The sum of the squares of the intercepts on t | vedclass

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(C) Given the curve $x^{2/3} + y^{2/3} = a^{2/3}$.Let the point of tangency be $(x_1, y_1)$.Differentiating with respect to $x$:$\frac{2}{3}x^{-1/3} +

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$a^2$

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(C) Given the curve $x^{2/3} + y^{2/3} = a^{2/3}$.Let the point of tangency be $(x_1, y_1)$.Differentiating with respect to $x$:$\frac{2}{3}x^{-1/3} + \frac{2}{3}y^{-1/3} \frac{dy}{dx} = 0$$\frac{dy}{dx} = -\left(\frac{y_1}{x_1}\right)^{1/3}$.The equation of the tangent at $(x_1, y_1)$ is:$y - y_1 = -\left(\frac{y_1}{x_1}\right)^{1/3}(x - x_1)$$\frac{y}{y_1^{1/3}} + \frac{x}{x_1^{1/3}} = \frac{y_1}{y_1^{1/3}} + \frac{x_1}{x_1^{1/3}} = y_1^{2/3} + x_1^{2/3} = a^{2/3}$.Dividing by $a^{2/3}$:$\frac{x}{a^{2/3}x_1^{1/3}} + \frac{y}{a^{2/3}y_1^{1/3}} = 1$.The $x$-intercept is $OA = a^{2/3}x_1^{1/3}$ and the $y$-intercept is $OB = a^{2/3}y_1^{1/3}$.The sum of the squares of the intercepts is:$OA^2 + OB^2 = (a^{2/3}x_1^{1/3})^2 + (a^{2/3}y_1^{1/3})^2$$= a^{4/3}x_1^{2/3} + a^{4/3}y_1^{2/3}$$= a^{4/3}(x_1^{2/3} + y_1^{2/3})$$= a^{4/3}(a^{2/3}) = a^2$.

The sum of the squares of the intercepts on the axes made by a tangent at any point on the curve $x^{2/3} + y^{2/3} = a^{2/3}$ is -

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