Let x=2t, y=frac{t^2}{3} be a conic. Let S be | vedclass

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(D) The given conic is $x=2t, y=\frac{t^2}{3}$. Squaring $x$,we get $x^2 = 4t^2$. Since $y = \frac{t^2}{3}$,$t^2 = 3y$. Thus,$x^2 = 4(3y) = 12y$. This

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$\frac{13}{18}$

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(D) The given conic is $x=2t, y=\frac{t^2}{3}$. Squaring $x$,we get $x^2 = 4t^2$. Since $y = \frac{t^2}{3}$,$t^2 = 3y$. Thus,$x^2 = 4(3y) = 12y$. This is a parabola with focus $S(0, 3)$.Given $SA \perp BA$,the product of the slopes of $SA$ and $BA$ is $-1$.Slope of $SA = \frac{\frac{t^2}{3} - 3}{2t - 0} = \frac{t^2 - 9}{6t}$.Slope of $BA = \frac{\frac{t^2}{3} - \alpha}{2t - 0} = \frac{t^2 - 3\alpha}{6t}$.Since $SA \perp BA$,$\left(\frac{t^2 - 9}{6t}\right) \cdot \left(\frac{t^2 - 3\alpha}{6t}\right) = -1$.$(t^2 - 9)(t^2 - 3\alpha) = -36t^2$.$t^4 - 3\alpha t^2 - 9t^2 + 27\alpha = -36t^2$.$27\alpha - 3\alpha t^2 = -36t^2 - t^4 + 9t^2 = -27t^2 - t^4$.$3\alpha(9 - t^2) = -(27t^2 + t^4)$.$3\alpha = \frac{27t^2 + t^4}{t^2 - 9}$.The centroid of $\Delta SAB$ with vertices $S(0, 3)$,$A(2t, \frac{t^2}{3})$,and $B(0, \alpha)$ has ordinate $k = \frac{3 + \frac{t^2}{3} + \alpha}{3} = 1 + \frac{t^2}{9} + \frac{\alpha}{3}$.Substituting $3\alpha = \frac{27t^2 + t^4}{t^2 - 9}$,we get $\frac{\alpha}{3} = \frac{27t^2 + t^4}{9(t^2 - 9)}$.$k = 1 + \frac{t^2}{9} + \frac{27t^2 + t^4}{9(t^2 - 9)} = \frac{9(t^2 - 9) + t^2(t^2 - 9) + 27t^2 + t^4}{9(t^2 - 9)} = \frac{9t^2 - 81 + t^4 - 9t^2 + 27t^2 + t^4}{9(t^2 - 9)} = \frac{2t^4 + 27t^2 - 81}{9(t^2 - 9)}$.As $t \rightarrow 1$,$k \rightarrow \frac{2(1)^4 + 27(1)^2 - 81}{9(1^2 - 9)} = \frac{2 + 27 - 81}{9(-8)} = \frac{-52}{-72} = \frac{13}{18}$.

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