Find the sum of the intercepts on the axes of coordinates by any tangent to the curve $\sqrt{x}+\sqrt{y}=2$

Here, equation of curve is $\sqrt{x}+\sqrt{y}=2$. 

Whose parametric coordinates are given by,

$\sqrt{x}=2{\cos }^2\theta$

and

$\sqrt{y}=2{\sin }^2\theta$

i.e. $x=4{\cos }^4\theta$

and $y=4{\sin }^4\theta$

$\therefore \frac{dy}{dx}=\frac{4\times 4{\sin }^3\theta \cos \theta }{4\times 4{\cos }^3\theta \left(-\sin \theta \right)}$

Now, equation of tangent is $\frac{y-4{\sin }^4\theta }{x-4{\cos }^4\theta }=-{\tan }^2\theta$

$\therefore x$-intercept, $\left|4{\cos }^4\theta \frac{4{\sin }^4\theta }{{\tan }^2\theta }\right|=4{\cos }^2\theta$

and $y$-intercept, $\left|4{\sin }^4\theta \frac{4{\sin }^4\theta }{{\tan }^2\theta }\right|=4{\sin }^2\theta$

Hence, the sum of intercepts made on the axes of coordinates is,

$4{\cos }^2\theta +4{\sin }^2\theta =4\left({\cos }^2\theta +{\sin }^2\theta \right)=4$

hence, the correct option is $4.$