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

2024
Views:

450

Pages:

6
Author:

Shahbaz A.

6663/01 Edexcel GCE Core Mathematics C1 Advanced Subsidiary shahbaz ahmed July 2024 Formulas Used x 1 y1 1 Area= 12 x2 y2 1 x 3 y3 1 Where A(x1 , y! ), B(x2 , y2 ), and C(x3 , y3 ) are vertices of the triangle ABC. Q 3. The curve C has equation 8 y = 9 − 4x − , x > 0 x 1 (1) The point P on C has x-coordinate equal to 2. (a) Show that the equation of the tangent to C at the point P is y=1 − 2x (b) Find an equation of the normal to C at the point P. The tangent at P meets the x-axis at A and the normal at P meets the x-axis at B. (c) Find the area of triangle APB. Solution(a) Equation of the curve C y = 9 − 4x − x8 , x > 0 Putting x=2 2 y = 9 − 4(2) − 8 2 = 9 − 8 − 4 = −3 The point P has coordinates x = 2, y = −3 dy dx = d(9−4x− x8 ) dx dy dx = d dx (9) dy dx = 0 − 4 dx dx − 8 dy dx = −4 − 8 d(xdx dy dx = −4 − 8(−1)x−2 dy dx = −4 + − d(4x) dx −1 − d( x8 ) dx d( x1 ) dx ) 8 x2 Putting x=2 dy dx = −4 + 8 22 = −4 + 2 = −2 = m Where m is the slope of tangent through the point P (2, −3) Equation of the line through the point P (2, −3) with m=-2 y − (−3) = −2(x − 2) y + 3 = −2x + 4 y = −2x + 4 − 3 y = −2x + 1 y = 1 − 2x (b) 3 Let m1 be the slope of the normal through the point P (2, −3) Since tangent and normal are mutually perpendicular mm1 = −1 Putting m = −2 −2m1 = −1 m1 = 1 2 Equation of the normal through the point P (2, −3) y − (−3) = 12 (x − 2) y + 3 = 12 (x − 2) 2y + 6 = x − 2 2y = x − 2 − 6 2y = x − 8 (c) Equation of tangent through the point P (2, −3) y = 1 − 2x Putting y = 0 in the equation 0 = 1 − 2x 1 − 2x = 0 −2x = −1 2x = 1 4 x= 1 2 Hence A( 12 , 0) is the point where tangent intersects x-axis Similarly, Equation of normal through the point P (2, −3) 2y = x − 8 Putting y = 0 x=8 Hence B(8, 0) is the point where normal intersects x-axis Also P (2, −3) Using the formula x 1 y1 1 Area= 21 x2 y2 1 x 3 y3 1 Here B(8, 0) = B(x1 , y1 ), A( 21 , 0) = A(x2 , y2 ) and P (2, −3) = P (x3 , y3 ) Area= 21 8 0 1 1 2 0 1 2 −3 1 Expanding Area= 12 [8[(0 − (−3)] − 0[ 12 − 2] + 1[ −3 − 0] 2 Area= 12 [24 − 32 ] 5 Area= 12 [ 45 2] = 45 4 = 11.25 6