Anyone who knows about Descartes' Method??? Find the equation of the tangent line using Descartes' Method!!!!!

Question

please help me with Descartes' Method!!!! thx~~
Find the equation of the tangent line using Descartes' Method!!! plz~ help? Make sure have to use Descartes' Method
i knows the way to do with derivative
Find the equation of the tangent line to the following functions,
using Descartes' Method.

#1 f(x) = x^2, at, (2,4)
#2 f(x) = x^3+7x, at, (1,8)
#3 f(x) = x, at, (-3,3)

Could u guys help me to get tangent line of equation above
3 problems??? have test help thx!!!

about Descartes' Method.
1. (x-xo)^2+y^2 = (x1-xo)^2+(y1)^2
substituting f(x) for y in the above equation, we get:
(x-xo)^2+(f(x))^2 = (x1-xo)^2 + (y1)^2

Answer 1

First, a warning: I had never heard of Descartes's method before (after two math degrees at university!). So, you better check that I do describe the method you seek.

I read a brief description in section 13.4 of the document at the web site (see source). The idea of the method is to find an equation that simultaneously solve two equations: that of the curve and that of the straight line we are trying to fit tangent to the curve. Then we find the condition(s) for the simultaneous solution to have only one solution in the neighbourhood of the point of contact.

In equation #1 we are lucky because f(x) is a parabola, which is a convex figure (the tangent will never cross the curve at any other point; there are equations where a local tangent may cross the curve at some far point -- see an example at 13.3 in the same document)

f(x) = x^2 is a parabola. The point (2,4) belongs to the parabola (because 4 = 2^2).

A straight line is represented by equation g(x) = mx + c where m is the constant slope and c is the intersection of the y axis. If the line goes through (2,4), then c can be expressed as a function of m (for example, if m = 0 -- a horizontal line -- then c = 4; if m = 1 -- a 45 degree slope -- then c = 2; if m = 2, then c = 0; and so on). In general, c = 4 - 2 m.

g(x) = mx + (4 - 2 m) for a line through (2,4)

At intersection points, the distance between the two lines is zero. So, let us subtract one line from the other and find out where the distance is zero:

f(x) - g(x) = x^2 - mx -(4 - 2 m) = 0
x^2 -mx +2m -4 = 0

roots: ( m +/- SQRT ( m^2 - 4(2m - 4)) )/ 2

What value of m will yield only one solution? The value such that the content of the square root (SQRT) is zero.

m^2 - 8m +16 = 0
(m-4)(m-4) = 0

m = 4
(which, substituting in f(x)-g(x), leaves us with x=2 as the point of contact... good!)

Thus our tangent is g(x) = 4 x - 4

Answer 2

using Reference a million as my practise guide and using a circle targeted on the y-axis truly than than the x-axis .. the conception being that the gap from the centre to the point the position the circle cuts the curve equals the gap from the centre to the point. in the adventure that they're a ideal sq. then the circle purely touches the curve and so the radius is perpendicular to the tangent to the curve #a million f(x) = x² at (2,4) (x)² + (y - yo)² = (x1)² + (y1 - yo)² the position the circle is targeted at (0, yo) At (2, 4) (x)² + (y - yo)² = (2)² + (4 - yo)² Now x² = y therefore y + y² - 2yoy + yo² - (4 + 16 - 8yo + yo²) = 0 ie y² - (2yo - a million)y + 4(2yo - 5) = 0 For there to be purely one cost for y (or truly 2 values a similar) the discriminant of this quadratic has to equivalent 0 ie (-(2yo - a million))² - 4 * a million * 4(2yo - 5) = 0 ie 4yo² - 4yo + a million + 80 - 32yo = 0 4yo² - 36yo + 80 one = 0 ie (2yo - 9)² = 0 yo = 9/2 So the centre of the circle is (0, 9/2) So the slope of the radius is (4 - 9/2)/(2 - 0) ie -a million/4 So slope of tangent is 4 and its equation is y = 4x - 4 #2 y = x³ + 7x at (a million, 8) pick circle targeted (Xo, 0) Then (x - Xo)² + (x³ + 7x)² = (a million - Xo)² + 8² So x² - 2Xox + Xo² + x^6 + 14x^4 + 49x² = a million - 2Xo + Xo² + sixty 4 x^6 + 14x^4 + 50x² -2Xox + 2Xo - sixty 5 = 0 ie YUCK!!!!!!!! I even do not comprehend the position to bypass from right here!!!