Answer:
[tex]y=\sqrt{1-x^{2} } +3\sqrt{2x} \\y=-\sqrt{1-x^{2} } +3\sqrt{2x}[/tex]
Step-by-step explanation:
Given the equation is
[tex]x^{2} +(y-3\sqrt{2x} )^{2} =1[/tex]
Subtract [tex]x^{2}[/tex] from both sides of the equation
[tex](y-3\sqrt2{x} )^{2} +x^{2} -x^{2} =1-x^{2}[/tex]
Subtracting [tex]x^{2}[/tex] from itself leaves [tex]0[/tex]
[tex](y-3\sqrt{2x} )^{2} =1-x^{2}[/tex]
Take the square root of both sides of the equation
[tex]\sqrt{(y-3\sqrt{2x} )^{2} } =\sqrt{1-x^{2} }[/tex]
The square root operation always gives one positive number, and one negative number.
[tex](y-3\sqrt2{x} )=\sqrt{1-x^{2} } \\(y-3\sqrt{2x} )=-\sqrt{1-x^{2} }[/tex]
take[tex]-3\sqrt{2x}[/tex] on the right side then the value of [tex]y[/tex] is
[tex]y=\sqrt{1-x^{2} } +3\sqrt2{x}[/tex]
[tex]y=-\sqrt{1-x^{2} } +3\sqrt{2x}[/tex]
Then equation is solved and
[tex]y=\sqrt{1-x^{2} } +3\sqrt{2x} \\y=-\sqrt{1-x^{2} } +3\sqrt{2x}[/tex]