Ex. 2. What is the transverse of the hyperbola whose conjugate is 36; the less absciss being 20, and its ordinate 24?Ans. 60.PROBLEM XIII. To find the Length of an Hyperbolic Curve.1. To 21 times the square of the conjugate, add 9 times the square of the transverse; and to the same 21 times the square of the conjugate, add 19 times the square of the transverse; and multiply each sum by the absciss.2. To each of these two products add 15 times the product of the transverse and square of the conjugate.3. Then as the less sum is to the greater, so is the double ordinate to the length of the curve, nearly.EXAMPLES.1. Required the length of the curve AVC to the ab∣sciss VB 20 and ordinate AB 24; the two axes being 60 and 36.0