Find the coordinates of the point on the curv | vedclass

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(A) Given the curve equation: $\sqrt{x}+\sqrt{y}=4$  $(i)$Differentiating both sides with respect to $x$:$\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \c

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($4$,$4$)

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(A) Given the curve equation: $\sqrt{x}+\sqrt{y}=4$  $(i)$Differentiating both sides with respect to $x$:$\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx} = 0$Solving for $\frac{dy}{dx}$:$\frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}}$Since the tangent is equally inclined to the axes,the slope of the tangent must be $\pm 1$. Thus,$\frac{dy}{dx} = \pm 1$.Case $1$: $-\sqrt{\frac{y}{x}} = 1 \Rightarrow \sqrt{\frac{y}{x}} = -1$ (Not possible as square root is non-negative).Case $2$: $-\sqrt{\frac{y}{x}} = -1 \Rightarrow \sqrt{\frac{y}{x}} = 1 \Rightarrow y = x$.Substituting $y = x$ into equation $(i)$:$\sqrt{x} + \sqrt{x} = 4$$2\sqrt{x} = 4$$\sqrt{x} = 2$$x = 4$.Since $y = x$,we have $y = 4$.Therefore,the required coordinates are $(4, 4)$.

Find the coordinates of the point on the curve $\sqrt{x}+\sqrt{y}=4$ at which the tangent is equally inclined to the axes.

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