Rotation of Axes for a hyperbola?

Question

Rotate the axes to sliminate the xy-term in the equations
xy-1=0. Then write the equation in standard form.

okay I really really don't undertand this chapter so I really need help. this problem is from an example from the book but I don't get it. please can you explain it because im completely lost. I would imensenly appreciate your help.

Answer 1

The general form of a quadratic equation is:

Ax² + Bxy + Cy² + Dx + Ey + F = 0

We want to rotate the axes to get rid of the xy term. Suppose we want to rotate the coordinate axes by an angle of θ. Then we introduce x and y in terms of the coordinates of the rotated system.

x = x'cosθ + y'sinθ
y = x'sinθ - y'cosθ

If we substitute these equations into the original equation we get:

xy - 1 = 0
(x'cosθ + y'sinθ)(x'sinθ - y'cosθ) - 1 = 0

For the case at hand θ = π/4.

[x'cos(π/4) + y'sin(π/4)][x'sin(π/4) - y'cosθ(π/4)] - 1 = 0
[(1/√2)x' + (1/√2)y'][(1/√2)x' - y'(1/√2)] - 1 = 0
x'²/2 - x'y'/2 + x'y'/2 - y'²/2 = 1

x'²/2 - y'²/2 = 1

This is the equation of the hyperbola in terms of the rotated coordinate system where the angle of rotation is π/4 radians.

In this case it was pretty obvious that the angle of rotation required to eliminate the xy term was θ = π/4. However if it wasn't obvious, it can be calculated by the formula:

cot(2θ) = (A - C)/B = (0 - 0)/1 = 0
2θ = π/2
θ = π/4

Answer 2

OK, I dont have much time to reply,but here is the basic,therefore make into standard form: xy=1

now draw the standard graph! remember the firdt rule is to always get the equation into standard form!