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2014-12-03T01:02:05+05:30

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\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\\

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\\

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)

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