Question

Use Newton's method to find the coordinates, correct to six decimal places, of the point on the parabola y = (x − 7)2 that is closest to the origin.

Answer 1

let that point be P(a,b)
then b = (a-7)^2

dy/dx = 2(x-7)
at P, dy/dx = 2(a-7) <---- the slope of the tangent at P

Join PO. If P is the closest to O, then the slope of PO must be
the negative reciprocal of the slope of the tangent.
That is, at the point closest to the origin PO is the normal to the tangent at P

2(a-7) = a/b = -a/(a-7)^2
expanding and simplifying we get
2a^3 - 42a^2 + 295a - 686 = 0

it is this equation we want to solve using Newton's Method
let y = 2a^3 - 42a^2 + 295a - 686
dy/da = 6a^2 - 84a + 295

a(new) = a - (2a^3 - 42a^2 + 295a - 686)/(6a^2 - 84a + 295)

a ..... a(new) , start with a = 5 and use your calculator
5 5.44
5.44 5.57798...
5.57798..... 5.591158509
5.591158509 5.591272805
5.591272805 5.591272814 <---- looks like we got it to more than 6 decimal places

so the x of P is 5.591273 correct to 6 decimals
sub the above into y = (x-7)^2 to find the y