Answer
$r \cos (\theta -\dfrac{\pi}{4})=3$
Work Step by Step
The conversion of polar coordinates and Cartesian coordinates are described as follows:
1. $r^2=x^2+y^2 \implies r=\sqrt {x^2+y^2}$
2. $\tan \theta =\dfrac{y}{x} \implies \theta =\tan^{-1} (\dfrac{y}{x})$
3. $x=r \cos \theta$ and 4. $y=r \sin \theta$
Given: $\sqrt 2 x+\sqrt y=6$
This implies that $\cos (\dfrac{\pi}{4})r \cos \theta+\sin (\dfrac{\pi}{4}) r \sin \theta=3$
or, $r[\cos (\dfrac{\pi}{4}) (\cos \theta)+\sin (\dfrac{\pi}{4}) (\sin \theta)]=3$
Hence, $r \cos (\theta -\dfrac{\pi}{4})=3$
