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A simple modification leads to a construction for a tangent to a parabola 
from any external point G. We have only to replace the directing circle by the 
directrix of the parabola. 

The Hyperbola. 

Proposition VI. If at two fixed points P and P, three lines A, B, and G, 
be pivoted, A at one point revolving in one direction at any velocity; B and G 
at the other pivot revolving in an opposite direction, G at such a rate that it constantly 
intersects A in the circumference of a directing circle described with P as a center, B 
at such a rate that the angle BG is constantly equal to the angle GA, then the locus of 
the intersection M of A and B is a hyperbola. 

Let the angle AG be denoted by (j> and BG by 8. 
Since ^=#, the segment JOf =segment PM in any j 
position. Therefore PM - PM = PM— NM=PW= | 
constant. Therefore the locus of M is a hyperbola. 

Proposition VII. If a circle with center G be I 
described in the plane of a hyperbola passing through one 
focus P and intersecting the directing circle at H, and the 

other focal radius P' Mbe drawn through this point H to 
meet the curve, at M, the line GM is tangent to the 

Draw MP and HP. The triangle BMP is isos- 
celes and GM is perpendicular to the base PU at its 
mid-point A. Therefore it passes through the vertex 
M and is tangent to the hyperbola. 


By ARCHIBALD HENDERSON, Ph. D„ Associate Professor of Mathematics, University of North Carolina, 

Chapel Hill, N. C. 

Consider any circle, whose center is the point (0, y g ) and whose radius is 
the distance from this point to the focus £i/(a 2 +& 3 ), 0] of an hyperbola. The 
equation of this circle is 

or^+y 8 -2y y-(a*+& 8 )=0....(l). 
Now we may represent any point on an asymptote to the hyperbola 


a 2 b* i—W 

by introducing the parameter t. Thus (a;,, y i )=(at 1 , bt^) represents any point 
on the asymptote 


and (x. x , y t )= (— at it bt t ) represents any point on the asymptote 


If the circle (1) cuts the asymptotes (3) and (4) in the specified points 
(^n tti)> ( x i> Hz)) respectively, we have 

(a*+b*Xt?-l)=2by t 1 ....(5), 
(a 2 +&*)(* 8 2 -l)=2ty * 8 ....(6). 
By division we obtain 

which may be written 

(< 1 -< i )(«,< i +l)=0....(7). 
The solution 

<,-< t =0....(8) 

shows that, for one position of (a 2 , # 2 ), the line joining (a;, , y, ) and (x 2 , y 2 ) is 
parallel to the z-axis. Discarding this case, let us consider the solution 

* 1 «„+1=0....(9). 

Since (x t , «/ 2 )=(-^-, —. — ), the equation of the line joining («,,y,) and 

y—bt 1 x—at-i 


But this line touches the hyperbola (2), since 





&«.... (ii). 

TTie lines joining the pairs of points {right hand, say) in which a system of co- 
axial circles, passing through the foci of an hyperbola, cuts the asymptotes, envelope 
that hyperbola. 

Since, moreover, the middle point 
of the line joining (x x , «/,), (a; 2 , t/ 8 ) lies 
on the hyperbola, we have the theorem : 

The middle points of the line? join- 
ing the pairs of points in which a system of I 
co-axial circles, passing through the foci of j 
an hyperbola, cuts the asymptotes, describe 
that hyperbola. 

These two theorems give two 
methods for constructing an hyperbola, 
the one by lines, the other by points, 
when the asymptotes and a focus are known.* Other constructions might readily 
have been given, but those given above seem the most instructive. 

The University of Chicago, November, 1902. 

* Compare the November number of the Monthly for a note by the writer on the converse of this 




163. Proposed by CHRISTIAN HORNUNGf, A.M.. Professor of Mathematics, Heidelberg University, Tiffin. 0. 

Three Dutchmen and their wives went to market to buy hogs. The names of the 
men were Hans, Klaus, and Hendricks, and of the women, Gertrude, Anna, and Katrine; 
but it was not known which was the wife of each man. They each bought as many hogs as 
each man or woman paid shillings for each hog, and each man spent three guineas more 
than his wife. Hendricks bought 23 hogs more than Gertrude, and Klaus bought 11 more 
than Katrine. What was the name of each man's wife? 

Solution by J. SCHEFFER. A. M„ Hagerstown, Md.. and M. E. GEABER, Heidelberg University, Tiffin, 0. 
Let x represent the number of one of the women's hogs, and?/ the number 
of her husband's; then by the conditions of the problem j/ a =a; 2 +63. Conse- 
quently a; 2 +63 must be an integer, since |/(a: 2 +63) represents the number of 
hogs. The equation y 2 — <c s =63 or (y-\- x)(y— x)—63 admits of three solutions, 
viz., 63x1, 21x3, and 9x7.