Honoured Sr

I returned but last week to Oxford, but whilst I was at \in/ Northampton Shire I began to think on Mr Leibnitz's problem, and I think I have hit upon the Solution of it, it not being difficult to any that understand Fluxions, The method I used was this

Suppose {sic} series of curves of the same nature which have all the same Axis AB and Vertex A. Let AI Figure be one of them and BI the curve which cuts it at right angles, Because the curve AI is given its subnormal GD will be given, But this subnormal is the subtangent of the curve BI. therefore FEEIIGGD. that is calling AG x and IG y: y·x·yGD. (here x· must be negative for when y increases x decreases, by this Analogy I got an equation and from the equation I obtain a value of the Latus Rectum. This value of the latus Rectum I put in its Room in the equation wch expresses the nature of the curve proposed and I have a new exquation {sic} expressed by x y and their fluxions which gives the nature of the curve required. For example suppose the curve AI a Parabola whose equation is 2lx=yy then GD=l and y·x·yx·yy=l
this value of l being put in the equation for the parabola gives 2xx·yy·=yy and 2xx·=yy· hence a2x2=12y2 or 2a22x2=y2 Hence the curve BI is an Ellipse whose Greater Axis is double in Power to its lesser. If the curves Proposed are Hyperbolas, whose center is C transverse Axis AM=2a, their equation is 2lax+lx22a=y2 and the Subnormal is la+lx2a=y2 and y·x·yla+lx2a hence l=2ayx·a+x×y· wch value of l being put in the equation of the Hyperbola gives 4a2yxx·2ayx2x·2a×a+x×y·=y2 and x2x·2axx·x+a=yy· that is xx·ax·+a2x·a+x=yy· or 2a2x·a+x=2xx·+2ax·+2yy· and taking the Fluents we have Fluent 2a2x·a+x + a Given quantity =x2+2ax+y2. From wch we draw the followin construction. at the center C Asymptotes CA CE describe the hyperbola Figure LGH whose rectangle inscribed is =2a2. make BF perpendicular to BA and = to it. and make up the rectang. MF in AB take any point N and let NO be a mean proportional {between} NA and NM, at the center O with the Radius OP whose Square is equal to the rectangle MF minus the Hyperbolick space BNGR describe the Arch PQ and let NG produced meet with it in P the point P will be the curve required.


The curve will be an oval Figure whose greatest Ordinate is at A. Mr Stirling an undergraduate here has likewise solved the Problem, I have also received the solution of this and several other Problems from Mr Pemberton. If you please you may put mine in the Transactions I am Sr

your most obliged Humble Servant

John Keill

The Honoured
Sr Isaac Newton
at his house in St Martins Street
near Leicester Fields


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