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

The answer to solvefor y,x=sqrt(1+y^2) is y=sqrt(x^2-1) {x>= 1},y=-sqrt(x^2-1) {x>= 1}

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solvefor y,x=sqrt(1+y^2)

solve for

y, x = 1 + y^{2}

y = x^{2} - 1 {x \geq 1}, y = - x^{2} - 1 {x \geq 1}

Solution steps

x = 1 + y^{2}

Square both sides:

x^{2} = 1 + y^{2}

x^{2} = 1 + y^{2}

Solve

x^{2} = 1 + y^{2} : y = x^{2} - 1, y = - x^{2} - 1

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

Verify Solutions:

y = x^{2} - 1 {x \geq 1}, y = - x^{2} - 1 {x \geq 1}

The solutions are

y = x^{2} - 1 {x \geq 1}, y = - x^{2} - 1 {x \geq 1}

3 x + 8 - 2 x - 4 = 8

x - 2 x - 8 = 8

2 x + 5 = 7

x, f = 2 x - 1 + 3

14 - 16 - x^{2} = x^{2}

What is the answer to solvefor y,x=sqrt(1+y^2) ?
The answer to solvefor y,x=sqrt(1+y^2) is y=sqrt(x^2-1) {x>= 1},y=-sqrt(x^2-1) {x>= 1}