Find the equation of the tangent to the curve
Question:

Find the equation of the tangent to the curve $\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}}=\mathrm{a}$, at the point $\left(\mathrm{a}^{2} / 4, \mathrm{a}^{2} / 4\right)$

Solution:

finding slope of the tangent by differentiating the curve

$\frac{1}{2 \sqrt{x}}+\frac{1}{2 \sqrt{y}}\left(\frac{d y}{d x}\right)=0$

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

at $\left(\frac{\mathrm{a}^{2}}{4}, \frac{\mathrm{a}^{2}}{4}\right)$ slope $\mathrm{m}$, is $-1$

the equation of the tangent is given by $y-y_{1}=m\left(x-x_{1}\right)$

$\mathrm{y}-\frac{\mathrm{a}^{2}}{4}=-1\left(\mathrm{x}-\frac{\mathrm{a}^{2}}{4}\right)$

$x+y=\frac{a^{2}}{2}$