# The Mathematical Principles of Natural Philosophy (1729)/Book 1/Section 4

*Of the finding of elliptic, parabolic, and hyperbolic orbits, from the focus given.*

Lemma XV.

*If from the two foci* S, H, (Pl. 7. Fig, 1.) *of any ellipisis or hyperbola, we draw to any third point* V *the right lines* SV, HV, *whereof one* HV *is equal to the* *principal axis of the figure, that is, to the axis in which the foci are situated, the other* SV *is bisected in* T *by the* *perpendicular* TR *let fall upon it; that perpendicular* TR *will somewhere touch the conic section: and vice versa if it does touch it*, HV *will be equal to the*
principal axis of the figure.

For, let the perpendicular *TR* cut the right line
*HV*, produced if need be, in *R*; and join *SR*. Because

*RS*, *TV* are equal, therefore the right lines *SR*,
*VR*, as well as the angles *TRS*, *TRV*, will be
also equal. Whence the point *R* will be in the conic
section, and the perpendicular *TR* will touch the same:
and the contrary. *Q. E. D.*

Proposition XVIII. Problem X.

*From a focus and the principal axes given, to describe elliptic and hyperbolic trajectories, which shall pass through given points, and touch right lines given by position.* Pl. 7. Fig. 2.

Let *S* be the common focus of the figures; *AB*
the length of the principal axis of any trajectory; *P* a
point through which the trajectory should pass; and
*TR* a right line which it should touch. About the
centre *P*, with the interval *AB* - *SP*, if the orbit
is an ellipsis, or *AB* + *SP* if the orbit is an
hyperbola, describe the circle *HG*. On the tangent
*TR* let fall the perpendicular *ST* and produce
the same to *V*, so that *TV* may be equal to *ST*; and
about *V* as a centre with the interval *AB* describe
the circle *FH*. In this manner whether two points
*P*, *p*, are given, or two tangents *TR*, *tr*, or a point *P*
and a tangent *TR*, we are to describe two circles. Let
*H* be their common intersection, and from the foci
*S*, *H* with the given axis describe the trajectory. I say the
thing is done. For (because *PH* + *SP* in the ellipsis,
and *PH* - *SP* in the hyperbola is equal to the
axis) the described trajectory will pass through the
point *P*, and (by the preceding lemma) will touch
the right line *TR*. And by the same argument it will either pass through the two points *P*, *p*, or touch
the two right lines *TR*, *tr*. *Q. E. F.*

Proposition XIX. Problem XI.

*About a given focus, to describe parabolic trajectory, which shall pass through given points, and touch right lines given by position*. Pl. 7. Fig. 3.

Let *S* the the focus, *P* a point, and *TR* a tangent
of the trajectory to be described. About *P* as a centre,
with the interval *PS*, describe the circle *FG*;
From the focus let fall *ST* perpendicular on the tangent,
and produce the same to *V*, so as *TV* may be equal
to *ST*. After the same manner another circle
*fg* is to be described, if another point *p* is given; or
another point *v* is to be found, if another tangent *tr*
is given; then draw the right line *IF*, which shall
touch the two circles *FG*, *fg*, if two points *P*, *p*
are given, or pass through the two points *V*, *v*,
if two tangents *TR*, *tr* are given, or touch the
circle *FG* and pass through the point *V*, if the point *P*
and the tangent *TR* are given. On *FI* let fall the
perpendicular *SL* and bisect the same in *K*; and with
the axis *SK*, and principal vertex *K* describe a parabola.
I say the thing is done. For this parabola (because
*SK* is equal to *IK*, and *SP* to *FP*) will pass through
the point *P*; and (by cor. 3. lem. 14.) because *ST* is
equal to *TV* and *STR*; a right angle, it will touch the
right line *TR*. *Q. E. F.*

Proposition XX. Problem XII.

*About a given focus to describe any trajectory given in specie, which shall pass thro' given points and touch right lines given by position.*

Case 1. About the focus (Pl. 7. *Fig.* 4.) it
is required to describe trajectory *ABC*, passing thro'
two points *B*, *C*. Because the trajectory is given in
specie, the ratio of the principal axe to the distance of
the foci will be given. In that ratio take *KB* to *BS*
and *LC* to *CS*. About the centres *B*, *C*, with the intervals
*BK*. *CL* describe two circles, and on the right
line *KL*, that touches the same in *K* and *L*, let fall the
perpendicular *SG*; which cut in *A* and *a*, so that *GA*
may be to *AS*, and *Ga* to *aS*, as *KB* to *BS*; and
with the axe *Aa*, and vertices *A*, *a*, describe a trajectory.
I say the thing is done. For let *H* be the other focus
of the described figure, and seeing *GA* is to *AS* as
*Ga* to *aS*, then by division we shall have *Ga* - *GA* or
*Aa* to *aS* - *AS* or *SH* in the same ratio, and therefore
in the ratio which the principal axe of the figure to be
described has to the distance of its foci; and therefore
the described figure is of the same species with the figure
which was to be described. And since *KB* to *BS*,
and *LC* to *CS* are in the same ratio, this figure will
pass thro' the points *B*, *C*, as is manifest from the conic
sections

Case 2. About the focus (Pl. 7. *Fig.* 5.) it is
required to describe a trajectory, which shall somewhere
touch two right lines *TR*, *tr*. From the focus
On those tangents let fall the perpendiculars *ST*, *St*, which produce to *V*, *v*, so that *TV*, *tv* may be
equal to *TS*, *tS*. Bifect *Vv* in O, and erect the indefinite
perpendicular *OH*, and cut the right Line
*VS* infinitely produced in *K* and *k*, so that *VK* be
to *KS*, and *Vk* to *kS* as the principal axe of the
trajectory to be described is to the distance of it's
foci. On the diameter *Kk* describe a circle cutting
*OH* in *H*; and with the foci *S*, *H*, and principal
axe equal to *VH*, describe a trajectory. I say the
thing is done. For, bisecting *Kk* in *X*, and joining
*HX*, *HS*, *HV*, *Hv*, because *VK* is to *KS*, as
*Vk* to *kS*; and by composition, as *Vk* + *Vk* to
*KS* + *kS*; and by division as *Vk* - *VK* to *kS* - *KS*
that is, as 2*VX* to 2*KX* and 2*KX* to 2*SX*, and therefore
as *VX* to *HX* and *HX* to *SX* the triangles *VXH*,
*HXS* will be similar; Therefore *VH* will be to *SH*,
as *VX* to *XH*; and therefore as *VK* to *KS*. Wherefore
*VH* the principal axe of the described trajectory
has the same ratio to *SH* the distance of the foci,
as the principal axe of the trajectory which was to be
described has to the distance of its foci; and is therefore
of the same species. And seeing *VH*, *vH*, are
equal to the principal axe, and *VS*, *vS* are perpendicularly
bisected by the right lines *TR*, *tr*; 'tis evident
(by lem. 15.) that those right lines touch the
described trajectory. *Q. E. F.*

Case 3.. About the focus *S* (Pl. 7. *Fig.* 6.) it
is required to describe a trajectory, which shall touch
a right line *TR* in a given point *R*. On the right line
*TR* let fall the perpendicular *ST*; which produce to
*K* so that *TV* maybe equal to *ST*, join *VR*, and
cut the right line *VS* indefinitely produced in *K* and
*k*, so that *VK* may be to *SK*, and *Vk* to *Sk' as*
the principal axe of the ellipsis to be described, to
the distance of itsfoci; and on the diameter *Kk* describing a circle, cut the right line *VR* produced
in *H*, then with the foci *S*, *H* and principal axe
equal to *VH*, describe trajectory. I say the thing
is done. For *VH* is to *SH* as *VK* to *SK*, and
therefore as the principal axe of the trajectory which
was to be described to the distance of its foci, (as appears
from what we have demonstrated in Case 1.)
and therefore the described trajectory is of the same
species with that which was to be described; but that
right line *TR*, by which the angle *VRS* is bisected,
touches the trajectory in the point *R*, is certain from
the properties of the conic sections. *Q. E. F.*

Case 4. About the focus *S* (Pl. 7. *Fig*. 7.)
it is required to describe a trajectory *APB* that shall
touch a right line *TR*. and pass thro' any given point
*P* without the tangent, and shall be similar to the
figure *apb*, described with the principal axe *ab*, and
foci *s*, *h*. On the tangent *TR* let fill the perpendicular
*ST*; which produce to *V*, so that *TV* may be
equal to *ST*. And making the angles *hsq*, *shq* equal
to the angles *VSP*, *SVP*; about *q* as a centre, and with
an interval which shall be to *ab* as *SP* to *VS* describe
a circle cutting the figure *apb* in *p*: join *sp*, and
draw *SH*, such that it may be to *sh*, as *SP* is to
*sp*, and may make the angle *PSH* equal to the angle
*psh*; and the angle *VSH* equal to the angle
*psq*. Then with the foci *S*, *H*, and principal axe *AB*
equal to the distance *VH*, describe a conic section.
I say the thing is done. For if *sv* is drawn so that it
shall be to *sp* as *sh* is to *sq*, and shall make the angle
*vsp* equal to the angle *hsq*, and the angle *vsh* equal
to the angle *psq*, the triangles *svh*, *spq*, will
be similar, and therefore *vh* will be to *pq*, as *sh* is
to *sq*, that is, (because of the similar triangles *VSP*,
*hsq*) as *VS* is to *SP* or as *ab* to *pq*. Wherefore *vh* and *ab* are equal. But because of the similar triangles
*VSH*, *vsh*, *VH* is to *SH* as *vh* to *sh*;
that is, the axe of the conic section now described is
to the distance of its foci, as the axe *ab* to the distance
of the foci *s*, *h*; and therefore the figure now
described is similar to the figure *aph'. But, because*
the triangle *PSH* is similar to the triangle *psh*, this
figure passes through the point *P*, and because *VH* is
equal to its axis, and *VS* is perpendicularly bisected
by the right line *TR*, the said figure touches the
right line *TR*. *Q. E. F.*

Lemma XVI.

*From three given point: to draw to a fourth point that is not given three right lines whose differences shall be either given or none at all.*

Case 1.. Let the given points be *A*, *B*, *C*
(Pl. 8. *Fig.* 1.) and *Z* the fourth point which we are to
find; because of the given difference of the lines *AZ*,
*BZ*, the locus of the point *Z* will be an hyperbola.
whose foci are *A* and *B*, and whose principal axe is
the given difference. Let that axe be *MMN*. Taking
*PM* to *MA*, as *MN* is to *AB*, erect *PR*
perpendicular to *AB*, and let all *ZR* perpendicular
to *PR*; then, from the nature of the hyperbola,
*ZR* will be to *AZ* as *MN* is to *AB*.
And by the like argument, the locus of the point *Z*
will be another hyperbola, whose foci are *A*, *C*, and
whose principal axe is the difference between *AZ* and
*CZ*; and *QS* a perpendicular on *AC* may be drawm
to which (*QS*) if from any point *Z* of this hyperbola

a perpendicular *ZS* is let fall, this (*ZS*) shall
be to *AZ* as the difference between *AZ* and *CZ'*
is to *AC*. Wherefore the ratio's of *ZR* and *ZS*
to *AZ* are given, and consequently the ratio of *ZR*
to *ZS* one to the other; and therefore if the right
lines *RP* *SQ* meet in *T*, and *TZ* and *TA* are
drawn, the figure *TRZS* will be given in specie,
and the right line *TZ*, in which the point *Z* is somewhere
placed, will be given in position. There
will be given also the right line *TA*, and the
angle *ATZ*; and because the ratio's of *AZ* and
*TZ* to *ZS* are given, their ratio to each other is
given also; and thence will be given likewise the triangle
*ATZ* whose vertex is the point *Z*. *Q. E. I.*

Case 2. If two of the three lines, for example
*AZ* and *BZ*, are equal, draw the right line
*TZ* so as to bisect the right line *AB*; then find
the triangle *ATZ* as above. *Q. E. I.*

Case 3. If all the three are equal, the point *Z*
will be placed in the centre of a circle that passes thro'
the points *A*, *B*, *C*. *Q. E. I.*

This problematic lemma is likewise solved in *Apollonui'*s Book of Tactions restored by *Victa*.

Proposition XXI. Problem XIII.

*About a give focus to describe a trajectory,*
that shall pass through given points
and touch right lines given by position.

Let the focus *S*, (Pl. 8. *Fig.* 2.) the point *P*, and
the tangent *TR* be given, suppose that the other
focus *H* is to be found. On the tangent let fall the perpendicular
*ST*, which produce to T, so that *TY* may be equal to *ST*; and *YH* will be equal to the
principal axe. Join *SP*, *HP*, and *SP* will be the
difference between *HP* and the principal axe. After
this manner if more tangents *TR* are given, or more
points *P*, we shall always determine as many lines
*YH* or *PH*, drawn from the said points *T* or *P*, to
the focus *H*, which either shall be equal to the axes,
or differ from the axes by given lengths *SP*; and
therefore which shall either be equal among themselves,
or shall have given differences; from whence
(by the preceding lemma) that other focus *H* is
But having the foci and the length of the axe (which
is either *TH*; or, if the trajectory be an ellipsis, *PH*
+ *SP*, or *PH* - *SP* if it be an hyperbola) the trajectory
is given. *Q. E. I.*

Scholium.

When the trajectory is an hyperbola, I do not comprehend its conjugate by hyperbola under the name of this trajectory. For a body going on with a continued motion can never pass out of one hyperbola into its conjugate hyperbola.

The case when three points are given is more readily
solved thus. Let *B*, *C*, *D* (PL 8. *Fig.* 3.) be
the given points. join *BC*, *CD*, and produce them
to *E*, *F*; s as *EB* may be to *EC*, as *SB* to *SC*;
and *FC* to *FD*, as *SC* to *SD*. On *EF* drawn
and produced let fall the perpendiculars *SG*, *BH* and
in *GS* produced indefinitely take *GA* to *AS*,
and *Ga* to *aS*, as *HB* is to *BS*; then *A* will be
the vertex, and *Aa* the principal axe of the trajectory:
Which, according as *GA* is greater than, equal
to, or less than *AS,* will be either an ellipsis, a parabola or an hyperbola; the point *a* in the first case falling on the same side of the line *(GF* as the point *A*; in the second, going off to an infinite distance; in the third, falling on the other side of the line *GF*. For if on *GF*, the perpendiculars *Ct*, *DK* are let fall, *IC* will be to *HB* as *EC* to *EB*; that is, as *SC* to *SB*; and by permutation *IC* to *SC* as *HB* to *SB*, or as *GA* to *SA*. And, by the like argument, we may prove that *KD* is to *SD* in the same ratio. Wherefore the points *B*, *C*, *D* lie in a conic section described about the focus *S*, in such manner
that all the right lines drawn from the focus *S* to the several points of the section, and the perpendicular: let fall from the same points on the right line *GF* are in that given ratio.

That excellent geometer M. *De la Hire* has solved this problem much after the same way in his conics, prop. 25. lib. 8.