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2. From E, with any radius, cut the produced long diameter

both ways in F 1, F 2, the foci, and join them to one angle of the rectangle, as C.

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3. Draw any straight line, as FG, equal to the sum of CF 1,

CF 2, and bisect it in H (Pr. 1). 4. Make EK and El equal to HF or HG; then KL is the

transverse aris. 5. From F 1, with half the transverse axis, as EK, as radius,

cut the line perpendicular to KL in M and N; then

MN is the conjugate axis.
6. Complete the required ellipse by Problem 80.

Problem 89.

To describe an oval by arcs of circles.

1. Draw any straight line AB, and describe a semicircle

CD equal in diameter to the proposed oval. 2. From C and D, with the radius of the semicircle, cut the

straight line in A and B. 3. From A and B, with radius BC, describe the arcs DF 4. From A or B, draw a straight line through the trans

and CE.

verse diameter, cutting it in G, and meeting the opposite arc in E or F.

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5. From G, with radius GE, describe arc EF, which will

complete the required oval CDFE. NOTE.—The oval may be made longer or shorter by increasing or diminishing the transverse diameter.

Problem 90.

To construct a parabola, its ordinate AB and abscissa BC being given.

6.

6B

1. Through C draw a line CD, parallel and equal to AB

(Pr. 8), and join AD. 2. Tide AB and AD into the same number of equal parts

(say six).

3. From C, draw lines to the points of division in AD.
4. From the points of division in AB, draw lines parallel to

BC, till each meets the .corresponding line from AD.
The points of intersection will be in the curve of the re-

quired parabola.
NOTE.—By repeating the process in the other half, the curve
of the required parabola will be completed.

Problem 91.

To construct a hyperbola, its diameter AB, abscissa BC, and ordinate CD being given.

A

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1. Through B draw a line BE, parallel and equal to CD

(Pr. 8), and join DE. 2. Divide DC and DE into the same number of equal parts

(say four). 3. From B, draw lines to the points of division in DE. 4. From A, draw lines to the points of division in DC.

The points of intersection will be in the curve of the re

quired hyperbola. NOTE.—By repeating the process in the other half, the curve of the required hyperbola will be completed.

.

SECTION VII.-INSCRIBED FIGURES.

DEFINITIONS.

1. Inscribed 'figures. Inscribed figures are either rectilineal or circular. (a.) A rectilineal figure is said to be inscribed in another

rectilineal figure, when all the angles of the inscribed figure are upon the sides of the figure in which it is inscribed. Ex. ABCD

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(6.) A rectilineal figure is said to be inscribed in a circle,

when all the angles of the inscribed figure are upon the circumference of the circle. Ex. ABCD

B

NOTE.-A circle is said to be inscribed in a rectilineal figure,

when the circumference of the circle touches each side of the figure. Ex. A–

A

2. A sector of a circle is a figure contained by two radii and the

intercepted arc. Ex. A

A

Problem 92.
To inscribe an equilateral triangle within a given circle A.

B

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1. Find the centre of the circle A (Pr. 45), and draw a dia

meter BC. 2. From C as centre, with radius CA, describe an arc

cutting the circumference in D and E. 3. Join DE, EB, BD; then DEB is the required equilateral

triangle inscribed within the given circle A.

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