The tangent to the curve $\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}}=\sqrt{\mathrm{a}}$ cuts the axes at P and Q then OP + OQ =

$$\begin{gathered} \text{ Curve: }\sqrt{x}+\sqrt{y}=\sqrt{a} \\ \frac{dy}{dx}=\frac{-y^{1/2}}{x^{1/2}} \\ the point (x_1,y_1) is lies on the curve then, \\ \sqrt{{\mathrm{x}}_1}+\sqrt{{\mathrm{y}}_1}=\sqrt{\mathrm{a}} \\ \text{ Slope }=m={\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}=\frac{-y_1^{1/2}}{x_1^{1/2}} \end{gathered}$$
$$\begin{gathered} equation of \tan gent at (x_1,y_1) \\ \begin{array}{r}y-y_1=\frac{-y_1^{1/2}}{x_1^{1/2}}\left(x-x_1\right)\\ \frac{y}{\sqrt{a}{y}_1^{1/2}}+\frac{x}{\sqrt{a}{x}_1^{1/2}}=1\\ P=\left(\sqrt{a}{x}_1^{1/2},0\right)\\ Q=\left(0,\sqrt{a}{y}_1^{1/2}\right)\end{array} \end{gathered}$$
$\begin{array}{l}\mathrm{OP}+\mathrm{OQ}=\sqrt{{\left(\sqrt{a}{x}_1^{1/2}\right)}^2+0}+\sqrt{0+{\left(\sqrt{a}{y}_1^{1/2}\right)}^2}\\ =\sqrt{a{x}_1}+\sqrt{a{y}_1}\\ =\sqrt{a}\left(\sqrt{x_1}+\sqrt{y_1}\right)\\ =\sqrt{a}\cdot \sqrt{a}\\ =\mathrm{a}\end{array}$