hyperbolae वाक्य

Pierre de Fermat also computed these areas ( except for the exceptional case of & minus; 1 ) by an algebraic trick he computed the quadrature of the higher hyperbolae via dividing the line into equal intervals, and then computed the quadrature of the higher parabolae by using a division into " unequal " intervals, presumably by inverting the divisions he used for hyperbolae.
Maybe it should also include " rotundae " as a British plural-- I don't know-- but American English tends to avoid "-ae " plurals, preferring e . g . " hyperbolas " over " hyperbolae " etc . ( although AmEng uses " alumnae " as the plural for " alumna " ) . talk ) 14 : 34, 20 December 2012 ( UTC)