If the tangent to the curve y=x^{3} at the po | vedclass

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Question

(A) The slope of the tangent to the curve $y=x^{3}$ at $P(t, t^{3})$ is given by $\frac{dy}{dx} = 3x^{2}$.At $x=t$,the slope is $3t^{2}$.The equation

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$-2t^{3}$

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(A) The slope of the tangent to the curve $y=x^{3}$ at $P(t, t^{3})$ is given by $\frac{dy}{dx} = 3x^{2}$.At $x=t$,the slope is $3t^{2}$.The equation of the tangent at $P(t, t^{3})$ is $y - t^{3} = 3t^{2}(x - t)$.To find the intersection with $y=x^{3}$,substitute $y=x^{3}$ into the tangent equation:$x^{3} - t^{3} = 3t^{2}(x - t)$.$(x - t)(x^{2} + xt + t^{2}) = 3t^{2}(x - t)$.Since $x 
eq t$ for point $Q$,we have $x^{2} + xt + t^{2} = 3t^{2}$,which simplifies to $x^{2} + xt - 2t^{2} = 0$.Factoring gives $(x - t)(x + 2t) = 0$,so $x = -2t$.The coordinates of $Q$ are $(-2t, (-2t)^{3}) = (-2t, -8t^{3})$.The point dividing $PQ$ in the ratio $1:2$ has an ordinate given by the section formula:$y = \frac{1 	imes (-8t^{3}) + 2 	imes (t^{3})}{1 + 2} = \frac{-8t^{3} + 2t^{3}}{3} = \frac{-6t^{3}}{3} = -2t^{3}$.

If the tangent to the curve $y=x^{3}$ at the point $P(t, t^{3})$ meets the curve again at $Q$,then the ordinate of the point which divides $PQ$ internally in the ratio $1:2$ is

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