At which points on the curve sqrt{x} + sqrt{y | vedclass

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(A) Differentiating the equation of the curve $\sqrt{x} + \sqrt{y} = \sqrt{a}$ with respect to $x$,we get $\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \

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$(a, 0), (0, a), (a/4, a/4)$

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(A) Differentiating the equation of the curve $\sqrt{x} + \sqrt{y} = \sqrt{a}$ with respect to $x$,we get $\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \frac{dy}{dx} = 0$,which simplifies to $\frac{dy}{dx} = -\sqrt{\frac{y}{x}}$.$(i)$ If the tangent is parallel to the $x$-axis,then $\frac{dy}{dx} = 0$. This implies $y = 0$. Substituting $y = 0$ into the curve equation,we get $\sqrt{x} = \sqrt{a}$,so $x = a$. The point is $(a, 0)$.$(ii)$ If the tangent is parallel to the $y$-axis,then $\frac{dy}{dx} = \infty$. This implies $x = 0$. Substituting $x = 0$ into the curve equation,we get $\sqrt{y} = \sqrt{a}$,so $y = a$. The point is $(0, a)$.$(iii)$ If the tangent makes equal angles with both axes,its slope $\frac{dy}{dx} = \pm 1$. Thus,$-\sqrt{\frac{y}{x}} = \pm 1$,which implies $\frac{y}{x} = 1$,or $y = x$. Substituting $y = x$ into the curve equation,we get $2\sqrt{x} = \sqrt{a}$,so $\sqrt{x} = \frac{\sqrt{a}}{2}$,which gives $x = \frac{a}{4}$. Since $y = x$,$y = \frac{a}{4}$. The point is $(\frac{a}{4}, \frac{a}{4})$.Thus,the points are $(a, 0), (0, a), (a/4, a/4)$.

At which points on the curve $\sqrt{x} + \sqrt{y} = \sqrt{a}$,is the tangent line: $(i)$ parallel to the $x$-axis? $(ii)$ parallel to the $y$-axis? $(iii)$ making equal angles with both axes?

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