Answer :
[tex]\frac{x^2}{9}+ \frac{y^2}{4}=1\\\\x=3Cos\theta,\ \ y=2Sin\theta\\Radius\ of\ curvature\ R=| \frac{[(x'(\theta))^2+(y'(\theta))^2]^{3/2}}{x'(\theta)*y"(\theta)-x"(\theta*y'(\theta)} |\\\\At\ A(3,0),\theta_A=0.,\ \ \ AtB(0,2),\ \ \theta_B=\pi/2\\\\x'(\theta)=-3Sin\theta,\ \ \ y'(\theta)=2Cos\theta\\x"(\theta)=-3Cos\theta,\ \ \ y"(\theta)=-2sin\theta\\\\At\ Point\ A(3,0),\ \ x'(0)=0,\ y'(0)=2,\ x"(0)=-3,\ y"(0)=0\\\\R_A=| \frac{[0^2+2^2]^{3/2}}{0-(-3)*2} |=8/6=4/3\ units\\[/tex]
[tex]At\ point\ B(0,2),\ x'(\pi/2)=-3,x"(\pi/2)=0\\.\ \ \ and,\ y'(\pi/2)=0,y"(\pi/2)=-2\\\\R_B=| \frac{[ -3^2+0^2 ]^{3/2}}{(-3*-2)-0*0} |=3^3/6=9/2\ units\\[/tex]
So the radii of curvature are found using the formula above as 4/3 and 9/2 respectively.
Osculating circle at A(3,0) is
(x-5/3)² + (y-0)² = (4/3)², as the center is (3-4/3, 0)
At B(0,2) is
(x-0)² + (y+5/2)² = (9/2)², as the center is (0, 2-9/2)
[tex]At\ point\ B(0,2),\ x'(\pi/2)=-3,x"(\pi/2)=0\\.\ \ \ and,\ y'(\pi/2)=0,y"(\pi/2)=-2\\\\R_B=| \frac{[ -3^2+0^2 ]^{3/2}}{(-3*-2)-0*0} |=3^3/6=9/2\ units\\[/tex]
So the radii of curvature are found using the formula above as 4/3 and 9/2 respectively.
Osculating circle at A(3,0) is
(x-5/3)² + (y-0)² = (4/3)², as the center is (3-4/3, 0)
At B(0,2) is
(x-0)² + (y+5/2)² = (9/2)², as the center is (0, 2-9/2)