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

2024
Views:

250

Pages:

5
Author:

Shahbaz A.

6663- 01 Edexcel GCE Core Mathematics C1 Advanced Subsidiary Differentiation: Tangents and Normal shahbaz ahmed July 2024 C1 Differentiation – Tangents and Normals Q 4. The curve C has equation 4 y = x2 (x − 6) + , x > 0. x The points P and Q lie on C and have x-coordinates 1 and 2 respectively. (a) Show that the length of PQ is √ 170. (b) Show that the tangents to C at P and Q are parallel. (c) Find an equation for the normal to C at P, giving your answer in the form ax + by + c = 0, where a, b and c are integers. Solution 1 (a) y = x2 (x − 6) + 4 x Putting x=1 y = 12 (1 − 6) + 4 1 y = −5 + 4 = −1 HenceP (1, −1) is a point on C. Putting x=2 y = 22 (2 − 6) + 4 2 = −16 + 2 = −14 Similarly Q(2, −14) is a point on C. p Length of PQ=|P Q| = (2 − 1)2 + [−14 − (−1)]2 2 |P Q| = p 12 + (−14 + 1)2 |P Q| = p 1 + (−13)2 |P Q| = |P Q| = √ √ 1 + 169 170 (b) Since y = x2 (x − 6) + 4 x dy dx = d[x2 (x−6)+ x4 )] dx dy dx = d[x2 (x−6)] dx dy dx ) = x2 d[(x−6)] + (x − 6) d(x dx dx + 4 dy dx = x2 (1) + 2x(x − 6) + 4 d(xdx dy dx = x2 + 2x2 − 12x + 4(−1)x−2 dy dx = x2 + 2x2 − 12x − dy dx = 3x2 − 12x − + d( x4 ) dx 2 −1 d( x1 ) dx ) 4 x2 4 x2 Putting x=1 dy dx = 3(1)2 − 12(1) − 4 12 = 3 − 12 − 4 = −13=slope at the point 4 22 = 12 − 24 − 1 = −13=slope at the point P (1, −1) Putting x=2 dy dx = 3(2)2 − 12(2) − 3 Q(2, −14) Since slopes are same,tangents to C at P and Q are parallel m=-13 Let slope of the normal to C at the point P=m1 mm1 = −1 −13m1 = −1 m1 = 1 13 Equation of the normal to C at P (1, −1) y − (−1) = y+1= 1 13 (x 1 13 (x − 1) − 1) 13y + 13 = x − 1 x − 1 − 13y − 13 = 0 x − 13y − 14 = 0 4 5