Question

Write the equation in terms of a rotated x'y'-system using θ, the angle of rotation. write the equation involving x' and y' in standard form. xy +16 = 0; θ = 45°

Answer 1

(y'^2)/32 - (x'^2)/32 = 1

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.

Answer 2

xy + 16 = 0 is an equation for rectangular hyperbola. A clockwise rotation by 45 degrees yields a hyperbola of the form:

x^2/^2 – y^2/b^2 = 1

Hence from the given choices of this problem, the answer is:

y'^2/32 - x'^2/32 =1