#### Find the co-ordinates of the point on the curve $\sqrt x +\sqrt y =4$ at which the tangent is equally inclined to the axes.

Given: curve $\sqrt x +\sqrt y =4$

To find: point coordinates on which tangent is equally inclined to the axis on the curve

Explanation: given $\sqrt x +\sqrt y =4$

After differentiating with respect to,

$\frac{\mathrm{d}(\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}})}{\mathrm{d} \mathrm{x}}=\frac{\mathrm{d}(4)}{\mathrm{dx}}$
Now using the sum rule of differentiation

$\\ \frac{\mathrm{d}(\sqrt{\mathrm{x}})}{\mathrm{d} \mathrm{x}}+\frac{\mathrm{d}(\sqrt{\mathrm{y}})}{\mathrm{d} \mathrm{x}}=0 \\\frac{\mathrm{d}\left(\mathrm{x}^{\frac{1}{2}}\right)}{\mathrm{dx}}+\frac{\mathrm{d}\left(\mathrm{y}^{\frac{1}{2}}\right)}{\mathrm{d} \mathrm{x}}=0$
Then, by differentiating the equation, we get

$\\ \Rightarrow \frac{1}{2} x^{\frac{1}{2}-1}+\frac{1}{2} y^{\frac{1}{2}-1} \frac{d(y)}{d x}=0 \\ \Rightarrow \frac{1}{2} x^{\frac{-1}{2}}+\frac{1}{2} y^{\frac{-1}{2}} \frac{d(y)}{d x}=0 \\ \Rightarrow \frac{1}{2} x^{\frac{-1}{2}}=-\frac{1}{2} y^{\frac{-1}{2}} \frac{d(y)}{d x} \\ \Rightarrow \frac{d y}{d x}=-\frac{x^{\frac{-1}{2}}}{y^{\frac{-1}{2}}} \\ \Rightarrow \frac{d y}{d x}=-\sqrt{\frac{y}{x}} \ldots . . \text { (i) }$

The given curve has this tangent

As mentioned in the question tangent is equally inclined to the axis,

$\\ \frac{d y}{d x}=\pm 1 \\\\ -\sqrt{\frac{y}{x}}=\pm 1 \\\\ \frac{y}{x}=1 \Rightarrow y=x$

Substituting values in the curve equation from equation (ii)

$\\ \sqrt y+\sqrt y=4 \\2 \sqrt y=4 \\ \sqrt y=2 \\\Rightarrow y=4$

When y = 4, then x = 4 from equation (ii)

Show the points on the curve $\sqrt{x}+\sqrt {y}=4$  at which the tangent equally inclined to the axis has the coordinates (4,4).