Lecture 16 Descartes method of normals
Descarte's Method of
finding tangents, from "La Geometrie" , an appendix to " Discours de la Methode."
[2] Let CE be the curved line. It is desired to draw a straight line at right angles to it through the point C. I suppose the problem to have been solved, and that the sought for line is CP, which I prolong to the point P where it meets the straight line GA. GA is the line to whose points all those of CE are referred; so that putting MA or CB equal to x, I have some equation showing the relation between x and y. Then I put PC = x and PA = v, whence PM = v - y. Since the triangle PMC is right-angled, the square on the hypotenuse s2 is equal to x2 + v2 - 2vy + y2, the sum of the squares of the two sides. That is to say, x = sqroot{s2 - v2 + 2vy - y2} or equally, y = v + sqroot{s2 - x2}. By this means I can get rid of one of the two unknown quantities x or y from the equation relating the points of the curve CE to those of the straight line GA. This is easily done by putting throughout sqroot{s2 - v2 + 2vy - y2} in place of x, the square of this in place of x2, its cube in place of x3 and so on. That is, if it is x I want to get rid of. If it is y, I put in its place x + sqroot{s2 - x2} and its square, cube etc., in place of y2, y3 etc. After this process there always remains an equation in only one unknown quantity x or y.
[3.] For example, if CE is an ellipse, MA the segment of its diameter on which CM is ordinate, and which has r for its latus rectum and q its major axis, then by I, 13 of Apollonius, we have x2 = ry - ry2/q. Getting rid of x2 from this gives s2 - v2 + 2vy - y2 = ry - ry2/q, or
y2 + (qry - 2qvy + qv2 - qs2)/(q - r) = 0
For it is better here to consider the wholw together in this way, than as one part equal to the other.
[4.] Such an equation having been found it is to be used, not to determine x, y or z, which are known since the point C is given, but to find v or s, which determine the required point P. With this in view, observe that if the point P fulfills the required conditions, the circle round P as centre passing through the point C will touch but not cut the curve CE; but if this point P be ever so little nearer to or father from A than
it should be, this circle must cut the curve not only at C but also in another point. Now if this circle cuts CE, the equation involving x and y as unknown quantities (supposing PA and PC known) must have two unequal roots.
Suppose, for example, that the circle cuts the curve in the points C and E. Draw EQ parallel to CM. Then x and y may be used to represent EQ and QA respectively in just the same way as they were used to represent CM and MA; since PE is equal to PC (being radii of the same circle), if we seek EQ and QA (supposing PE and PA given) we shall get the same equation that we should obtain by seeking CM and MA (supposing PC and PA given). It follows that the value of x, or y, or any other such quantity, will be two-fold in this equation, that is, the equation will have two unequal roots. If the value of x be required, one of these roots will be CM and the other EQ; while if y be required, one root will be MA and the other QA. It is true that if E is not on the same side of the curve as C, only one of these wiII be a true root, the other being drawn in the opposite direction, or less than nothing. The nearer together the points C and E are taken however, the less difference there is between the roots; and when the points coincide, the roots are exactly equal, that is to say, the circle through C will touch the curve CE at the point C without cutting it.
[5] Furthermore, it is to be observed that when an equation has two equal roots, its left-hand member must be similar in form to the expression obtained by multiplying by itself the difference between the unknown quantity and a known quantity equal to it; and then, if the resulting expression is not of as high a degree as the original equation, multiplying it by another expression which will make it of the same degree. This last step makes the two expressions correspond term by term.
[6] For example, I say that the first equation found in the present discussion, namely
y2 + (qry - 2qvy + qv2 - qs2)/(q-r)
must be of the same form as the expression obtained by making e = y and multiplying y - e by itself, that is, as y2 - 2ey + e2. We may then compare the two expressions term by term, thus:
Since the first term, y2, is the same in each, the second term, (qry - 2qvy)/(q - r) of the first is equal to -2ey, the second term of the second; whence, solving for v, or PA, we have v = e - (r/q)e + (1/2) r; or, since we have assumed e equal to y, v = y - (r/q)y + (1/2)r. In the same way, we can find s from the third term, e2 = (qv2 - qs2)/(q-r); but since v completely determines P, which is all that is required, it is not necessary to go further.
Explanation
Given a point C on a curve, drawn with respect to an axis, it is enough to find the subnormal, that is the distance from a reference point on the axis to the point where the normal at C meets the axis. Descartes regarded his method of finding the subnormal as the most important result of his book and he criticized Fermat for his mysterious 'e'. Descartes' method has no such problems. The radius of a circle is normal to the circumference; so find a circle which is tangent to the curve at C, and its radius at C is normal to the curve. To construct such a circle, consider any circle which passes through C. It crosses the original curve twice. Find a quadratic whose two roots are the x--coordinates of these two points. When the coefficients are changed to make the roots coincide, the circle is tangent to the curve.
[In modern notation, let C(x,y) be a point on the graph of y=f(x) , and it is required to find the normal at C. Let P be the point at which the normal meets the x--axis. Let the curve meet this axis at A, let M be the foot of the perpendicular at C. Suppose the circle through C with centre P meets the curve again at E, and let Q be the foot of the perpendicular at E.
Let CP = s, and PA = v so PM = v-x . The equation of the circle is
f is known (in Descartes' case it is an ellipse), x is known so the equation can be solved for v in terms of s. There will be two solutions, say M and Q. When E is near C, the difference of the roots is small, and when the circle is a tangent, the roots are equal, ie the equation has a double root, say x0. But this means (x-x0)2 is a factor:
By equating like powers of x, solve for v.
In the example, the equation of the ellipse is x2 - ry + (r/q) y2 = 0 , the major axis is q and the equation above (with x and y reversed) is
Finding either r or s gives P. When there is a double root for y, the equation is divisible by (y - e)2 = y2 - 2ey + e2, so the other factor has no term in y. Hence the coefficient of y is (qr - 2qv)/(q - r) = -2e , so qr - 2qv = -2eq + 2er and hence v = (qr + 2eq + 2er)/(2q) = e-r/q + r/2 = y - r/qy + r/2 , which is what you need.