4. The base of a cone is the circle which is described by the revolving leg of the triangle.

5. If a cone be cut through the vertex, by a plane which also cuts the base, the section will be a triangle.

6. If a cone be cut into two parts, by a plane parallel to the base, the section will be a circle.

7. If a cone be cut by a plane which passes through its two slant sides in an oblique direction, the section will be an ellipsis.

8. The longest straight line that can be drawn in an elipsis is called the transverse axis; and a line drawn perpendicular to the transverse axis, passing through the centre of the ellipse, and terminated both ways by the circumference, is called the conjugate axis.

Ө

9. An ordinate is a right line drawn from any point of the curve, perpendicular to either of the diameters.

10. An abscissa is that part of the diameter which is contained between the vertex and the ordinate.

11. If a cone be cut by a plane, which is parallel to either of its slant sides, the section will be a parabola.

12. The axis of a parabola is a right line drawn from the vertex, so as to divide the figure into two equal parts.

13. The ordinate is a right line drawn from any point in the curve perpendicular to the axis.

14. The abscissa is that part of the axis which is contained between the vertex and the ordinate.

15. *If a cone be cut into two parts, by a plane, which, being continued, would meet the opposite cone, the section is called an hyperbola.

16. The transverse diameter, or axis of an hyperbola, is that part of the axis intercepted between the two opposite

cones.

17. The conjugate diameter is a line drawn through the centre perpendicular to the transverse.

18. An ordinate is a line drawn from any point in the curve perpendicular to the axis; and the abscissa is the distance intercepted between that ordinate and the vertex.

PROBLEM I.

To describe an ellipsis, the transverse and conjugate diameters being given.

* The two opposite cones, in this definition, are supposed to be generated together, by the revolution of the same line.

All the figures which can possibly be formed by the cutting of a cone, are mentioned in these definitions, and are the five following

Construction.* 1. Draw the transverse and conjugate diameters, AB, CD, bisecting each other perpendicularly in the centre o.

2. With the radius Ao, and centre C, describe an arc cutting AB in Ff; and these two points will be the foci of the ellipse.

3. Take any number of points nn, &c. in the transverse diameter AB, and with the radii An, nB, and centres Ff, describe arcs intersecting each other in s, s, &c.

4. Through the points 8, s, &c. draw the curve A&CBD, and it will be the circumference of the ellipse required.

PROBLEM II.

In an ellipsis, any three of the four following terms being given, viz. the transverse and conjugate diameters, an ordinate and its abscissa, to find the fourth.

CASE I.

When the transverse, conjugate, and abscissa are known, to find the ordinate.

ones: viz. a triangle, a circle, an ellipsis, a parabola, and an hyperbola; but the last three only are usually called the conic sections.

* It is a known property of the ellipse, that the sum of two lines drawn from the foci, to meet in any point in the curve, is equal to the transverse diameter, and from this the truth of the construction is evident.

From the same principle is also derived the following method of describing an ellipse, by means of a string and two pins.

Having found the foci F, f, as before, take a thread of the length of the transverse diameter, and fasten its ends with two pins in the points F,f; then stretch the thread F sf to its greatest extent, and it will reach to the point s in the curve; and by moving a pencil round within the thread, keeping it always stretched, it will trace out the curve required.

RULE.*

As the transverse diameter is to the conjugate,

So is the square root of the rectangle of the two abscissas,

To the ordinate which divides them.

EXAMPLES.

1. In the ellipsis ADBC, the transverse diameter AB is 120, the conjugate diameter CD is 40, and the abscissa BF 24: what is the length of the ordinate EF?

40

Here 120 (AB): 40 (CD) :: √96×24 (AF× FB) : √96×24= √2304=1×48=16=EF the ordinate

120

required.

2. If the transverse diameter be 35, the conjugate 25, and the abscissa 28; what is the ordinate?

Ans. 10.

CASE II.

When the transverse, conjugate, and ordinate are known, to find the abscissa.

* Let t the transverse diameter, c=conjugate, x=any abscissa, and y=ordinate. Then will the general equation expressing the property of the ellipse, be t2: c2:: x × (t−x) : y; and from this the four rules here given are deduced,