[
\sqrt{x^2 - y^2} = 3y - x
]



Step 1: Isolate the square root

We have:

[
\sqrt{x^2 - y^2} = 3y - x
]

Since the left side is a square root, it must be **non-negative**, so:

[
3y - x \ge 0 \implies x \le 3y
]

Step 2: Square both sides

Squaring both sides gives:

[
x^2 - y^2 = (3y - x)^2
]

Expand the right side:

[
x^2 - y^2 = 9y^2 - 6xy + x^2
]

Subtract (x^2) from both sides:

[

* y^2 = 9y^2 - 6xy
]

Combine like terms:

[

* y^2 - 9y^2 + 6xy = 0
]

[
-10y^2 + 6xy = 0
]

Step 3: Factor

[
2y(3x - 5y) = 0
]

Since (y > 0), we must have:

[
3x - 5y = 0 \implies 3x = 5y \implies x = \frac{5}{3}y
]

Step 4: Solve for (xy)

[
xy = x \cdot y = \frac{5}{3}y \cdot y = \frac{5}{3} y^2
]

We can leave the answer in terms of (y^2), or if the problem wants **just the product in terms of numbers**, it is:

[
xy = \frac{5}{3} y^2
]


[
\sqrt{x^2 - y^2} = \sqrt{\left(\frac{5}{3}y\right)^2 - y^2} = \sqrt{\frac{25}{9}y^2 - y^2} = \sqrt{\frac{16}{9}y^2} = \frac{4}{3} y
]

Right-hand side:

[
3y - x = 3y - \frac{5}{3}y = \frac{4}{3}y
]



Answer:

[
\boxed{xy = \frac{5}{3} y^2}
]