Application of Derivatives — Class 12 Maths Solution

We have, $\sqrt x + \sqrt y = 4$ …….(i)

$\Rightarrow$ ${x^{1/2}} + {y^{1/2}} = 4$

$\Rightarrow$ $\frac{1}{2} \cdot \frac{1}{{{x^{1/2}}}} + \frac{1}{2} \cdot \frac{1}{{{y^{1/2}}}} \cdot \frac{{dy}}{{dx}} = 0$

Therefore,$\frac{{dy}}{{dx}} = - \frac{1}{2} \cdot {x^{ - 1/2}}2 \cdot {y^{1/2}}$

$= - \sqrt {\frac{y}{x}}$

Since, tangent is equally inclined to the axes.

Therefore, $\frac{{dy}}{{dx}} = \pm 1$

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

$\Rightarrow$ $\frac{y}{x} = 1 \Rightarrow y = x$

From Eq. (i), $\sqrt y + \sqrt y = 4$
$\Rightarrow$ $2\sqrt y = 4$

$\Rightarrow$ $4y = 16$

Therefore,$y = 4$ and $x = 4$

When $y = 4$, then $x = 4$

So, the required coordinates are (4,4).