Mr. Burch's Method of Drawing Hyperbolas. 375 



points on the axis of x ; slide the triangle along this edge 

 until the point / falls on the corresponding line from a, and 



f/c.3. 



mark the position of the vertex d of the triangle (see fig. 4) . 

 This method is^ perhaps even more rapid and convenient than 

 the first, as it involves the drawing of only one set of lines 

 ac, ac f , &c. Like the first, it involves the use of only a triangle 

 and a T-square. 



3rd, method. — Let the vertices of the two similar triangles 

 coincide at b (fig. 5). Then if be = unity we have as before 



To determine the points on the curve graphically in this 

 case, we need a triangle or bevel-gauge of an angle equal to 

 yox and a parallel ruler. The side be of the triangle or bevel- 



gauge should be of unit length and a distance fb= a 



laid off on the side bq. The triangle is placed with its side 

 bq coinciding with the axis of a, and one edge of the parallel 

 ruler is brought against the point c of the triangle and a pin 

 placed at the origin. The other blade of the parallel ruler is 

 then moved out until it passes through the point / on the 



