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

2023
Views:

280

Pages:

4
Author:

Shahbaz A.

Q32 Ex 3.7 page 154 THOMAS’ Calculus Twelfth Edition shahbaz ahmed August 2024 Q 32 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. y(x − 2)(x − 3) − x + 7 = 0, (7, 0) Solution Given equation y(x − 2)(x − 3) − x + 7 = 0 Putting x=7,y=0 0(7 − 2)(7 − 3) − 7 + 7 = 0 0=0 1 L.H.S = R.H.S Hence ,the point (7,0) is on the curve. Now differentiating implicitly d d [y(x − 2)(x − 3) − x + 7 =] = (0) dx dx d dx d d [y(x − 2)(x − 3)] − + (7) =] = (0) dx dx dx dx d dy y [(x − 2)(x − 3)] + (x − 2)(x − 3) − 1 + 0 = 0 dx dx d dy y [(x − 2)(x − 3)] + (x − 2)(x − 3) − 1 = 0 dx dx dy d y [x2 − 5x + 6] + (x − 2)(x − 3) − 1 = 0 dx dx dy y(2x − 5) + (x − 2)(x − 3) − 1 = 0 dx dy 2xy − 5y + (x − 2)(x − 3) − 1 = 0 dx dy = 1 − 2xy + 5y (x − 2)(x − 3) dx dy (x2 − 5x + 6) = 1 − 2xy + 5y dx dy 1 − 2xy + 5y = 2 dx (x − 5x + 6) 2 Putting x=7,y=0 dy 1 − 2(7)(0) + 5(0) = dx (72 − 5(7) + 6) 1 dy = =m dx 20 Where m is the slope of tangent at the point (7,0). Hence equation of tangent. y−0= 1 (x − 7) 20 y= 1 (x − 7) 20 y= 1 7 x− 20 20 Let m1 be the slope of the normal at the point (7,0) mm1 = −1 1 20 m1 = −1 m1 = −20 Equation of normal at the point (7,0) with slope m1 = −20 y − 0 = −20(x − 7) y = −20x + 140 3 Tangent and normal at the point (7,0) is shown here in figure below. 4