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

2023
Views:

223

Pages:

4
Author:

Shahbaz A.

Q33 Ex 3.7 page 154 THOMAS’ Calculus Twelfth Edition shahbaz ahmed August 2024 Q 33 Verify that the given point is on the curve and find the lines that are (a) tangent and(b) normal to the curve at the given point. 6x2 + 3xy + 2y 2 + 17y − 6 = 0 , (-1,0) Solution 6x2 + 3xy + 2y 2 + 17y − 6 = 0 Putting x=-1.y=0 6(−1)2 + 3(−1)(0) + 2(0)2 + 17(0) − 6 = 0 6+0+0+0−6=0 0=0 1 L.H.S = R.H.S Point (-1,0) is on the curve. d d [6x2 + 3xy + 2y 2 + 17y − 6] = (0) dx dx d d d d d d (6x2 ) + (3xy) + (2y 2 ) + (17y) − (6) = (0) dx dx dx dx dx dx d d d d 6 (x2 ) + 3 (xy) + 2 (y 2 ) + 17 (y) − 0 = 0 dx dx dx dx dy dx d dy dy 6(2x) + 3[x + y ] + 2 (y 2 ) + 17 =0 dx dx dy dx dx dy dy dy + y] + 2(2y) + 17 =0 dx dx dx dy dy dy 12x + 3x + 3y + 4y + 17 =0 dx dx dx dy dy dy 3x + 4y + 17 = −12x − 3y dx dx dx dy [3x + 4y + 17] = −12x − 3y dx dy −12x − 3y = dx 3x + 4y + 17 12x + 3[x Putting x=-1,y=0 −12(−1) − 3(0) dy = dx 3(−1) + 4(0) + 17 2 dy 12 = dx −3 + 0 + 17 12 dy = dx 14 dy 6 = =m dx 7 Where, m is the slope of the tangent at the point (-1,0) Let m1 be the slope of the normal at the same point. Then mm1 = −1 6 7 m1 = −1 m1 = − 67 Hence equation of the tangent to the curve at the point (-1,0) with slope m = 6 7 y − 0 = 67 [x − (−1)] y = 67 x + 6 7 Similarly, equation of the normal to the curve at the point (-1,0) with slope m1 = − 67 y − 0 = − 67 [x − (−1)] y = − 76 [x + 1] 3 y = − 76 x − 7 6 The tangent and normal at the point (-1,0) are shown here in figure below. 4