 The symptom of a parabola leads to a useful result that two ordnance to a parabola be given. The squares on the ordnance are to each other as the lines cut off on the diameter. So if we take our conic section, we'll draw some parallel lines through the section, and remember the diameter will bisect all these parallel lines. So let's draw a few lines ordnance-wise. Then the squares on the ordnance are going to be to each other as the lines cut off from the diameters. So here we might take our ordnance as x2, and the line cut off from the diameter is y2. And here the ordnance will be x1, and the line cut off from the diameter y1. And so we have x2 squared is to x1 squared, as y2 is to y1. Now in Latin, to cut off is apsindere, so we can refer to the cut off lines as the abscises. Because these symptoms provide a relationship on lines within the figure, Apollonius can use them to find the tangent to the curve. Apollonius proves, conics book 1, proposition 17, if a straight line is drawn from the vertex that is parallel to an ordnance, that line will be outside the conic section at the vertex. This is equivalent to giving the tangent to the curve. We leave the proof for the viewer. What if you wanted to find the tangent at a non-vertex point? For a parabola, Apollonius proves, let AC be a parabola with diameter AE and vertex A. If CE is a line drawn ordnance-wise, and AE equals AF, then CF touches the parabola at C. To prove this, Apollonius supposes that CH somehow sneaks back under the parabola. Let's extend GH ordnance-wise to D on the parabola. And to avoid getting lost in the geometric argument, we'll let AE equal AF equal A, AG we'll call E, we'll call GHH EC y1, and GD y2. Now because ACD is a parabola, the squares of the lines drawn ordnance-wise are to each other as the abscissus. So E is to A as y2 squared is to y1 squared. But since GD is greater than GH, then y2 squared to y1 squared is greater than H squared to y1 squared. Which means that E is to A is greater than H squared is to y1 squared. Now because CE and GD are parallel, we have a couple of similar triangles. And so by similar triangles, we have H is to y1 as A plus E is to 2A. And so H squared is to y1 squared as A plus E squared is to 4A squared. And so that tells us that E is to A is greater than A plus E squared is to 4A squared. Now we can rewrite this proportionality as E over A is greater than A plus E squared over 4A squared. Then cross multiplying and simplifying. But note that E is greater than A. And this means that A plus E is greater than 2A. So A plus E squared is greater than 4A squared. Again, since A is greater than E, 4A squared must be greater than 4A E. And so A plus E squared must be greater than 4A E. And so this inequality is impossible. And that means this line can't drop under the parabola. And a similar proof shows that the line will not sneak under the parabola before it arrives at the point C.