise ( )( )( ) = + 7. 7 9 0 0 + + =, and hence full Since = 5 is a root, + 5 is a factor. ( )( ) + + = + + 7 9 0 5 6 This does not factorise an further since the quadratic has b ac < 0. In the eamples above, we have been given a root or factor to help factorise polnomials. However, we can still use snthetic division if we do not know a factor or root. Provided that the polnomial has an integer root, it will divide the constant term eactl. So b tring snthetic division with all divisors of the constant term, we will eventuall find the integer root. 5. Full factorise 5 8 5 +. Numbers which divide 5: ±, ±, ± 5, ± 5. Tr = : ( ) + 5( ) 8( ) 5 = + 5 8 5 0. Tr = : ( ) + 5( ) 8( ) 5 = + 5 + 8 5 0. Note For ±, it is simpler just to evaluate the polnomial directl, to see if these values are roots. Page 80 HSN000

Unit Polnomials and Quadratics Tr = : 5 8 5 6 5 5 0 Since = is a root, is a factor. So + 5 8 5 = ( )( + + 5) = ( )( + )( + 5 ). Using snthetic division to solve equations We can also use snthetic division to help solve equations. EXAMPLE 6. Find the solutions of 5 6 0 + + =. Numbers which divide : ±, ±, ±, ±, ± 6, ±. Tr = : ( ) 5( ) + 6( ) + = 5 + 6 + 0. Tr = : ( ) 5( ) + 6( ) + Tr = : = 5 6 + 0. 5 6 6 0 Since = is a root, is a factor: = 0 = The Factor Theorem and Remainder Theorem For a polnomial f ( ): If f ( ) is divided b h or + = 0 = then the remainder is f ( ) f ( h) = 0 h is a factor of f ( ). h, and As we saw, snthetic division helps us to write f ( ) in the form where ( ) ( h) q ( ) + f ( h) q is called the quotient and f ( ) 5 + 6 + = 0 ( )( ) 6 = 0 ( )( + )( 6) = 0 h the remainder. or 6 = 0 = 6. Page 8 HSN000

Unit Polnomials and Quadratics EXAMPLE 7. Find the quotient and remainder when f ( ) = + is divided b +, and epress ( ) The quotient is ( ) ( )( ) f = + +. f as ( ) q( ) f ( h) + +. + and the remainder is, so 0 Finding Unknown Coefficients Consider a polnomial with some unknown coefficients, such as + p p +, where p is a constant. If we divide the polnomial b h, then we will obtain an epression for the remainder in terms of the unknown constants. If we alread know the value of the remainder, we can solve for the unknown constants. EXAMPLES. Given that is a factor of is a factor = is a root. p 6 8 + p 6 + p + p + +, find the value of p. p Since = is a root, the remainder is zero: + p = 0 p = p =. Note This is just the same snthetic division procedure we are used to. Page 8 HSN000

Unit Polnomials and Quadratics. When f ( ) = p + q 7 + q is divided b, the remainder is 6, and is a factor of f ( ). Find the values of p and q. When f ( ) is divided b, the remainder is 6. p q 7 q p p + q 8 p + q p p + q p + q 7 8 p + 8q Since the remainder is 6, we have: 8 p + 8q = 6 8 p + 8q = 0 p + q = 5. Since is a factor, f ( ) = 0 : f ( ) = p( ) + q ( ) 7( ) + q = p + q 7 + q = p + 5q 7 i.e. p + 5q = 7. Note There is no need to use snthetic division here, but ou could if ou wish. Solving and simultaneousl, we obtain: : q = q =. Put q = into : p + = 5 Hence p = and q =. p =. Page 8 HSN000

Unit Polnomials and Quadratics Finding Intersections of Curves We have alread met intersections of lines and parabolas in this outcome, but we were mainl interested in finding equations of tangents We will now look at how to find the actual points of intersection and not just for lines and parabolas; the technique works for an polnomials. EXAMPLES. Find the points of intersection of the line = and the parabola =. To find intersections, equate: = 6 8 = 0 = 0 ( )( ) + = 0 = or =. Find the -coordinates b putting the -values into one of the equations: when =, = ( ) = = 8, when =, = = 6 =. So the points of intersection are (, 8) and (, ).. Find the coordinates of the points of intersection of the cubic = 9 + 0 0 and the line = + 5. To find intersections, equate: 9 + 0 0 = + 5 9 + 5 = 0 ( )( ) 8 + 5 = 0 ( )( )( 5) = 0 = or = or = 5. Find the -coordinates b putting the -values into one of the equations: when =, = + 5 = + 5 =, when =, = + 5 = 9 + 5 =, when = 5, = 5 + 5 = 5 + 5 = 0. So the points of intersection are ( ),, (, ) and ( 5, 0). Remember You can use snthetic division to help with factorising. Page 8 HSN000

Unit Polnomials and Quadratics. The curves = + and = 6 + are shown below. = 6 + A O B C = + Find the -coordinates of A, B and C, where the curves intersect. To find intersections, equate: + = + ( )( ) ( )( )( ) 6 5 + + 8 = 0 + 6 + 8 = 0 + = 0 = or = or =. Remember You can use snthetic division to help with factorising. So at A, = ; at B, = ; and at C, =.. Find the -coordinates of the points where the curves and = 0 + 7 + 5 intersect. = 0 To find intersections, equate: 0 = 0 + 7 + 5 7 + 7 + 5 = 0 ( )( ) + 8 + 5 = 0 ( + )( )( 5) = 0 = or = or = 5. So the curves intersect where =,,5. Page 85 HSN000

Unit Polnomials and Quadratics Determining the Equation of a Curve Given the roots, and at least one other point ling on the curve, we can establish its equation using a process similar to that used when finding the equation of a parabola. EXAMPLE. Find the equation of the cubic shown in the diagram below. 6 O 6 Step Write out the roots, then rearrange to get the factors. Step The equation then has these factors multiplied together with a constant, k. Step Substitute the coordinates of a known point into this equation to find the value of k. Step Replace k with this value in the equation. = 6 = = + 6 = 0 + = 0 = 0. = k ( + 6)( + )( ). Using ( 0, 6) : k ( 0 + 6)( 0 + )( 0 ) = 6 k ( )( )( 6) = 6 8k = 6 = ( + 6)( + )( ) k =. ( )( ) ( ) = + + 5 6 = + 5 6 + + 5 8 = + + 6 8 6. Page 86 HSN000

Unit Polnomials and Quadratics Repeated Roots If a repeated root eists, then a stationar point lies on the -ais. Recall that a repeated root eists when two roots, and hence two factors, are equal. EXAMPLE. Find the equation of the cubic shown in the diagram below. 9 O = = = + = 0 = 0 = 0. So = k( + )( ). Use ( 0, 9 ) to find k : 9 = k ( 0 + )( 0 ) 9 = k 9 k =. Note = is a repeated root, so the factor ( ) appears twice in the equation. = + + = + + + = + 9. So = ( + )( ) ( )( 6 9) ( 6 9 8) Page 87 HSN000

Unit Polnomials and Quadratics Approimating Roots Polnomials have the special propert that if f ( a ) is positive and f ( b ) is negative then f must have a root between a and b. We can use this propert to find approimations for roots of polnomials to an degree of accurac b repeatedl zooming in on the root. EXAMPLE Given f ( ) = + 7, show that there is a real root between = and =. Find this root correct to two decimal places. Evaluate f ( ) at = and = : f ( ) = ( ) ( ) + 7 = f ( ) = ( ) ( ) + 7 = 5 Since f ( ) > 0 and f ( ) < 0, f ( ) has a root between these values. Start halfwa between = and =, then take little steps to find a change in sign: f (. 5) =. 65 < 0 f (. ) = 0. 896 < 0 f (. ) = 0. 6 < 0 f (. ) = 0. 568 > 0. Since f (. ) > 0 and f (. ) < 0, the root is between =. and =.. Start halfwa between =. and =. : f (. 5) = 0. 05 > 0 f (. 6) = 0. 9976 > 0 f (. 7) = 0. 05678 > 0 f (. 8) = 0. 068 < 0. Since f (. 7) > 0 and f (. 8) < 0, the root is between these values. Finall, f (.75) = 0.007875 > 0. Since f (. 75) > 0 and f (. 8) < 0, the root is between =. 75 and =. 8. Therefore the root is =. 8 to d.p. f ( a) a b f ( b) Page 88 HSN000

Unit Integration OUTCOME Integration Indefinite Integrals In integration, our aim is to undo the process of differentiation. Later we will see that integration is a useful tool for evaluating areas and solving a special tpe of equation. We have alread seen how to differentiate polnomials, so we will now look at how to undo this process. The basic technique is: n+ n d = + c n, c is the constant of integration. n + Stated simpl: raise the power (n) b one (giving n + ), divide b the new power ( n + ), and add the constant of integration ( c ). EXAMPLES. Find. Find. Find d. d = + c = + c. d. d = + c = + c 5 d. 9 5 9 d = + = 9 c 9 + c.. We use the smbol for integration. The must be used with d in the eamples above, to indicate that we are integrating with respect to. The constant of integration is included to represent an constant term in the original epression, since this would have been zeroed b differentiation. Integrals with a constant of integration are called indefinite integrals. Page 89 HSN000

Unit Integration Checking the answer Since integration and differentiation are reverse processes, if we differentiate our answer we should get back to what we started with. For eample, if we differentiate our answer to Eample above, we do get back to the epression we started with. + c differentiate integrate Integrating terms with coefficients The above technique can be etended to: n+ n n a a d = a d = + c n, a is a constant. n + Stated simpl: raise the power (n) b one (giving n + ), divide b the new power ( n + ), and add on c. EXAMPLES 6 d.. Find 6 6 d = + c = + c. 5. Find d. d = + c = 8 + c 8 = + c. Note It can be eas to confuse integration and differentiation, so remember: d = + c k d = k + c. Page 90 HSN000

Unit Integration Other variables Just as with differentiation, we can integrate with respect to an variable. EXAMPLES 5 p dp. 6. Find p = + c. p 5 p dp = + c 7. Find p d. p d = p + c. Integrating several terms The following rule is used to integrate an epression with several terms: ( f ( ) + g ( )) d = f ( ) d + g ( ) d. Stated simpl: integrate each term separatel. EXAMPLES 8. Find ( ) ( ) d. d = + c 5 9. Find 8 ( + + 7 ) = + c = + c. d. 5 8 ( ) 8 + + 7 d = + + 7 + c 8 8 = 8 + + 7 + c 8 = + + 7 + c. Note dp tells us to integrate with respect to p. Note Since we are integrating with respect to, we treat p as a constant. Page 9 HSN000

Unit Integration Preparing to Integrate As with differentiation, it is important that before integrating all brackets are multiplied out and there are no fractions with an term in the denominator (bottom line), for eample: = EXAMPLES = = d. Find for 0. d is just a short wa of writing d, so:. Find d d = d = d = + c = + c. d for > 0. = = d d = + c = + c. 5 = 5 5 = 5. 7. Find p dp where p 0. 7 p 7 dp = p dp = 7 p + c 7 = + c. p Page 9 HSN000

Unit Integration. Find 5 5 d. 5 5 = ( 5 5 ) d d 6 5 = + c 6 6 = 5 8 + c 6 = 5 8 8 + c. Differential Equations A differential equation is an equation d involving derivatives, e.g. d =. A solution of a differential equation is an epression for the original function; in this case = + c is a solution. + c differentiate integrate In general, we obtain solutions using integration: d = d or f ( ) = f ( ) d. d This will result in a general solution since we can choose the value of c, the constant of integration. O The general solution corresponds to a famil of curves, each with a different value for c. The graph to the left illustrates some of the curves = + c for particular values of c. If we have additional information about the function (such as a point its graph passes through), we can find the value of c and obtain a particular solution. Page 9 HSN000

Unit Integration EXAMPLES = passes through the point (, ). The graph of f ( ) d If = 5, epress in terms of. d d = d d ( ) = 5 d = 5 + c. We know that when =, = so we can find c: So = + c 5 = ( ) 5( ) + c = 9 5 + c c = = +. 5.. The function f, defined on a suitable domain, is such that f ( ) = + +. Given that f ( ) =, find a formula for f ( ) in terms of. f ( ) = f ( ) d = + + ( ) d = + + d = + + c = + + c. We know that f ( ) =, so we can find c: ( ) = f + + c = ( ) + ( ) + c = + + c c =. So ( ) f = + +. Page 9 HSN000

Unit Integration Definite Integrals If F( ) is an integral of f ( ), then we define: b b [ ] ( ) ( ) f ( ) d = F ( ) = F b F a a a where a and b are called the limits of the integral. Stated simpl: Work out the integral as normal, leaving out the constant of integration. Evaluate the integral for = b (the upper limit value). Evaluate the integral for = a (the lower limit value). Subtract the lower limit value from the upper limit value. Since there is no constant of integration and we are calculating a numerical value, this is called a definite integral. EXAMPLES. Find 5 d. 5 5 d = 5( ) 5( ) = = 5 5 = 5 5 =. 0. Find ( + ) ( ) 0 d. + d = + = + 0 0 0 = + 0 + = 6 + 8 0 = + 8 =. Page 95 HSN000

Unit Integration. Find d. d = d = = = ( ) = + = 7 6 8. 5 Geometric Interpretation of Integration We will now consider the meaning of integration in the contet of areas. Consider ( ) d = 8 ( ) = 8 0 = 5. 0 0 On the graph of = : = O 0 The shaded area is given b ( ) d. Therefore the shaded area is 5 square units. In general, the area enclosed b the graph = f ( ) and the -ais, between = a and = b, is given b b f ( ) d. a Page 96 HSN000

Unit Integration EXAMPLE. The graph of = is shown below. Calculate the shaded area. = O 5 d = 5 ( ) 5 5 = ( 5) ( ) = 5 6 50 + = 6 8 =. So the shaded area is square units. Areas below the -ais Care needs to be taken if part or all of the area lies below the -ais. For eample if we look at the graph of = : = The shaded area is given b d = ( ) O = = 6 + = 6 0 = 0 = 9. In this case, the negative indicates that the area is below the -ais, as can be seen from the diagram. The area is therefore 9 square units. ( ) ( ) Page 97 HSN000

Unit Integration Areas above and below the -ais Consider the graph from the eample above, with a different shaded area: O From the working above, the total shaded area is: Area + Area = + = 9 square units. Using the method from above, we might tr to calculate the shaded area as follows: d = 5 ( ) 5 = 5 = 5 50 + = = 8 6. 5 ( ) ( ) Clearl this shaded area is not 6 square units since we alread found it to be square units. This problem arises because Area is above the -ais, while Area is below. To find the true area, we needed to evaluate two integrals: ( ) d and ( ) = Area 5 Area 5 d. We then found the total shaded area b adding the two areas together. We must take care to do this whenever the area is split up in this wa. Page 98 HSN000

Unit Integration EXAMPLES. Calculate the shaded area shown in the diagram below. O = To calculate the area from = to = : d = ( ) = = ( ( ) ( ) ( ) ) ( ) ( ) ( ) ( ) ( 9 9 9) ( ) = + = + 9 = 0 So the area is 0 square units. We have alread carried out the integration, so we can just substitute in new limits to work out the area from = to = : ( ) d = = ( ( ) ( ) ( ) ) ( ) ( ) ( ) ( ) ( 6 8 ) ( ) = = 8 + Remember =. So the area is square units. So the total shaded area is 0 + = square units. The negative sign just indicates that the area lies below the ais. Page 99 HSN000

Unit Integration. Calculate the shaded area shown in the diagram below. = + O 5 First, we need to calculate the root between = and = 5 : + = 0 ( )( + 6) = 0 = or = 6. So the root is = To calculate the area from = to = : + d = + ( ) = + = ( ( ) + ( ) ( ) ) ( ) + ( ) ( ) ( ) 6 ( 6 96 8 ) ( 8) = + + To calculate the area from = to = 5 : 5 ( ) = 56 6 = 7 So the area is 7 square units. + d = + = ( ( 5) + ( 5) ( 5) ) ( ) + ( ) ( ) 5 ( ) 5 ( 5 0 6 ) ( 6 96) = + + = 6 5 = 5 So the area is 5 square units. So the total shaded area is 7 + = 5 square units. Page 00 HSN000

Unit Integration 6 Areas between Curves The area between two curves between b ( lower curve) upper curve a So for the shaded area shown below: = f ( ) = a and = b is calculated as: d square units. a O = g ( ) b The area is b a ( f ( ) g ( )) d square units. When dealing with areas between curves, areas above and below the -ais do not need to be calculated separatel. However, care must be taken with more complicated curves, as these ma give rise to more than one closed area. These areas must be evaluated separatel. For eample: = g ( ) O a b c = f ( ) b In this case we appl ( lower curve) upper curve a So the shaded area is given b: b c ( g ( ) f ( )) d + ( f ( ) g ( )) d d to each area.. a b Page 0 HSN000

Unit Integration EXAMPLES. Calculate the shaded area enclosed b the curves with equations = 6 and =. = 6 O = To work out the points of intersection, equate the curves: 6 = 6 + + = 0 9 = 0 ( + )( ) = 0 = or =. Set up the integral and simplif: ( upper curve lower curve) (( 6 ) ( )) = d ( 6 ) ( 9 ) d. = + + d = Carr out integration: 9 d = 9 ( ) d ( ) ( ) = 9( ) 9( ) ( 7 7 7 ) ( 7 ) = + = 7 9 + 7 9 = 6. Therefore the shaded area is 6 square units. Page 0 HSN000

Unit Integration. Two functions are defined for R b f ( ) = 7 + 8 + 6 and g ( ) = +. The graphs of = f ( ) and = g ( ) are shown below. = f ( ) = g ( ) O 6 Calculate the shaded area. b Since the shaded area is in two parts, we appl ( upper lower) Area from = to = : ( upper lower) ( 7 8 6 ( ) ) = + + + d ( 7 ) = + + d 7 = + + d twice. 7 ( ) 7( ) = + + + ( ) + ( ) ( 56 8 7 ) ( ) = + + + + = 99 =. d So the first area is square units. a Note The curve is at the top of this area. Page 0 HSN000

Unit Integration Area from = to = 6 : 6 ( upper lower) ( ( 7 8 6) ) = + + d 6 6 ( 7 ) = + d 7 = + 6 7 6 6 7 = + 6 + = ( + 50 7 7) ( + 56 8 ) = 60 = 5. d So the second area is 5 square units. So the total shaded area is + = 5 78 square units. 6 Note The straight line is at the top of this area. Page 0 HSN000

Unit Integration. A trough is metres long. A cross-section of the trough is shown below. = + 5 = O The cross-section is part of the parabola with equation = + 5. Find the volume of the trough. To work out the points of intersection, equate the curve and the line: + 5 = + = 0 ( )( ) = 0 so = or =. Set up the integral and integrate: ( upper lower) = ( ( )) + 5 = ( + ) d d d = + ( ) ( ) = + ( ) ( ) + ( ) ( ) ( 9 8 9) ( ) = + + = 0 + + = =. Therefore the shaded area is square units. Volume = cross-sectional area length = = 8 =. Therefore the volume of the trough is cubic units. Page 05 HSN000

Unit Integration 7 Integrating along the -ais For some problems, it ma be easier to find a shaded area b integrating with respect to rather than. EXAMPLE The curve with equation = 9 9 = is shown in the diagram below. =6 = O Calculate the shaded area which lies between = and = 6. We have = 9 9 = = ± 9 = ±. = The shaded area in the diagram to the right is given b: 6 6 d = d 6 = = 6 = 6 = 6 8 =. = Since this is half of the required area, the total shaded area is square units. O =6 Page 06 HSN000

Unit Trigonometr OUTCOME Trigonometr Solving Trigonometric Equations You should alread be familiar with solving some trigonometric equations. EXAMPLES. Solve sin = for 0 < < 60. sin = 80 = 0 or 80 0 = 0 or 50.. Solve cos = for 0 < < 60. 5 cos = 80 5 S A T C 80 + 60 First quadrant solution: = sin 80 + = 0. S A T C ( ) = 80 6 5 or 80 + 6 5 = 6 565 or 5.. Solve sin = for 0 < < 60. 60 ( ) = cos 5 = 6 5 (to d.p.). There are no solutions since sin. Since sin is positive Since cos is negative Note that cos, so cos = also has no solutions. Remember The eact value triangle: 60 0 Page 07 HSN000

Unit Trigonometr. Solve tan = 5 for 0 < < 60. Note tan = 5 80 = 80 78 690 or 60 78 690 = 0 0 or 8 0. S A T C 80 + 60 = tan ( 5) = 78 690 (to d.p.). Since tan is negative All trigonometric equations we will meet can be reduced to problems like those above. The onl differences are: the solutions could be required in radians in this case, the question will not have a degree smbol, e.g. Solve tan = rather than tan = ; eact value solutions could be required in the non-calculator paper ou will be epected to know the eact values for 0, 0, 5, 60 and 90 degrees. Questions can be worked through in degrees or radians, but make sure the final answer is given in the units asked for in the question. EXAMPLES 5. Solve sin = 0 where 0 60. sin = 80 S A 0 60 sin = T C 0 70 = 0 or 80 0 80 + 60 = sin = 0. or 60 + 0 or 60 + 80 0 or 60 + 60 + 0 = 0 or 50 or 90 or 50 = 5 or 75 or 95 or 55. ( ) Remember The eact value triangle: Note There are more solutions ever 60, since sin(0 ) = sin(0 +60 ) = So keep adding 60 until > 70. 60 0 Page 08 HSN000

Unit Trigonometr 6. Solve cos = where 0 π. cos = π S A T C = π or π π or π + π = π or 7π = π 7 8 or π 8. 7. Solve cos = where 0 < < π. ( cos ) = cos = ± cos = ± π + π = cos = π. ( ) 0 π 0 π S A Since cos can be positive or negative T C Remember = cos The eact value triangle: = π 6. ( ) = π 6 or π π 6 or π + π 6 or π π 6 or π + π 6 = π 5 7 6 or π 6 or π 6 or π 6. 8. Solve tan( 0 ) = 5 where 0 60. tan( 0 ) = 5 tan( 0 ) = 5 S A T C 0 60 0 080 0 0 060 ( ) 0 = tan 5 = 59 06 (to d.p.). 0 = 59 06 or 80 + 59 06 or 60 + 59 06 or 60 + 80 + 59 06 or 60 + 60 + 59 06 or 60 + 60 + 80 + 59 06 or 60 + 60 + 60 + 59 06 Remember The eact value triangle: π π π 6 π Page 09 HSN000

Unit Trigonometr 0 = 59 06 or 9 06 or 9 06 or 599 06 or 779 06 or 959 06 = 79 06 or 59 06 or 9 06 or 69 06 or 799 06 or 979 06 = 6 5 or 86 5 or 6 5 or 06 5 or 66 5 or 6 5. 9. Solve ( π ) ( π ) cos + = 0 8 for 0 < < π. cos + = 0 8 S A T C 0 < < π 0 < < π π < + π < π + π 07 < + π < 6 (to d.p.) π + = cos ( 0 8 ) = 0 6 (to d.p.). + π = 0 6 or π 0 6 or π + 0 6 or π + π 0 6 or π + π + 0 6 or π + π + π 0 6 + π.... = 5 660 or 6 906 or 9 or 89 =. 6 or 5. 859 or 0. 896 or. =. 07 or. 90 or 5. 8 or 6. 07. Trigonometr in Three Dimensions It is possible to solve trigonometric problems in three dimensions using techniques we alread know from two dimensions. The use of sketches is often helpful. The angle between a line and a plane The angle a between the plane P and the line ST is calculated b adding a line perpendicular to the plane and then using basic trigonometr. T Remember Make sure our calculator uses radians. P a S Page 0 HSN000

Unit Trigonometr EXAMPLE. The triangular prism ABCDEF is shown below. E A 0 cm Calculate the acute angle between: (a) The line AF and the plane ABCD. (b) AE and ABCD. (a) Start with a sketch: A Opposite tana = = Adjacent ( ) 0 a = tan 0 = 6. 699 (or 0. 9 radians) (to d. p.). Note Since the angle is in a right-angled triangle, it must be acute so there is no need for a CAST diagram. (b) Again, make a sketch: b angle to be A calculated We need to calculate the length of AC first using Pthagoras s Theorem: C AC = 0 + 6 6 cm A = 6. 0 cm D Therefore: E Opposite tanb = = cm Adjacent 6 A a 0 cm b F 6 cm cm D B C E cm C F D C 6 cm cm ( ) b = tan 6 =. 6 (or 0. 5 radians) (to d. p.). Page HSN000

Unit Trigonometr The angle between two planes The angle a between planes P and Q is calculated b adding a line perpendicular to Q and then using basic trigonometr. P Q a EXAMPLE. ABCDEFGH is a cuboid with dimensions 8 8 cm as shown below. H G E F 8 cm 8 cm A cm B (a) Calculate the size of the angle between the planes AFGD and ABCD. (b) Calculate the size of the acute angle between the diagonal planes AFGD and BCHE. (a) Start with a sketch: F A Opposite tana = = Adjacent ( ) 8 a = tan =. 690 (or 0. 588 radians) (to d. p.). (b) Again, make a sketch: Let AF and BE intersect at T. E F ATB is isosceles, so TAB = TBA ɵ =. 690. A a cm T 8 cm B B D C ATB = 80 (. 690 +. 690 ) =. 60. Note Angle GDC is the same size as angle FAB. Note The angle could also have been calculated using rectangle DCGH. So the acute angle is: BTF = ATE = 80. 60 = 67. 80 (or. 76 radians) (to d. p.). Page HSN000

Unit Trigonometr Compound Angles When we add or subtract angles, the result is called a compound angle. For eample, 5 + 0 is a compound angle. Using a calculator, we find: sin( 5 + 0 ) = sin( 75 ) = 0. 966 ; sin( 5 ) + sin( 0 ) =. 07 (both to d.p.). This shows that sin ( A + B) is not equal to sin A + sinb. Instead, we can use the following identities: sin( A + B) = sin AcosB + cos Asin B sin( A B) = sin AcosB cos Asin B cos( A + B) = cos AcosB sin Asin B cos( A B) = cos AcosB + sin Asin B. These are given in the eam in a condensed form: sin( A ± B ) = sin AcosB ± cos Asin B cos( A ± B ) = cos AcosB sin Asin B. EXAMPLES. Epand and simplif cos( + 60 ). cos( + 60 ) = cos cos60 sin sin 60 = cos sin. Remember The eact value triangle: 60 0. Show that ( ) a = π 6 and b = π. RHS = sin a cosb + cosa sinb = sinπ 6 cosπ + cosπ 6 sinπ sin a + b = sina cosb + cos a sinb for ( a b) LHS = sin + π π ( 6 ) π ( ) = sin + = sin =. Since LHS ( ) ( ) = + = RHS, the claim is true for = + =. a = π 6 and b = π. Remember The eact value triangle: π π 6 Page HSN000

Unit Trigonometr. Find the eact value of sin75. ( ) sin 75 = sin 5 + 0 = sin5 cos0 + cos5 sin0 ( ) ( ) = + + = + = = 6+. Finding Trigonometric Ratios You should alread be familiar with the following formulae (SOH CAH TOA). Opposite sin a = Hpotenuse Adjacent cos a = tan a Hpotenuse = Opposite. Adjacent p If we have sin a = q where 0 < a < π, then we can form a right-angled triangle to represent this ratio. a q p Hpotenuse a Adjacent Opposite Opposite p Since sina = = q then: Hpotenuse the side opposite a has length p; the hpotenuse has length q. The length of the unknown side can be found using Pthagoras s Theorem. Once the length of each side is known, we can find cosa and tana using SOH CAH TOA. The method is similar if we know cosa and want to find sina or tana. Page HSN000

Unit Trigonometr EXAMPLES. Acute angles p and q are such that sin p = 5 and sinq = 5 sin 6 ( p + q) = 65.. Show that 5 p sin p = 5 cos p = 5 q 5 sinq = 5 cosq = ( ) sin p + q = sin pcosq + cos psinq 5 ( ) ( ) = 5 + 5 = 8 5 65 + 65 = 6 65. Note Since Show that is used in the question, all of this working is required. Confirming Identities EXAMPLES 5. Show that sin( π ) sin( π ) = cos. = sin cosπ cos sinπ = sin 0 cos = cos. sin( s + t ) 6. Show that = tan s + tant for cos s 0 and cost 0. cos s cost sin( s + t ) sin s cost + cos s sint = cos s cost cos s cost sin s cost cos s sint = + cos s cost cos s cost sin s sint = + cos s cost = tan s + tan t. Remember sin tan. cos = Page 5 HSN000

Unit Trigonometr Double-Angle Formulae Using the compound angle identities with A = B, we obtain epressions for sina and cosa. These are called double-angle formulae. sin A = sin A cos A Note that these are given in the eam. cos A = cos A sin A = cos A = sin A. EXAMPLES. Given that tanθ =, where 0 < θ < π, find the eact value of sinθ and cosθ. 5 θ sinθ = cosθ = 5 sin θ = sinθ cosθ = 5 5 = 5. cos cos sin θ = θ θ ( ) ( ) = 5 5 = 9 6 5 5 = 7 5.. Given that cos = 5, where 0 < < π, find the eact values of sin and cos. Since cos sin sin = sin sin =, = = = 5 8 8 6 sin = ±. We are told that 0 < < π, so onl sin = is possible. Note An of the cosa formulae could have been used here. a So cos =. a = = = 9 =. Page 6 HSN000

Unit Trigonometr 5 Further Trigonometric Equations We will now consider trigonometric equations where double-angle formulae can be used to find solutions. These equations will involve: sin and either sin or cos; cos and cos; cos and sin. Solving equations involving sin and either sin or cos EXAMPLE. Solve sin = sin for 0 < 60. sin cos = sin sin cos + sin = 0 sin = 0 sin ( cos + ) = 0 cos + = 0 Replace sin using the double angle formula Take all terms to one side, making the equation equal to zero Factorise the epression and solve = 0 or 80 or 60 cos = = 80 60 or 80 + 60 So = 0 or 0 or 80 or 0. Solving equations involving cos and cos EXAMPLE. Solve cos = cos for 0 π. = 0 or 0. Remember The double-angle formulae are given in the eam. S A T C = cos = 60 ( ) cos = cos cos = cos cos cos 0 = ( cos + )( cos ) = 0 Replace cos b cos Take all terms to one side, making a quadratic equation in cos Solve the quadratic equation (using factorisation or the quadratic formula) cos + = 0 cos = = π π or π + π = π or π So = 0 or π or π or π. S A T C = cos = π ( ) cos = 0 cos = = 0 or π. Page 7 HSN000

Unit Trigonometr Solving equations involving cos and sin EXAMPLE. Solve cos = sin for 0 < < π. cos = sin sin = sin sin + sin = 0 ( sin )( sin + ) = 0 Replace cos b sin Take all terms to one side, making a quadratic equation in sin Solve the quadratic equation (using factorisation or the quadratic formula) sin = 0 sin = = π 6 or π π 6 = π 6 or 5π 6 So = π 5 6 or π 6 or π. S A T C = sin = π 6 ( ) sin + = 0 sin = = π. Page 8 HSN000

Unit Circles OUTCOME Circles Representing a Circle The equation of a circle with centre ( a, b ) and radius r units is ( a) + ( b) = r. This is given in the eam. For eample, the circle with centre (, ) and radius units has equation: ( ) ( ) + + = ( ) ( ) + + = 6. Note that the equation of a circle with centre ( 0, 0 ) is of the form + = r, where r is the radius of the circle. EXAMPLES. Find the equation of the circle with centre (, ) and radius units. ( ) ( ) a + b = r ( ) ( ) ( ) + ( ) = ( ) ( ) + + =.. A is the point (,) and B( 5, ). Find the equation of the circle which has AB as a diameter. The centre of the circle is the midpoint of AB; 5 + C = midpoint ( ) AB =, =,. The radius r is the distance between A and C: r = ( ( ) ) + ( ) = + = 7. So the equation of the circle is ( ) ( ) + = 7. Note You could also use the distance between B and C, or half the distance between A and B. Page 9 HSN000

Unit Circles Testing a Point Given a circle with centre ( a, b ) and radius r units, we can determine whether a point ( p, q ) lies within, outwith or on the circumference using the following rules: ( ) ( ) ( ) ( ) ( ) ( ) p a + q b < r the point lies within the circle p a + q b = r the point lies on the circumference of the circle p a + q b > r the point lies outwith the circle. EXAMPLE A circle has the equation ( ) ( ) + + 5 = 9. Determine whether the points (, ), ( 7, ) and (, ) lie within, outwith or on the circumference of the circle. Point (, ) : ( ) + ( + ) = ( ) + ( + 5) = 0 + 6 = 6 > 9 So outwith the circle. Point ( 7, ) : ( ) + ( + ) = ( 7 ) + ( + 5) = 5 + = 9 So on the circumference. Point (, ) : ( ) + ( + ) ( ) ( 5) = + + = + = < 9 So within the circle. The General Equation of a Circle The equation of an circle can be written in the form where the centre is ( g, f ) This is given in the eam. g f c + + + + = 0 and the radius is g + f c units. Note that the above equation onl represents a circle if since: g f c + > 0, if g + f c < 0 then we cannot obtain a real value for the radius, since we would have to square root a negative; if g + f c = 0 then the radius is zero the equation represents a point rather than a circle. Page 0 HSN000

Unit Circles EXAMPLES. Find the radius and centre of the circle with equation + + 8 + 7 = 0. Comparing with g = so g = f = 8 so f = c = 7 g f c + + + + = 0: Centre is, ( g f ) = (, ) Radius is. Find the radius and centre of the circle with equation + 6 + 0 = 0. The equation must be in the form each term b : + + 5 = 0 Now compare with g = so g = f = 5 so f = 5 c =. Eplain wh g f c g + f c ( ) 7 = + = + 6 7 = g f c + + + + = 0: Centre is, ( g f ) 5 (, ) = units. + + + + = 0, so divide Radius is g + f c 5 ( ) ( ) = + + = 9 5 + + = = 8 8 units. + + 8 + 9 = 0 is not the equation of a circle. Comparing with + + g + f + c = 0 : g = so g = g + f c = + ( ) 9 f = 8 so f = = 9 < 0. c = 9 The equation does not represent a circle since satisfied. g f c + > 0 is not Page HSN000

Unit Circles. For which values of k does a circle? Comparing with k k k g f c c k k + + + = 0 represent + + + + = 0 : g = k so g = k To represent a circle, we need f = so f = = +. Intersection of a Line and a Circle g f c ( ) k k k + > 0 + + > 0 k + 8 > 0 k < 8. A straight line and circle can have two, one or no points of intersection: two intersections one intersection no intersections If a line and a circle onl touch at one point, then the line is a tangent to the circle at that point. To find out how man times a line and circle meet, we can use substitution. EXAMPLES. Find the points where the line with equation = intersects the circle with equation + = 0. + = 0 + ( ) = 0 + 9 = 0 0 = 0 = = = ± = = ( ) = = ( ) =. So the circle and the line meet at (, ) and (, ). Remember m m m ( ab) = a b. Page HSN000

Unit Circles. Find the points where the line with equation = + 6 and circle with equation + + + 8 = 0 intersect. Substitute = + 6 into the equation of the circle: + ( + 6) + + ( + 6) 8 = 0 + ( + 6)( + 6) + + + 8 = 0 + + + 6 + + + 8 = 0 5 + 0 + 0 = 0 ( ) 5 + 6 + 8 = 0 ( )( ) + + = 0 + = 0 + = 0 = = = ( ) + 6 = = ( ) + 6 =. So the line and circle meet at (, ) 5 Tangents to Circles and (, ). As we have seen, a line is a tangent if it intersects the circle at onl one point. To show that a line is a tangent to a circle, the equation of the line can be substituted into the equation of the circle, and solved there should onl be one solution. EXAMPLE Show that the line with equation + = is a tangent to the circle with equation + + 6 + = 0. Substitute using the equation of the straight line: + + 6 + = 0 ( ) ( ) + + 6 + = 0 ( )( ) ( ) + + 6 + = 0 + 6 8 + + 6 + 8 = 0 + = 0 ( ) + = 0 + = 0 Page HSN000

Unit Circles Then (i) factorise or (ii) use the discriminant + = 0 ( )( ) = 0 = 0 = = 0 =. Since the solutions are equal, the line is a tangent to the circle. + = 0 a = b ac b = = ( ) ( ) c = = = 0. Since b ac = 0, the line is a tangent to the circle. Note If the point of contact is required then method (i) is more efficient. To find the point, substitute the value found for into the equation of the line (or circle) to calculate the corresponding -coordinate. 6 Equations of Tangents to Circles If the point of contact between a circle and a tangent is known, then the equation of the tangent can be calculated. If a line is a tangent to a circle, then a radius will meet the tangent at right angles. The gradient of this radius can be calculated, since the centre and point of contact are known. Using mradius mtangent =, the gradient of the tangent can be found. The equation can then be found using b = m( a), since the point is known, and the gradient has just been calculated. Page HSN000

Unit Circles EXAMPLE Show that A (, ) lies on the circle equation of the tangent at A. Substitute point into equation of circle: + + 6 + = + + 6( ) + ( ) = + 9 + 6 + 6 + + 6 + = 0 and find the = 0. Since this satisfies the equation of the circle, the point must lie on the circle. Find the gradient of the radius from (, ) to (, ): mradius = + = + =. So m = since m m =. tangent radius tangent Find equation of tangent using point (, ) and gradient m = : b = m( a) = ( ) = + = + + = 0. Therefore the equation of the tangent to the circle at A is + = 0. Page 5 HSN000

Unit Circles 7 Intersection of Circles Consider two circles with radii r and r with r > r. Let d be the distance between the centres of the two circles. r d r d > r + r d The circles do not touch. d = r + r r r < d < r + r d d The circles touch eternall. The circles meet at two distinct points. Note Don t tr to memorise this, just tr to understand wh each one is true. d = r r d The circles touch internall. d < r r d The circles do not touch. EXAMPLES. Circle P has centre (, ) and radius units, circle Q has equation + + 6 + = 0. Show that the circles P and Q do not touch. To find the centre and radius of Q: Compare with + + g + f + c = 0: g = so g = Centre is ( g, f ) Radius rq = g + f c f = 6 so f = = (, ). = + 9 c =. = 9 = units. Page 6 HSN000

Unit Circles We know P has centre (, ) and radius r P = units. So the distance between the centres d = ( + ) + ( + ) = 5 + ( ) = 9 = 5. 9 units (to d.p.). Since r P + r Q = + = 5 < d, the circles P and Q do not touch.. Circle R has equation equation ( ) ( ) eternall. + = 0, and circle S has + 6 =. Show that the circles R and S touch To find the centre and radius of R: Compare with g = so g = f = so f = g f c c =. + + + + = 0: Centre is, To find the centre and radius of S: compare with ( ) ( ) r a = b = 6 = so r =. ( g f ) = (, ). + =. a b r Centre is, = ( a b) ( ), 6. Radius rr = g + f c S = ( ) + ( ) + = 9 = units. Radius r = units. So the distance between the centres d = ( ) + ( 6) ( ) ( ) = + 5 = 5 units. Since r R + r S = + = 5 = d, the circles R and S touch eternall. = Page 7 HSN000