How do I find the angle of rotation of 153x^2 - 192xy + 97y^2 - 30x -40y - 200 = 0153x2−192xy+97y2−30x−40y−200=0, given cot 2theta = (A - C)/Bcot2θ=A−CB?

1 Answer
Jim H
Aug 5, 2015

theta = 1/2cot^-1(-56/192) ~~ 53.13^@θ=12cot−1(−56192)≈53.13∘ (Using a calculator.)

Explanation:

cot 2theta = (153-97)/(-192) = -56/192cot2θ=153−97−192=−56192

So 2theta = cot^-1(-56/192)2θ=cot−1(−56192) and

theta = 1/2cot^-1(-56/192)θ=12cot−1(−56192) Get out a calculator or trig tables. That is not a special angle we know.

But wait, there's more.

Perhaps we are really interested in eliminating the xyxy term from the equation. In which case, we don't really need to know thetaθ, we just need the sine and cosine of thetaθ. Those we can get without trig tables or calculators.

-56/192 = -28/96 = -7/24−56192=−2896=−724 AHA! (or perhaps not)
Even if we don't immediately see it, that is a 7, 24, 257,24,25 right triangle.
Or use trigonometry: if the cotangent of an angle is -7/24−724, find the cosine of that angle.

If cot 2theta = -7/24cot2θ=−724, the cos 2 theta = -7/25cos2θ=−725.

Now we can use half-angle formulas to get sin thetasinθ and cos theta#

And then we can do the substitution to eliminate the xyxy term.

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