 In his new methods of conic sections, John Wallace extended the analytic geometry of Fermat and Descartes. Like Fermat, Wallace allowed algebraic relationships to define curves, and like Descartes, he translated geometric problems into algebraic equations. For example, Wallace constructs a cubicle parabolaid where the intercepted parts of the diameter are to each other as the cubes of the lines drawn ordinate-wise. Now Wallace's figure is not very readable, so we'll redraw it with the important components. So, if at some point on the curve alpha, the line drawn ordinate-wise is p-alpha, which intercepts p-a on the diameter, and at o, another point on the curve, the line drawn ordinate-wise d-o, which cuts off d-a on the diameter, and the defining property of this curve is the ratio p-a to d-a is the same as p-alpha cubed to d-o cubed. Wallace also presented a new method of finding tangents. So, again, let the curve be given and let the tangent at alpha meet the axis at some point f. So, Wallace let p-f equal f, the sub-tangent, p-a equal to d, d-p equal to a, and p-alpha equal to p. Now, by the defining property of the curve, the ratio p-a to d-a is p-alpha cubed to d-o cubed. Now, p-a is d, and in our figure d-a is drawn as being past p, but remembering Wallace's original diagram, it could be on either side, and so Wallace wrote it as d-plus or minus a, and p-alpha is p, so p-alpha cubed is p-cubed. And so we can solve for d-o cubed, which will be... Now, by the properties of similar triangles, the ratio p-f to p-alpha is d-f to d-t. Remember, p-f is f, p-alpha is p, d-f, again, is going to be f-plus or minus some amount, and so now we can solve for d-t and then for d-t cubed. Notice the tangent line is outside the curve, so d-t is greater than d-o, which means that d-t cubed must be greater than d-o cubed. Substituting in our values, and we can simplify this to get, and if we let a vanish, in other words, we let a be zero we get, so our sub-tangent f will be equal to 3-d. So going back to our picture, remember our sub-tangent is this length here, p-f, and Wallace provides a way of finding the point where our tangent line intersects the axis of the curve. Now Wallace's procedure glosses over some important details. However, we can make it rigorous with a little bit of work. So to begin with, let's think about the curve that Wallace is considering. Suppose alpha is a specific point on the curve, so p-a and p-alpha have definite lengths. Let p-a be, say, k and p-alpha be h. If we let d-a equal y and d-o equal x, the defining property of the curve is that p-a is to d-a as p-alpha cubed is to d-o cubed. And so we can write that as this equation, and we simplify we get. So Wallace is talking about the curve we describe as y equals c-x cubed. Now remember Wallace's graph looked like this, so that Wallace's graph is very different from our modern graph of y equals c-x cubed. So let's draw a more familiar version of y equals x cubed, let alpha be the point of tangency, and extend the tangent line to f on the y-axis. Now our goal is to find where this point f is, because if it's too close or too far away the line clearly isn't tangent. So let's draw our line drawn ordinate-wise, and consider some point t beyond alpha. And we'll draw a line ordinate-wise to the curve as well, and label some points. Now as before we'll let p-f be f, p-a be d, d-p equal to a, and p-alpha equal to p. Since the curve is y equal to x cubed, then p-alpha cubed is p-a, and so p cubed equals d. d-o cubed, that's d to this point o on the curve, well that's going to be d-a, and so d-o cubed is d plus a. Again, by similar triangles, p-f is to p-alpha as d-f is to d-t. And we know p-f and p-alpha, and if we look at our picture d-f is going to be f plus a, and d-t is, well let's leave that alone and solve for it, and d-cube will be, but our tangent line is outside of the curve, so d-t has to be greater than d-o, which means that d-t cubed must be greater than d-o cubed. So we could substitute in our values, and do a little bit of algebra. What's important is to recognize that this inequality must be true for all a greater than zero. So if we let a equal to zero, then we're able to write the inequality, and so 3-d must be greater than f. We can interpret this result as follows. Provided the intersection point is no further than 3-d away, the line through alpha will be beyond the curve for all points beyond alpha. However, it's possible the line might intersect the curve before alpha, so we need to analyze that situation as well. So again, as before, we'll let p-f equal f, p-a equal d, d-p equal a, p-alpha equal to p, and again, since the curve is y equals x cubed, p-alpha cubed is p-a, so p-cubed equals d. d-o cubed is d-a, but this time d-a is d-minus a. And again, by similar triangles, we have d-t cubed. And again, if our tangent is outside of the curve, we need d-t greater than d-o, so d-t cubed is greater than d-o cubed, and so we find... Note that f is greater than a, so this quantity we're subtracting is a product of two positive numbers. Since the sine quality must be true for all a greater than zero, we need... and so f is greater than or equal to 3-d. And as before, we can interpret this result as follows. Provided the intersection point is at least 3-d away. The line through the alpha will be beyond the curve for all points up to alpha. Put together, the point f needs to be no further than 3-d away, so that we don't intersect beyond alpha, but it also needs to be at least 3-d away, so we don't intersect before alpha. And so that means our point f should be exactly 3-d away from p in order for the line f-alpha to be tangent to the curve.