University:
University of Cambridge
Course:
0580 IGCSE | Mathematics
Academic year:

2024
Views:

317

Pages:

4
Author:

Shahbaz A.

6663-01 Edexcel GCE Core Mathematics C1 Advanced Subsidiary shahbaz ahmed July 2024 Differentiation:Tangents and Normals Q 8. The curve C has equation y = 31 x3 − 4x2 + 8x + 3 The point P has coordinates (3, 0). (a) Show that P lies on C. (b) Find the equation of the tangent to C at P, giving your answer in the form y = mx + c, where m and c are constants. Another point Q also lies on C. The tangent to C at Q is parallel to the tangent to C at P (c) Find the coordinates of Q. Solution 1 (a) Since y = 31 x3 − 4x2 + 8x + 3 Putting x=3,y=0 0 = 13 (33 ) − 4(32 ) + 8(3) + 3 0 = 9 − 36 + 24 + 3 0=0 L.H.S=R.H.S P lies on C. (b) y = 13 x3 − 4x2 + 8x + 3 dy dx = d( 13 x3 −4x2 +8x+3) dx dy dx = d(( 31 x3 ) dx − dy dx = 1 d((x3 ) 3 dx ) dx − 4 d(x dx + 8 dx + 0 dy dx = 1 d((x3 ) 3 dx ) − 4 d(x dx + 8 dy dx = 31 (3x2 ) − 4(2x) + 8 dy dx = x2 − 8x + 8 d(4x2 ) dx + d(8x) dx + d dx (3) 2 2 Putting x=3 dy dx = 32 − 8(3) + 8 = −7 = m 2 Where m is the slope of the tangent to C at P Equation of the tangent through the point (3,0) with slope m=-7 y − 0 = −7(x − 3) y = −7x + 21 (c) Since the tangent to C at Q is parallel to the tangent to C at P m=-7 for tangent to C at Q , Let tangent at Q passes through the point (x1 , y1 ) =⇒ dy dx = x21 − 8x1 + 8 = −7 x21 − 8x1 + 8 = −7 x21 − 8x1 + 15 = 0 x21 − 5x1 − 3x1 + 15 = 0 (x21 − 5x1 ) − (3x1 − 15) = 0 x1 (x1 − 5) − 3(x1 − 5) = 0 (x1 − 3)(x1 − 5) = 0 (x1 − 3) = 0 =⇒ x1 = 3 (3, 0) is the point P (x1 − 5) = 0 =⇒ x1 = 5 3 Putting x1 = 5 in the equation y1 = 31 x3 − 4x2 + 8x + 3 y1 = 13 (5)3 − 4(5)2 + 8(5) + 3 y1 = 13 (125) − 100 + 40 + 3 y1 = − 46 3 The coordinates of Q are (5, − 46 3 ). 4