Question
Show that the numerical value of the radius of curvature at the point (x1, y1) on the parabola y^2=4ax is [2(a+x1)^3/2)]/a^1/2. If c is the centre of the curvature at the origin O and S is the point (a, 0), show that OC=2(OS)
plz show me working plz
plz show me working plz
Answers
Steve
r = (1+y'^2)^(3/2) / |y"|
y^2 = 4ax, so
y = 2√(ax)
y' = a/√(ax)
y" = -a^2/(2(ax)^(3/2))
= -a/(2x√(ax))
so just plug it in:
(1+y'^2) = 1+a/x
r = ((a+x)/x)^3/2 * 2x√(ax)/a
= 2(a+x)^(3/2) / √a
y^2 = 4ax, so
y = 2√(ax)
y' = a/√(ax)
y" = -a^2/(2(ax)^(3/2))
= -a/(2x√(ax))
so just plug it in:
(1+y'^2) = 1+a/x
r = ((a+x)/x)^3/2 * 2x√(ax)/a
= 2(a+x)^(3/2) / √a