The sum of the intercepts on the axes cut off | vedclass

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Question

(C) Given the curve $x^{2/3} + y^{2/3} = 2$. Differentiating with respect to $x$,we get $\frac{2}{3}x^{-1/3} + \frac{2}{3}y^{-1/3} \frac{dy}{dx} = 0$.

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) Given the curve $x^{2/3} + y^{2/3} = 2$. Differentiating with respect to $x$,we get $\frac{2}{3}x^{-1/3} + \frac{2}{3}y^{-1/3} \frac{dy}{dx} = 0$. At the point $(1, 1)$,the slope of the tangent is $\frac{dy}{dx} = -\frac{x^{-1/3}}{y^{-1/3}} = -\frac{1^{-1/3}}{1^{-1/3}} = -1$. The equation of the tangent at $(1, 1)$ is $y - 1 = -1(x - 1)$,which simplifies to $x + y = 2$. The intercept form of the line is $\frac{x}{2} + \frac{y}{2} = 1$. The $x$-intercept is $2$ and the $y$-intercept is $2$. The sum of the intercepts is $2 + 2 = 4$.

The sum of the intercepts on the axes cut off by the tangent to the curve $x^{2/3} + y^{2/3} = 2$ at the point $(1, 1)$ is:

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