202 SUCCESSORS OF THE GREAT GEOMETERS 
Let PN be the ordinate from P; draw A Y at right angles 
to the axis meeting PT in Y, and join SY. 
Now PA r2 = AL.AN 
= 4 AS. AN 
= 4AS. AT (since AN = AT). 
But PN — 2AY (since iA = AT); 
therefore A Y 2 = TA . AS, 
and the angle TYS is right. 
The triangles SYT, SYP being right-angled, and TY being 
equal to YP, it follows that SP = ST. 
K 
With the same figure, let BP be a ray parallel to AN 
impinging on the curve at P. It is required to prove that 
the angles of incidence and reflection (to S) are equal. 
We have SP = ST, so that ‘ the angles at the points T, P 
are equal. So says the author, ‘ are the angles TP A, KPR 
[the angles between the tangent and the curve on each side of 
the point of contact]. Let the difference between the angles 
be taken; therefore the angles SPA, RPB which remain 
[again ‘ mixed ’ angles] are equal. Similarly we shall show 
that all the lines drawn parallel to AS will be reflected at 
equal angles to the point S.’ 
The author then proceeds: ‘Thus burning-mirrors con 
structed with the surface of impact (in the form) of the 
section of a right-angled cone may easily, in the manner