Answer to Question #145439 in Differential Geometry | Topology for Dolly

Question

Find the centre of curvature of the curve
x=3t y=t^2-6 at (x,y)

Answer 1

Here $9(y+6)=x^2.$ Hence $y=\frac{x^2}{9}-6.$ Hence $y_1=\frac{2x}{9}, y_2=\frac{2}{9}.$

Hence center of curvature is ($x-\frac{y_1[1+y_1^2]}{y_2}, y+\frac{1+y_1^2}{y_2})$ . Hence required coordinate is

($x-\frac{2x/9}{2/9}(1+\frac{4x^2}{81})$, $y+\frac{1+\frac{4x^2}{81}}{2/9}$ )= ($-4x^3/81, y+9/2+\frac{2x^2}{9})$=($-4x^3/81, 3y+16.5$ ) .