PL0BLEM XXIV. To find the area of a hyperbola. RULE.* 1. To the product of the transverse and abscissa add of the square of the abscissa, and multiply the square root of the sum by 21. 2. Add 4 times the square root of the product of the transverse and abscissa to the last found product, and divide the sum by 75. * Demonstration. Let, transverse diameter, c= con x jugate, x= abscissa, y= ordinate, and z=. Then it is 21+x =c= conjugate axe, by the nature of the hyper And this, thrown into a series, will very nearly agree with the former, which shows the rule to be an approximation. Q. E. I. RULE 2. If 2Y, 2y= bases, V, and v their distances from the centre and the other letters as before, then will 3. Divide 4 times the product of the conjugate and abscissa by the transverse; and this last quotient, multipliedby the former, will give the area required, nearly. EXAMPLES. 1. In the hyperbola AVC, the transverse is 30, the conjugate is 18, and the abscissa or height is 10; what is the area? |