Cor. 2. SP and HP make equal angles with every tangent.

Cor. 3. Since HPK, the exterior angle of the triangle SPH, is bisected by the straight line Tt, cutting the base SH produced in T

Tangents drawn at the vertices of any diameter are parallel.

Let Tt, Ww, be tangents at P,p, the vertices of the diameter PCp.

Join S,P; P,H; S.p; p, H;

Then, by Prop. II, SH is a parallelogram, and since the opposite angles of parallelograms are equal, .. ang. SPH angle SpH supplement of ang. SPH supplement of ang. SpH

or,

=

ang. SPT+ang. HPt=ang. SpW+ ang. Hpw

But ang. SPT=ang. HPt

W

T

w

And ang. SpW=ang. Hp by Prop. IV. Cor. 2.

Hence, these four angles are all equal,

.. ang. SPT=ang. Hpw.

And since SP is parallel to Hp,

=

ang. SPp ang. PpH,

H

.. whole ang. TPp = whole ang. upP, and they are alternate angles,

.. Tt is parallel to Ww.

Cor. Hence, if tangents be drawn at the vertices of any two diameters, they will form a parallelogram circumscribing he ellipse.

PROPOSITION VI. THEOREM.

If straight lines be drawn from the foci to a vertex of any di ameter, the distance from the vertex to the insertion of the conjugate diameter, with either focal distance, is equal to the semi axis, major.

That is, if Dd be a diameter conjugate to Pp, cutting SP in E, and HP in e,

Perpendiculars, from the foci upon the tangent at any point intersect the tangent in the circumference of a circle, whose diameter is the major axis.

From S let fall SY perpendicular on

Tt a tangent.

Join S, P; H, P; produce HP to meet SY produced in K.

Join CY;

Then, since angle SPY=angle KPY

(Prop. IV.) and the angles at Y are right angles, and PY common to the A two triangles, SPY, KPY.

... SP=PK

K

H

And SY-YK.

And, since SY=YK, and HC=CS, CY cuts the sides of the triangle HSK proportionally,

.. CY is parallel to HK.

Also, since CY is parallel to HK, SYYK, HC=CS, .. CY=HK

Hence, a circle inscribed with Centre C and radius CA will pass through Y.

And in like manner, if HZ be drawn perpendicular to Tt, it may be proved that the same circle will pass through Z also.

PROPOSITION VIII. THEOREM.

The rectangle, contained by the perpendiculars, from the foci upon the tangent at any point, is equal to the square of the semi-axis, minor.

Let Tt be a tangent at any point P. On Aa describe a circle cutting Tt T

Since yYZ is a right angle, the seg

ment in which it lies is a semicircle, and Z, y, are the extremities of a diameter.

..yCZ is a straight line and a diameter.

Hence the triangles CSy, HCZ, are in every respect equal.

... Sy=HZ

.. SY. HZ=YS. Sy=AS. SA=BC2 Prop. III. Cor. I.

PROPOSITION IX. THEOREM.

Perpendiculars let fall from the foci upon the tangent at any point are to each other as the focal distance of the point of

contact.

That is,

SY: HZ:: SP: HP.

For the triangles SPY, HPZ, are

manifestly similar,

.. SY: HZ :: SP: HP.

TY

If a tangent be applied at any point, and from the same point an ordinate to the axis be drawn, the semi-axis major is a mean proportional between the distance from the centre to the intersection of the ordinate with the axis, and the distance from the centre to the intersection of the tangent with the axis.

K

That is, CT: CA:: CA: CM.

a

MH.

A

Since the exterior angle HPK is bisected by Tt, Prop. IV. ST HT SP: HP. (B. IV. Prop. XVI. El. Geom.) ST+HT: ST-HT ; : SP+HP : SP-HP

a

But since PM is drawn from the vertex of the triangle SPH perpendicular on base SH,

.. SM+HM : SP-HP :: SP+HP : SM-HM

or,

or,

(2.)

SH: SP-HP :: 2 AC : 2 CM Comparing this with the proportion marked (1,) we have, 2 CT: 2 AC :: 2 AC: 2 CM

CT:

AC:: CA :

СМ.

PROPOSITION XI. THEOREM.

If a circle be described on the major axis of an ellipse, and if any ordinate to this axis be produced to meet the circle, tangents drawn to the ellipse and circle, at points in which they are intersected by the ordinate, will cut the major axis in the same point.

Let AQa, be a circle described on Aa.

Take any point P in the ellipse, draw PM perpendicular to Aa, and produce MP to meet the circle in Q, join C, Q.

a

Draw PT a tangent to the ellipse at P cutting CA produced in T. Join TQ.

S

MH A

Then QT is a tangent to the circle at Q.

For if TQ be not a tangent, draw QT' a tangent to Q cutting CA in T'.

Then CQT' is a right angle.

.. Since QM is drawn from the right angle CQT' perpendicular on the hypothenuse.

.. CT': CQ:: CQ: CM. (Prop. XVII, Cor. 2. B. IV. E. G.) or, CT': CA: CA: CM, CQ=CA.

But, by the last proposition,

·

CT CA: CA: CM,

CT=CT',

which is absurd, therefore QT' is not a tangent at Q; in the same manner it may be proved that no line but QT can be a tangent at Q,

.. &c.