Question

find the radius of curvature for the ellipse x²/a²+y²/b²=1​

Answer 1

$\large\underline{\sf{Solution-}}$

Given equation of ellipse is

$\rm :\longmapsto\:\dfrac{ {x}^{2} }{ {a}^{2} } + \dfrac{ {y}^{2} }{ {b}^{2} } = 1$

Now, the parametric coordinates of ellipse are

$\rm :\longmapsto\:x = a \: cost$

and

$\rm :\longmapsto\:y = b \: sint$

Now,

Consider,

$\rm :\longmapsto\:x = a \: cost$

On differentiating x, w . r. t. t, we get

$\rm :\longmapsto\:x' = - a \: sint$

Again, on differentiating both sides w. r. t. t, we get

$\rm :\longmapsto\:x'' = - a \: cost$

Now, Consider

$\rm :\longmapsto\:y \: = \: b \: sint$

On differentiating both sides w. r. t. t, we get

$\rm :\longmapsto\:y' \: = \: b \: cost$

On differentiating both sides w. r. t. t, we get

$\rm :\longmapsto\:y'' \: = - \: b \: sint$

Now, we know,

Radius of curvature of curve in parametric coordinates is given by

$\boxed{ \bf{ \: \rho \: = \: \dfrac{ {\bigg( {(x')}^{2} + {(y')}^{2} \bigg) }^{\dfrac{3}{2} } }{ |x'y'' - y'x''| } }}$

So, on substituting the values evaluated above, we get

$\rm :\longmapsto\: \rho = \dfrac{ {\bigg( {( - asint)}^{2} + {(bcost)}^{2} \bigg) }^{ \dfrac{3}{2} } }{ |ab {sin}^{2}t + ab {cos}^{2}t | }$

$\rm :\longmapsto\: \rho = \dfrac{ {\bigg( {a}^{2} {sin}^{2}t + {b}^{2} {cos}^{2}t \bigg) }^{ \dfrac{3}{2} } }{ |ab( {sin}^{2}t +{cos}^{2}t ) | }$

$\rm :\longmapsto\: \rho = \dfrac{ {\bigg( {a}^{2} {sin}^{2}t + {b}^{2} {cos}^{2}t \bigg) }^{ \dfrac{3}{2} } }{ |ab( 1 ) | }$

$\rm :\longmapsto\: \rho = \dfrac{ {\bigg( {a}^{2} {sin}^{2}t + {b}^{2} {cos}^{2}t \bigg) }^{ \dfrac{3}{2} } }{ |ab| }$

$\rm :\longmapsto\: \rho = \dfrac{ {\bigg( {a}^{2} {sin}^{2}t + {b}^{2} {cos}^{2}t \bigg) }^{ \dfrac{3}{2} } }{ ab }$

Additional Information :-

1. Radius of curvature for cartesian curve y = f(x) is

$\boxed{ \bf{ \: \rho = \dfrac{ {\bigg( 1 + {(y')}^{2} \bigg) }^{\dfrac{3}{2} } }{ |y''| } \: }}$