If the tangent drawn to the curve y=x^3 at a | vedclass

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(C) Given the curve $y=x^3$.The slope of the tangent at point $(\alpha, \beta)$ is $\frac{dy}{dx} = 3x^2$. At $x=\alpha$,the slope is $3\alpha^2$.The

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-$8$

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(C) Given the curve $y=x^3$.The slope of the tangent at point $(\alpha, \beta)$ is $\frac{dy}{dx} = 3x^2$. At $x=\alpha$,the slope is $3\alpha^2$.The equation of the tangent line at $(\alpha, \beta)$ is $(y-\beta) = 3\alpha^2(x-\alpha)$.Since $(\alpha_1, \beta_1)$ lies on the curve,$\beta_1 = \alpha_1^3$ and $\beta = \alpha^3$.Substituting these into the tangent equation: $\alpha_1^3 - \alpha^3 = 3\alpha^2(\alpha_1 - \alpha)$.Dividing by $(\alpha_1 - \alpha)$ (assuming $\alpha_1 
eq \alpha$): $\alpha_1^2 + \alpha_1\alpha + \alpha^2 = 3\alpha^2$.$\alpha_1^2 + \alpha_1\alpha - 2\alpha^2 = 0$.Factoring the quadratic: $(\alpha_1 - \alpha)(\alpha_1 + 2\alpha) = 0$.Since $\alpha_1 
eq \alpha$,we have $\alpha_1 = -2\alpha$.Therefore,$\frac{\beta_1}{\beta} = \frac{\alpha_1^3}{\alpha^3} = \left(\frac{\alpha_1}{\alpha}ight)^3 = (-2)^3 = -8$.

If the tangent drawn to the curve $y=x^3$ at a point $(\alpha, \beta)$ cuts the curve again at another point $(\alpha_1, \beta_1)$,then $\frac{\beta_1}{\beta}=$

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