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hyperbola at the point f; then an ordinate Mƒ will, if produced, meet the horizontal h 2 in the point h, which will be a point on the required hyperbola.

130. To draw any hyperbola by means of a scale of parts.-Let A X (fig. 110) be the axis, A the vertex of the hyperbola. Measure the abscissæ by any scale from the vertex A, according to the numbers marked along the axis A X, or shown in the following table. Then by measuring

the ordinates on the same scale, we obtain points upon the rectangular hyperbola. If we use a different scale for the

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ordinates to that used for the abscissæ, we obtain points upon any other hyperbola we may require.

Table of Ordinates of a Rectangular Hyperbola to the Corresponding Values of Abscissa.

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By taking any number of points on the vertical CB, we can in the same manner find as many points as we please upon the required curve.

132. To find points on the contour of any hyperbola by the ruler only.-Let CA (fig. 112) be the axis, C the centre, and A the vertex of the required curve. Take CB equal to CA. Suppose the curve is required to pass through a given point D; draw DM at right angles to the axis, and draw A E parallel to MD, ED parallel to AM. Divide

ED and MD into the same number of equal parts by the points 1, 2, &c. Draw the lines B 1, B 2, &c., intersecting

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the lines A 1, A 2, &c., in the points a, b, &c., which will be upon the contour of the required curve.

133. To draw the tangent at any given point of a hyperbola.-Let HCS (fig. 106) be the axis, H and S the foci, P the given point on the curve. Draw SP, HP, and produce HP to K, making PK equal to PS. Join SK, then PT drawn parallel to S K is the tangent at the point P.

The tangent can also be drawn by bisecting the angle HPS by the line PT.

Another way is to find CT a third proportional (15) to CM and CA, PM being the ordinate at P perpendicular to the axis. Then TP will be the tangent at P.

134. To draw a tangent to a hyperbola from any point upon the axis outside the curve.-Let T (fig. 106) be the given point on the axis, C the centre, A the vertex. Find

CM a third proportional (15) to CT and CA; draw the ordinate M P at right angles to the axis, and meeting the curve at P. Then the line TP will be a tangent to the curve at the point P.

135. To draw a normal at any given point on a hyperbola.-Let P (fig. 106) be the given point on the hyperbola whose foci are H and S. Draw HP, SP, and take PZ equal to PS; join ZS, and draw Pn parallel to ZS; then Pn is the normal at P, and is perpendicular to the tangent PT.

Another way is to produce HP to K, making PK equal PS, joining KS, and bisecting KS at n, then Pn is the normal at P.

It can also be drawn by bisecting the angle SP K, by the line Pn.

Whenever a hyperbola is used as the outline of an arch, the joints of the voussoirs must be in the normals to the curve. 136. To find the foci of a given hyperbola.-Let HA (fig. 112) be the axis of the hyperbola, C the centre, A the vertex, D a given point on the curve, MD an ordinate perpendicular to A M. From C as a centre, with CA as radius, describe a circle. Find CT, a third proportional (15) to CM and CA, and draw the tangent TD, cutting the circle at the point I. Draw IS at right angles to TD, and cutting the axis at S; then S is a focus of the hyperbola. To find the other focus, take C H equal C S, and H is the further focus of the curve.

137. To approximate to the contour of a hyperbola by means of arcs of circles.-Let A C (fig. 112) be the axis of the hyperbola, A the vertex, S the focus, a, b, c, &c., points found upon the curve by the previous method (132).

Find the normals (135) at a, b, &c., by taking a d equal

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