Find, to the nearest degree, the acute angle of rotation that eliminates the xy term of -3x^2+3xy+5y^2-x+2y-6=0.

a.)81°
b.)41°
c.)40°
d.)80°

Question

Find, to the nearest degree, the acute angle of rotation  that eliminates the xy term of -3x^2+3xy+5y^2-x+2y-6=0.
 	
a.)81°
b.)41°
c.)40°
d.)80°

superdavesuper · Answer

the curve looks like this but no idea how to eliminate the xy term :(

ranga · Answer

If the general equation is:  Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0
you can eliminate the Bxy term by rotating an angle of theta where
tan(2*theta) = B / (A - C) 

Here, B = 3, A = -3, C = 5.  tan(2*theta) = 3 / (-3 - 5) = -3/8
2*theta = arctan (-3/8)
theta = 1/2 * arctan (-3/8)

ranga · Answer

arctan(-3/8) = -20.556 or +159.444
Choosing the positive angle, 2 * theta = 159.444
theta = 159.444 / 2 = 79.722
nearest degree is 80 degrees.

superdavesuper · Answer

Nice work @ranga!!

I was stumped yesterday but I was staring at the plot while you typed up ur solution n noticed that if I rotate them to the "East-West opening hyperbola" form, the xy term will disappear.  So an alternative way to find the angle is by solving for the slope of its transverse axis.

It comes out to be 5.513 n arctan(5.513) gives the same 79.7 angle :)

ranga · Answer

Exactly @superdavesuper   That is another way to do it.  Well done!