Let y/(x-2)=k, then y=k(x-2).

Brought:

x^2 +[k(x-2)]^2=3

x^2 +(k^2) x^2 -4(k^2)x+4k^2 -3=0

( 1+k^2)x^2 -(4k^2)x +(4k^2-3)= 0

Think of it as a univariate quadratic equation about x, because real numbers x and y exist,

Therefore, the equation has a real number solution, then:

△= 16(k^4)-4( 1+k^2)(4k^2-3)≥0

16(k^4)-(4+4k^2)(4k^2-3)≥0

16(k^4)-( 16k^4 +4k^2- 12)≥0

That is -4k 2+ 12 ≥ 0.

k^2 -3≤0

The solution is -√3≤k≤√3.

So the maximum value of y/(x-2) is √3.

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