6. Diameters which coincide with the diagonals of the parallelogram on the axes are equal and conjugate.

7. Also diameters which coincide with the diagonals of any parallelogram formed by tangents at the extremities of conjugate diameters are conjugate.

8. The angular points of these parallelograms lie on an ellipse similar to the given ellipse and of twice its area.

9. If from the extremities of the axes of an ellipse four parallel lines be drawn, the points in which they cut the curve are the extremities of conjugate diameters.

10. If from the extremity of each of two semi-diameters ordinates be drawn to the other, the two triangles so formed will be equal in area.

11. Also if tangents be drawn from the extremity of each to meet the other produced, the two triangles so formed will be equal in area.

12. If on the semi-axes a parallelogram be described, and about it an ellipse similar and similarly situated to the given ellipse be constructed, any chord PQR of the larger ellipse, drawn from the further extremity of the diameter CD of the smaller ellipse, is bisected by the smaller ellipse at Q.

13. If TP, TQ be tangents to an ellipse, and PCP' be the diameter through P, then P'Q is parallel to CT.

CHAPTER VII.

THE PARABOLA AND HYPERBOLA.

51. As already stated, most of the properties of the hyperbola are the same as the corresponding properties of the ellipse, and proved by the same process, e being greater than 1. There are, however, some properties both of it and of the parabola which may be conveniently developed by a process more analogous to that of the Cartesian geometry. This process we shall develope presently. In the meantime we proceed to give a brief outline of the application to the parabola of the method employed in the preceding Chapter for the ellipse.

If p' be another point in the parabola, p'-p=ẞ, the limit to which ẞ approaches is a vector along the tangent; so that if жß = π-ρ, π is the vector to a point in the tangent; this gives

hence the equation of the tangent becomes

Sπ (Þp + a ̄1) + Sa ̄1p = 1 ..............

а

From (2) it is evident that

Sapp=0

.........

..(5).

..(6), .

so that op is a vector perpendicular to the axis.

From the same equation

reads by (6), 'vector along NP=SP - vector along AN', which requires that

For the subtangent AT, put xa for π in (5), and there results

whence also the tangent bisects the angle SPQ; and SQ is perpendicular to and bisected by the tangent.

or (10) AY is parallel to, and equal to half of NP.

53. If now we substitute Cartesian co-ordinates, making -2ai;

p = xi + yj, a =

The locus of the middle points of parallel chords is thus found.

Let the chords be parallel to ß, π the vector of the middle point of one of the chords,

then

and

-1

S (π+xẞ) & (π+xß) + 2Sα ̄1 (π + xẞ) = 1; which, since the term involving x must disappear, gives

a straight line perpendicular to $ẞ, i.e. (6) parallel to the axis.

which shews (8) that the chords are perpendicular to the normal vector at the point where p =π, i.e. at the point where the locus of the chords meets the curve: in other words, the chords are parallel to the tangent at the extremity of the diameter which bisects them.

54. EXAMPLES.

Ex. 1. If two chords be drawn always parallel to given lines, and cut one another at points either within or without the parabola,