TANGENTS AND NORMALS

Size: px

Start display at page:

Download "TANGENTS AND NORMALS"

Kerry Glenn
6 years ago
Views:

1 Mathematics Revision Guides Tangents and Normals Page 1 of 8 MK HOME TUITION Mathematics Revision Guides Level: AS / A Level AQA : C1 Edecel: C OCR: C1 OCR MEI: C TANGENTS AND NORMALS Version : 1 Date: Eamples 6-7 are copyrighted to their respective owners and used with their permission

2 Mathematics Revision Guides Tangents and Normals Page of 8 Tangents and Normals Differentiation helps to find the gradient of the tangent to a curve, but we can use ideas learnt in the section of straight lines to find the actual equation of the tangent at a given point The normal at the same point is perpendicular to the tangent, therefore the product of their gradients is 1 Eample (1): Find the gradients, and hence the equations, of the tangent and the normal to the curve at the point (, -4) The tangent and normal have been shown here for reference (Note that the aes of the graph must be shown to uniform scales, or the normal and tangent might not appear at right angles) The derivative of is 4 7; substituting for = gives a gradient of 1 The equation of the tangent at (, -4) is therefore y- y 1 = m( - 1 ) y + 4 = 1( - ) In gradient-intercept form ( m + c ): y + 4 = y = - 6equation of tangent is y = - 6 In a + by + c = 0 form : y + 4 = 1( ) y + 4 = -y - 4 = 0 y - 6 = 0 equation of tangent is y - 6 = 0

3 Mathematics Revision Guides Tangents and Normals Page of 8 The product of the gradients of the tangent and the normal must be 1, and therefore the normal to the curve must have a gradient of 1 The equation of the normal at (, -4) is therefore y + 4 = -1( - ) In gradient-intercept form ( m + c ): y + 4 = -1( - ) y + 4 = y = - - equation of normal is y = - - In a + by + c = 0 form : y + 4 = -1( ) y + 4 = y = 0 + y + = 0 equation of normal is + y + = 0 See diagram below

4 Mathematics Revision Guides Tangents and Normals Page 4 of 8 Eample (): Find the equations of the tangent and normal to the curve at the point (, -4) Give the equations in gradient-intercept ( m + c ) form, and hence show that the tangent passes through the origin The derivative of is - 8+, and thus its value at (, -4) is ( ) (8 ) +, or - The tangent to the curve therefore has a gradient of The equation of the tangent will be therefore y- y 1 = m( - 1 ) y + 4 = -( - ) y + 4 = y = -equation of tangent is y = - This line has a y-intercept at the origin The gradient of the normal must be 1 (product of the gradients of tangent and normal must be 1) The equation of the normal is therefore 1 ( ) y y + 4 = ½ y = ½ 5 equation of normal is y = ½ - 5

5 Mathematics Revision Guides Tangents and Normals Page 5 of 8 Eample (): Find the equations of the tangent and normal to the curve Give the result in a + by + c = 0 form 4 at (4, 4 1 ) The tangent has a gradient of, so when = 4, its gradient is 1, and its equation is y- y 1 = m( - 1 ) or 1 4 y 8y = -(-4) 8y = 0 + 8y - 6 = 0equation of tangent is + 8y - 6 = 0 The gradient of the normal at (4, 4 1 ) is thus 8 by the rule of the product of gradients Its equation is therefore y = 8( 4) 4y 1 = (-4) y + 1 = 0-4y - 17 = 0 equation of normal is - 4y - 17 = 0 Eample (4): A curve has equation y = i) Find the equation of the tangent to the curve when = 5, in a + by + c = 0 form ii) Find the -coordinate of the point on the curve where the tangent is parallel to the one at = 5 i) If y = , then = d When = 5, y = = -5, so the curve passes through (5, -5) Also, when = 5, = = -1 d the equation of the tangent at (5, -5) is y + 5 = -1( 5) y + 5 = y - 40 = 0 ii) Since the gradient of the tangent in part i) is equal to -1, the gradient of the required parallel tangent will also be -1 Hence = = -1 d Rearranging as = 0, the resulting quadratic derivative factorises out as ( 5)( 5) The solutions of ( 5)( 5) = 0 are = 5 (corresponding to the result from i)) and = 5, the -coordinate of the point where the curve and the parallel tangent meet is 5

6 Mathematics Revision Guides Tangents and Normals Page 6 of 8 Eample (5): Two tangents are drawn to the curve at the points P (1, ) and Q (5, 10) Give the coordinates of their point of intersection The derivative of is - 4, and thus its values at P and Q are and 6 respectively The equation of the tangent at P is y- y 1 = m( - 1 ) y- = -( - 1) y- = -+ y- + = 0 + y 4 = 0 Similarly the equation of the tangent at Q is y - 10 = 6( - 5) y-10 = y + 10 = y - 0 = 0 We now have a pair of simultaneous linear equations which can be solved by elimination: + y 4 = y - 0 = = 0 A B A+B This gives =, and substituting into either equation gives y = - the two tangents meet at (, -)

7 Mathematics Revision Guides Tangents and Normals Page 7 of 8 Eample (6): The equation of a curve is given by i) Find y 16 and hence the coordinates of the stationary points on the curve d y 16 ii) Distinguish between the maimum and minimum points obtained in part i) iii) Given that the line 0 - y 144 = 0 is the equation of the tangent to the curve at the point (p, q), find the values of p and q (Copyright OCR, GCE Mathematics Paper 471, June 005, Q10) (altered), then 16 d i) If y 16 d The -coordinates of the stationary points are 4 and The two stationary points are 18 4, and 18 4, after substituting in y 16 ii) The second derivative, At d y d 18 d y 4,, 8 (ie > 0), hence d 18 d y 4,, 8 d On the other hand, at 18, 4 is a local minimum (ie < 0), hence 18, 4 is a local maimum iii) We can find the gradient of the line 0 - y 144 = 0 0 = y y = the gradient of the line 0 - y 144 = 0 is 0 (Or we could have used the fact that a line with equation a + by + c = 0 has a gradient of Net, we must find the points on the curve where the gradient is also 0, ie we solve 16 = 0 6 = 0 ( + 6)( - 6) = 0 When = 6, 16 = 7-96 = -4; similarly when = -6, 16 The two possible values for (p, q) are (6, -4) or (-6, 4) = = 4 a b We then substitute each pair of values into the epression 0 - y 144; the correct pair should give a result of 0 At (6, -4), 0 y 144 = 0; at (-6, 4), 0 y 144 = -88 the line 0 - y 144 = 0 is a tangent to the curve y 16 at the point (6, -4)

Mathematics Revision Guides Tangents and Normals Page 8 of 8 Eample (7): The point P on the curve at P is parallel to the line 5 + y = 0 y k has an -coordinate of 5 The normal to the curve i) Find

8 Mathematics Revision Guides Tangents and Normals Page 8 of 8 Eample (7): The point P on the curve at P is parallel to the line 5 + y = 0 y k has an -coordinate of 5 The normal to the curve i) Find the value of k and hence the coordinates of P ii) The normal to the curve meets the -ais at the point Q Calculate the area of the triangle OPQ where O is the origin (Copyright OCR, GCE Mathematics Paper 471, June 009, Q11) (altered) i) The gradient of the normal to the curve k 5 + y = 0 as y = -5 and y 5 y when = 5 can be found by rearranging 5 the gradient of the normal is when = 5 a (Or we could have used the fact that a line with equation a + by + c = 0 has a gradient of b Since a tangent and normal to a curve at a given point are perpendicular, the gradient of the tangent to the curve when = 5 is 5 Differentiating the function y k gives d k k The gradient of the tangent at = 5 is ie 5 5 k k We then rearrange to find k Hence y 4 and the coordinates of point P are (5, 0) d k 4 1 k ii) We know that the normal at P is parallel to the line 5 + y = 0, and also that it passes through the point (5, 0) Substituting for (, y) gives the equation of the normal as 5 + y = 165 The -coordinate of the -intercept at Q satisfies 5 + y = 165 for y = 0, so 5 = 165 and = The coordinates of Q are therefore (,0) See diagram for the calculation of the area of triangle OPQ 1

( ) 2. Integration. 1. Calculate (a) x2 (x 5) dx (b) y = x 2 6x. 2. Calculate the shaded area in the diagram opposite.

( ) 2. Integration. 1. Calculate (a) x2 (x 5) dx (b) y = x 2 6x. 2. Calculate the shaded area in the diagram opposite. Integration 1. Calculate (a) ( 5) d (b) 4 + 3 1 d (c) ( ) + d 1 = 6. Calculate the shaded area in the diagram opposite. 3. The diagram shows part of the graph of = 7 10. 5 = + 0 4. Find the area between