Find the required point be `P(x_1, y_1)dot` The tangent to the curve `sqrt(x)+sqrt(y)=4` at which tangent is equally inclined to the axes.

Answer

The required conditions are `(4,4)`

Explanation

We have, `\sqrt{x}+\sqrt{y}=4`
$$
\begin{aligned}
&\Rightarrow x^{1 / 2}+y^{1 / 2}=4 \\
&\Rightarrow \frac{1}{2 \cdot 1} /\left(x^{1 / 2}\right)+\frac{1}{2} \cdot\left(\frac{1}{y^{1 / 2}}\right) \cdot \frac{d y}{d x}=0 \\
&\frac{d y}{d x}=-\frac{1}{2} \cdot x^{-1 / 2} 2 \cdot y^{1 / 2} \\
&=-\sqrt{\frac{y}{x}}
\end{aligned}
...