If the tangent to the curve x^{2/3} + y^{2/3} | vedclass

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(C) Given the curve is $x^{2/3} + y^{2/3} = a^{2/3}$ ...$(i)$Let a point on the curve be $P(a \cos^3 	heta, a \sin^3 	heta)$.Differentiating the cur

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$a$

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(C) Given the curve is $x^{2/3} + y^{2/3} = a^{2/3}$ ...$(i)$Let a point on the curve be $P(a \cos^3 	heta, a \sin^3 	heta)$.Differentiating the curve $(i)$ with respect to $x$,we get:$\frac{2}{3} x^{-1/3} + \frac{2}{3} y^{-1/3} \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\left(\frac{y}{x}ight)^{1/3}$.Substituting the point $P$,the slope of the tangent $m$ is:$m = -\left(\frac{a \sin^3 	heta}{a \cos^3 	heta}ight)^{1/3} = -\frac{\sin 	heta}{\cos 	heta} = -	an 	heta$.The equation of the tangent at point $P$ is:$y - a \sin^3 	heta = -	an 	heta (x - a \cos^3 	heta)$.$y \cos 	heta - a \sin^3 	heta \cos 	heta = -x \sin 	heta + a \cos^3 	heta \sin 	heta$.$x \sin 	heta + y \cos 	heta = a \sin 	heta \cos 	heta (\cos^2 	heta + \sin^2 	heta) = a \sin 	heta \cos 	heta$.For point $A$ (where $y=0$),$x \sin 	heta = a \sin 	heta \cos 	heta \Rightarrow x = a \cos 	heta$. So $A = (a \cos 	heta, 0)$.For point $B$ (where $x=0$),$y \cos 	heta = a \sin 	heta \cos 	heta \Rightarrow y = a \sin 	heta$. So $B = (0, a \sin 	heta)$.The distance $AB = \sqrt{(a \cos 	heta - 0)^2 + (0 - a \sin 	heta)^2} = \sqrt{a^2 \cos^2 	heta + a^2 \sin^2 	heta} = \sqrt{a^2} = a$.Thus,$AB = a$.

If the tangent to the curve $x^{2/3} + y^{2/3} = a^{2/3}$ meets the $X$-axis at $A$ and $Y$-axis at $B$,then $AB =$

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