If the tangent drawn at the point (alpha, bet | vedclass

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(C) The equation of the curve is $x^{2/3} + y^{2/3} = 4$. Differentiating with respect to $x$,we get $\frac{2}{3}x^{-1/3} + \frac{2}{3}y^{-1/3} \frac{

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$28$

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(C) The equation of the curve is $x^{2/3} + y^{2/3} = 4$. Differentiating with respect to $x$,we get $\frac{2}{3}x^{-1/3} + \frac{2}{3}y^{-1/3} \frac{dy}{dx} = 0$. So,$\frac{dy}{dx} = -\frac{x^{-1/3}}{y^{-1/3}} = -\left(\frac{y}{x}\right)^{1/3}$. At the point $(\alpha, \beta)$,the slope of the tangent is $m = -(\beta/\alpha)^{1/3}$. The given line is $\sqrt{3}x + y = 1$,which can be written as $y = -\sqrt{3}x + 1$. Its slope is $-\sqrt{3}$. Since the tangent is parallel to the line,$-(\beta/\alpha)^{1/3} = -\sqrt{3}$,which implies $(\beta/\alpha)^{1/3} = \sqrt{3}$. Cubing both sides,we get $\beta/\alpha = 3\sqrt{3}$,so $\beta = 3\sqrt{3}\alpha$. Since $(\alpha, \beta)$ lies on the curve,$\alpha^{2/3} + (3\sqrt{3}\alpha)^{2/3} = 4$. $\alpha^{2/3} + (3^{3/2}\alpha)^{2/3} = 4 \implies \alpha^{2/3} + 3\alpha^{2/3} = 4$. $4\alpha^{2/3} = 4 \implies \alpha^{2/3} = 1 \implies \alpha^2 = 1$. Then $\beta^2 = (3\sqrt{3}\alpha)^2 = 27\alpha^2 = 27(1) = 27$. Therefore,$\alpha^2 + \beta^2 = 1 + 27 = 28$.

If the tangent drawn at the point $(\alpha, \beta)$ on the curve $x^{2/3} + y^{2/3} = 4$ is parallel to the line $\sqrt{3}x + y = 1$,then $\alpha^2 + \beta^2 =$

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