80 



PROB. To find the Length of the Arc of a Parabola. 



Let F be the focus and A the vertex of a 

 parabola AT : draw AK perpendicular to AF, 

 and let a tangent at the point T meet it in P ; 

 take p indefinitely near to P, and let Fp in- 

 tersect PT in q : then since FP and Fp are 

 perpendicular to the tangents PT and pf, it 

 follows (by the preceding lemma) that P(jr is 

 ultimately equal to the increment of the dif- 

 ference between the arc AT and the tangent 

 TP. With the centre F and semiaxis FA 

 describe the equilateral hyperbola GAH, and 



bisect the angles AFP, AFp by the straight lines FL, F/. Then 

 (since the square of the radius vector of an equilateral hyberbola 

 is inversely as the cosine of twice the angle which it makes with 

 the axis) the square of LF will be equal to AF x PF ; and because the 

 angle LF/ is half the angle PF^f, therefore the area LF/ will be equal 

 to one-fourth of the rectangle under AF and P^. 



Hence the hyperbolic sector AFL is equal to one-fourth of the 

 rectangle under AF and the difterence between the parabolic arc AT 

 and the tangent TP. 



Lemma 2. If on either axis of an ellipse 

 a semicircle be described, of which CD and 

 CH are two radii at right angles to each 

 other, and if DN and HM be drawn per- 

 pendicular to the axis oA, and meeting the 

 ellipse in E and L ; then CE and CL will be conjugate semi- 

 diameters. 



a N 



