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On the osculating plane z = the curve H consists of the x-axis and the para- 

 bola 3x- = 4y. It divides the plane into four regions: Ri, the top half plane; 

 Rj, that partof the lower left quarter plane which is "outside" the parabola; R3, 

 t liat part of the lower right quarter plane which is "ouside" the parabola; R4, 

 that part of the lower half j)lane which is "inside" the parabola. Throughout 

 tlie first three regions, D > 0, while in the fourth region, D < 0; everywhere 

 on H itself of course D = 0. 



Any other osculating plane 3 t- x — 3 t y + z — t ^ = 0, t ±0, cuts the plane 

 z = in the line 3 t x — 3 y — t- = 0. This ecjuation represents the one para- 

 meter family of lines which envelope the parabola 3x'- = 4y. The parameter t, 

 is in fact the slope of these tangents. 



From any point in Rus two tangents can be drawn to the parabola and 

 through each of these pass two osculating planes; from no point in R4 can a 

 tangent be drawn to the parabola and through this region of z = there passes 

 no line which is the intersection of two osculating planes. Through any point 

 on the parabola itself one and only one tangent can be drawn and (excepting the 

 tangent at the origin) this is the intersection of two osculating planes. Through 

 any point (except the origin) on the x-axis, which is a part of H, two tangents 

 can be drawn to the parabola but one of these is in all cases the x-axis itself, 

 through which passes no osculating plane distinct from z= 0. Therefore 

 through any point (except the origin) on the x-axis there passes one line which 

 is the intersection of two osculating planes. 



Examples on the osculating plane z = 0. 



a) Consider the point (0, —3, 0) in Ri 



3 t X — 3 y — f^ = gives ti = 3 to =— 3 

 Li: 3 X — y — 3 = through which pass osculating planes 



27x — 9y -H z = 27 and z = 

 L2: 3x -|- y -|- 3 = through which pass 



27x + 9y -F z -F 27 = and z = 



b) Consider the point (1/2, — 3/2, 0) in R, 



3 t X — 3 y — t= = gives ti = 3 t. =—3/2 

 Li: same as Li under a) 



L2: 6x + 4y-|-3 = through which pass 



54 X + 18 y + 4 z -f 27 = and z = 



c) Consider the point (2, 3, 0) on the parabola 



3tx — 3y — t= = gives t = 3 



L: same as Li under a) 



