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

2024
Views:

169

Pages:

5
Author:

Shahbaz A.

6663/01 Edexcel GCE Core Mathematics C1 Advanced Subsidiary shahbaz ahmed July 2024 Differentiation: Tangents and Normals C1 Differentiation – Tangents and Normals Q2. The curve C has equation y = x3 − 2x2 − x + 9, x > 0 The point P has coordinates (2, 7). 1 (a) Show that P lies on C Solution Putting x=2 in the equation y = x3 − 2x2 − x + 9, x > 0 y = 23 − 2(22 ) − 2 + 9 = 7, x > 0 Hence (2,7) lies on the curve 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. Solution 2 y = x3 − 2x2 − x + 9, x > 0 dy dx = d(x3 −2x2 −x+9) dx dy dx = d(x3 ) dx dy dx ) = 3x2 − 2 d(x dx − 1 + 0 dy dx = 3x2 − 2(2x) − 1 dy dx = 3x2 − 4x − 1 − d(2x2 ) dx − dx dx + d dx (9) 2 Putting x=2 dy dx = 3(22 ) − 4(2) − 1 = 3=m Where m is the slope of tangent to the curve C at P (2, 7) Now equation of line passing through the point P (2, 7) with slope 3 m=3 y − 7 = 3(x − 2) y − 7 = 3x − 6 y = 3x − 6 + 7 y = 3x + 1 The point Q also lies on C. Given that the tangent to C at Q is perpendicular to the tangent to C at P, (c) show that the x-coordinate of Q is 13 (2 + √ 6) Solution Let Q(x1 , y1 ) be the point Q on the curve satisfying the given condition. Now slope of tangent at any point on the curve C dy dx = 3x2 − 4x − 1 Putting x = x1 in the equation dy dx = 3x21 − 4x1 − 1 = m1 Where m1 is the slope of tangent to the curve C at Q(x1 , y1 ) Tangents are mutually perpendicular mm1 = −1 4 3(3x21 − 4x1 − 1) = −1 9x21 − 12x1 − 3 + 1 = 0 9x21 − 12x1 − 2 = 0 x1 = x1 = x1 = √ 12± 144+72 18 √ 12± 216 18 √ 2± 6 3 Hence x1 = 13 (2 + √ 6) is the x-coordinate of the point Q 5