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

2023
Views:

61

Pages:

4
Author:

Shahbaz A.

Q 29 Ex 3.7 THOMAS’ CALCULUS Twelfth Edition Page 154 shahbaz ahmed August 2024 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. x2 + xy − y 2 = 1, (2, 3) solution Since x2 + xy − y 2 = 1, (2, 3) Putting x=2,y=3 in equation (1) 22 + (2)(3) − 32 = 1 4+6−9=1 1 (1) 10 − 9 = 1 1=1 L.H.S = R.H.S Hence the point (2,3) lies on the curve Now differentiating equation(1) implicitly. d 2 d [x + xy − y 2 ] = (1) dx dx d 2 d d (x ) + (xy) − (y 2 ) = 0 dx dx dx 2x + x dy dx d dy + y − (y 2 ) =0 dx dx dy dx dy dy + y − 2y =0 dx dx dy dy x − 2y = −2x − y dx dx dy [x − 2y] = −[2x + y] dx dy 2x + y =− dx x − 2y 2x + x 2 Putting x=2,y=3 dy 2(2) + 3 =− dx 2 − 2(3) 7 dy = −( ) dx −4 dy 7 = =m dx 4 Where m is the slope of the tangent at the point (2,3) Let m1 be the slope of the normal at the same point Then mm1 = −1 Putting m = 7 4 m1 7 4 = −1 m1 = −1 × 4 7 m1 = − 74 Hence equation of tangent with slope m = y − 3 = 74 (x − 2) y = 74 x − 72 + 3 y = 47 x − 1 2 3 7 4 at the point (2,3) Similarly, equation of the normal with slope m1 = − 47 at the point (2,3) y − 3 = − 74 (x − 2) y = − 47 x + 78 + 3 y = − 47 x + 29 7 4