What is the answer to solvefor x,fx= 1/(sqrt(x^2-y^2)) ?

The answer to solvefor x,fx= 1/(sqrt(x^2-y^2)) is

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solvefor x,fx= 1/(sqrt(x^2-y^2))

solve for

x, f x = \frac{1}{x ^{2} - y ^{2}}

x = \frac{y ^{2} f + y ^{4} f ^{2} + 4}{2 f}, x = - \frac{y ^{2} f + y ^{4} f ^{2} + 4}{2 f}, x = \frac{y ^{2} f - y ^{4} f ^{2} + 4}{2 f}, x = - \frac{y ^{2} f - y ^{4} f ^{2} + 4}{2 f}

Solution steps

f x = \frac{1}{x ^{2} - y ^{2}}

Multiply both sides by

x^{2} - y^{2}

f x x^{2} - y^{2} = \frac{1}{x ^{2} - y ^{2}} x^{2} - y^{2}

Simplify

f x x^{2} - y^{2} = 1

Square both sides:

x^{4} f^{2} - x^{2} y^{2} f^{2} = 1

x^{4} f^{2} - x^{2} y^{2} f^{2} = 1

Solve

x^{4} f^{2} - x^{2} y^{2} f^{2} = 1 : x = \frac{y ^{2} f + y ^{4} f ^{2} + 4}{2 f}, x = - \frac{y ^{2} f + y ^{4} f ^{2} + 4}{2 f}, x = \frac{y ^{2} f - y ^{4} f ^{2} + 4}{2 f}, x = - \frac{y ^{2} f - y ^{4} f ^{2} + 4}{2 f}

x = \frac{y ^{2} f + y ^{4} f ^{2} + 4}{2 f}, x = - \frac{y ^{2} f + y ^{4} f ^{2} + 4}{2 f}, x = \frac{y ^{2} f - y ^{4} f ^{2} + 4}{2 f}, x = - \frac{y ^{2} f - y ^{4} f ^{2} + 4}{2 f}

x - 1 + 2 = x

3 x^{\frac{1}{3}} + 2 x^{\frac{2}{3}} = 5

\frac{2}{1 - x ^{2}} = π

x + 2 + 9 = 0

x^{- \frac{2}{3}} - 4 = 0

What is the answer to solvefor x,fx= 1/(sqrt(x^2-y^2)) ?
The answer to solvefor x,fx= 1/(sqrt(x^2-y^2)) is