Verified answer

x² - y² ≥ 1

transfer -y² to the right with changing its sign ( this is a rule )

transfer 1 to the left with changing its sign

x² - 1 ≥ y²

or

y² ≤ x² - 1

y ≤ l√(x² -1)l

-√(x² -1) ≤ y ≤ √( x² -1)

y = ± √(x² - 1)

if you draw these ... both starts at x = 1 one go up, the other down

for a specified x ( example x = 3 and y = ±√8 )

if you come back to x = 2, one reaches from √8 to √3 ( goes down )

the other reaches from -√8 to -√3 ( goes up ) but y² goes down (8 → 3)

You have some kind of typo but ok

Suppose you have all the numbers and you square each one.

When you square a number between -2 and 2, the result is less than 4.

When you square numbers larger than 2, or smaller than -2, the result is more than 4.

Correct?

So is it not true that if y^2 < 4, then y must be between -2 and 2?

And if y^2 > 4 then either y > 2 or y < -2.

So there's your reason. Not hard to figure out if you try thinking about the different cases.

Geometric interpretation

x^2 + y^2 ≥ 1 points outside of circle

y ≥ sqrt(1-x^2) stuff above x-axis without circle

y ≤ -sqrt(1-x^2) below x-axis without circle.

Same thing (for -1≤x≤1)

In case you mean x^2 + y^2 ≤ 1, then that is points inside the circle, so we have -sqrt(1-x^2) ≤ y ≤ sqrt(1-x^2) which are again the inequalities that define the two halves of the region inside a circle.

Rule:

y^2 ≤ A^2

<===> -A ≤ y ≤ A , with A>0

It's because after you add y^2 to both sides and subtract 1 from both, making it

x^2 - 1 ≥ y^2 and then reverse to get

y^2 ≤ x^2 - 1,

any positive value that works for y would also work if made negative

Like, say x = 3 and y = either 2 or -2, you still get 4 ≤ 8

so y can be anywhere from -√(x^2 - 1) to + √(x^2 - 1)