Find the equations for the osculating crircle of the parabol

Find the equations for the osculating crircle of the parabola y=0.5x^2 at the point (1, 1/2)

Solution

The circle (xx)²+(yy)² = r² meets y= ½x² when (xx)²+(½x²y)² = r²

Expanding gives f(x) = ¼x + x²(1y) – (2x)x + (x²+y²r²) = 0 … (i)

If circle is osculating then this equation has at least a triple root

For triple root f(x)=f’(x)=f’’(x)=0

f’(x) = x³ + 2x(1y) – 2x = 0 … (ii)

f’’(x) = 3x² + 2(1y) = 0 … (iii)

For point x=1 : (iii) gives y=5/2, (ii) gives x=1 and (i) gives r²=8

So at (1,½) osculating circle is (x+1)²+(y5/2)² = 8 x²+y²+2x5y¾ = 0