Answer:(y'^2)/32 - (x'^2)/32 = 1
Step-by-step explanation:
Given
xy = -16
θ = 45°
x and y are rewritten in terms of x' and y' as follows:
x = x' cos θ - y' sin θ
x = x' cos 45° - y' sin 45°
x = x' √2/2 - y' √2/2
x = √2/2 (x' - y')
y = x' sin θ + y' cos θ
y = x' sin 45° + y' cos 45°
y = x' √2/2 + y' √2/2
x = √2/2 (x' + y')
Replacing in the original formula:
[√2/2 (x' - y')] [√2/2 (x' + y')] = -16
2/4 (x'^2 - y'^2) = -16
(x'^2)/2 - (y'^2)/2 = -16
(y'^2)/32 - (x'^2)/32 = 1
Which is the standard form of a hyperbola.
