or $2\mathrm{yy}-2\mathrm{qy}=\frac{2\mathrm{b}}{\mathrm{a}}\mathrm{xx}+\mathrm{ax}+\frac{2\mathrm{b}}{\mathrm{a}}\mathrm{px}+\mathrm{ap}$ & substracting
$\mathrm{ax}+\frac{\mathrm{b}}{\mathrm{a}}\mathrm{xx}-\mathrm{yy}=0$ twice (viz so often as the Curve has
dimensions) there comes $\frac{2\mathrm{bpx}}{\mathrm{a}}-\mathrm{ax}+2\mathrm{qy}+\mathrm{ap}=0$, a
strait line w^ch drawn will cut y^e Curve in the
points desired & so of the rest.

Where note that this Problem in any Curve is al
ever solvable by a line of an inferior degree. And
also that a line drawn from a given point may touch
a Curve of two Dimensions in 1×2 points, of three
in 2×3 points of 4, in 3×4 p^ts. &c unless it be polar
&c.

Probl. 3.

From a given point (C) to draw a Perpend [i] ^r.
(CB) to any Curve (VB)

Make VE=r, EC=s and AD=v so is y−s∶r− [s] \x/ ∷
y∶v. or $\frac{\mathrm{ry}-\mathrm{xy}}{\mathrm{y}-\mathrm{s}}=\mathrm{v}$ put this valor
of v in the room of v found ꝑ Prob. 1.
And and you have a Curve w^ch
described will cut the propounded
Curve in the points to w^ch the
perpendicul^r may be drawn [but ever try if by means
of the nature of the given Curve you can reduce this
resulting to any simpler form or degree.

Ex^a. If ax+$\frac{\mathrm{b}}{\mathrm{a}}$xx−yy=0, then $\frac{1}{2}$a+$\frac{\mathrm{b}}{\mathrm{a}}$x=v, or
ry−xy=$\frac{1}{2}$y−$\frac{1}{2}$as+$\frac{\mathrm{b}}{\mathrm{a}}$xy−$\frac{\mathrm{b}}{\mathrm{a}}$sx, w^ch appears not
reducible to a simpler form.

Yet it may be worth while to try if it may be solved
by a Circle by assuming d+ex+fy=xx+yy and com-
paring these 3 equations of the given Conick, found [illeg] Hy-
perbola and assum'd Circle to find d, e & f w^ch if they
be found by plane Geometry that Circle described will
cut the Curve in the defined points. This, I say, might
be tried but it would be found impossible (unless the
Conic be a Parabola) because there are 4 given points
thró w^ch the Circle must pass, w^ch make the Probl. con-
tradicting, since 3 are [now] to determine a Circle.

This might be done by the Parabola, but the Hyper-
bola falling in so naturally, and being as easily described,
'tis not worth the while.

What is said of the Conics may be applied to any
other Curves.

From hence it appears that this Probl. is ever solvable
by a Curve of the same degree and sometimes ꝑhaps by
a Curve of an inferior Degree.