Answer
See explanations.
Work Step by Step
Step 1. Start with the original equation $\sqrt x+\sqrt y=1$, take the square on both sides, we have
$x+2\sqrt {xy}+y=1$. Rearrange the equation as $2\sqrt {xy}=1-(x+y)$ and take the square again to get
$4xy=1-2(x+y)+x^2+2xy+y^2$ or $x^2-2xy+y^2-2x-2y+1=0$.
Step 2. With coefficients $A=1, B=-2, C=1$, to remove the xy-term, the angle of axes rotation is given by $cot2\phi=\frac{0}{-2}=0$, thus $2\phi=\frac{\pi}{2}$ and $\phi=\frac{\pi}{4}=45^{\circ}$
Step3. Calculate the discriminant as $B^2-4AC=4-4=0$, thus the graph of the equation is a parabola.
Step 4. Conclusion: we confirmed that the original equation is part of a parabola by rotating the axes through an angle of $45^{\circ}$
