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Class-12-science»Maths

Find the radius of curvature at (a,a/2) on the curve 4ay² = (2a - x )^3.

$y^{2} = \frac{{(2 a - x)}^{3}}{4 a} y = \sqrt{\frac{{(2 a - x)}^{3}}{4 a}} = {[\frac{{(2 a - x)}^{3}}{4 a}]}^{\frac{1}{2}} y_{1} = \frac{1}{2} {[\frac{{(2 a - x)}^{3}}{4 a}]}^{\frac{1}{2} - 1} * \frac{1}{4 a} * 3 {(2 a - x)}^{2} * (- 1) y_{1} = \frac{- 3 {(2 a - x)}^{2}}{8 a} {[\frac{{(2 a - x)}^{3}}{4 a}]}^{- \frac{1}{2}} y_{1} = \frac{- 3 {(2 a - x)}^{2}}{8 a {[\frac{{(2 a - x)}^{3}}{4 a}]}^{\frac{1}{2}}} {(y_{1})}_{(a, \frac{a}{2})} = \frac{- 3 {(2 a - a)}^{2}}{8 a {[\frac{{(2 a - a)}^{3}}{4 a}]}^{\frac{1}{2}}} = \frac{- 3 a^{2}}{8 a {[\frac{a^{3}}{4 a}]}^{\frac{1}{2}}} = \frac{- 3 a}{8 {[\frac{a^{2}}{4}]}^{\frac{1}{2}}} = \frac{- 3 a}{8 (\frac{a}{2})} = \frac{- 3}{4} y_{2} = \frac{- 3 {(2 a - x)}^{2}}{8 a} * \frac{- 1}{2} {[\frac{{(2 a - x)}^{3}}{4 a}]}^{- \frac{1}{2} - 1} * 3 {(2 a - x)}^{2} * (- 1) + {[\frac{{(2 a - x)}^{3}}{4 a}]}^{- \frac{1}{2}} * \frac{- 3}{8 a} * 2 (2 a - x) * (- 1) y_{2} = \frac{- 9 {(2 a - x)}^{4}}{16 a} {[\frac{{(2 a - x)}^{3}}{4 a}]}^{- \frac{3}{2}} + \frac{3 (2 a - x)}{4 a} {[\frac{{(2 a - x)}^{3}}{4 a}]}^{- \frac{1}{2}} {(y_{2})}_{(a, \frac{a}{2})} = \frac{- 9 {(2 a - a)}^{4}}{16 a} {[\frac{{(2 a - a)}^{3}}{4 a}]}^{- \frac{3}{2}} + \frac{3 (2 a - a)}{4 a} {[\frac{{(2 a - a)}^{3}}{4 a}]}^{- \frac{1}{2}} = \frac{- 9 a^{4}}{16 a} {[\frac{a^{3}}{4 a}]}^{- \frac{3}{2}} + \frac{3 a}{4 a} {[\frac{a^{3}}{4 a}]}^{- \frac{1}{2}} = \frac{- 9 a^{3}}{16} {[\frac{a^{2}}{4}]}^{- \frac{3}{2}} + \frac{3}{4} {[\frac{a^{2}}{4}]}^{- \frac{1}{2}} = \frac{- 9 a^{3}}{16} * \frac{8}{a^{3}} + \frac{3}{4} * \frac{2}{a} = \frac{- 9}{2} + \frac{3}{2 a} ρ_{(a, \frac{a}{2})} = \frac{{(1 + {y_{1}}^{2})}^{\frac{3}{2}}}{y_{2}} = \frac{{(1 + \frac{9}{16})}^{\frac{3}{2}}}{\frac{- 9}{2} + \frac{3}{2 a}} = \frac{{(\frac{25}{16})}^{\frac{3}{2}}}{\frac{- 9}{2} + \frac{3}{2 a}} = \frac{{(\frac{5}{4})}^{3}}{\frac{- 9}{2} + \frac{3}{2 a}} = \frac{\frac{125}{64}}{\frac{3 (1 - 3 a)}{2 a}} = \frac{125}{64} * \frac{2 a}{3 - 9 a} = \frac{125 a}{32 (3 - 9 a)}$

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