Rotation of Axes
The point P is the same in both the xy and xy coordinate systems.
The point can be represented in the xy plane as xr cos and yrsin The point can be represented in the xy plane as xr cos and yr sin
Using the sum of angles identities from trigonometry gives us ... xrcos yrsin rrcos cos sin sin rrcos sin sin cos xycos sin xysin cos
Substituting x and y into the general form Ax22 Bxy Cy Dx Ey F 0 gives: Axcos y sin2 Bx cos y sin x sin y cos Cxsin y cos2 Dx cos y sin Ex sin y cos F 0
Expanding the factors gives: Ax 22cos 2 xy cos sin y 22 sin B x2222cos sin xy cos xy sin y cos sin Cx22sin 2 xy cos sin y 2 cos 2 Dxcos y sin Ex sin y cos F 0
Collecting the terms yields: xA22 cos B cos sin C sin 2 xy Bcos22 sin 2 A C cos sin yA22sin B cos sin C cos 2 xDcos E sin y D sin E cos F 0 The double angle identities cos 2 cos22 sin and sin 2 2cos sin allow us to rewrite the xy term. xA22 cos B cos sin C sin 2 xy Bcos2 A C sin 2 yA22sin B cos sin C cos 2 xDcos E sin y D sin E cos F 0
The general form can be written as Ax 22 Bxy Cy Dx Ey F 0 where AA cos22 B cos sin C sin BB cos 2 AC sin 2 CA sin22 B cos sin C cos DD cos E sin ED sin E cos FF
The goal is to find the angle of rotation Some people will find it easier to use such that B 0 . 1 1 B BACcos2 sin 2 0 tan BACcos2 sin 2 2 A C cos 2 AC We don't because it is possible that A C and you might get division by zero. sin 2 B A C cot 2 We can avoid this by using the inverse B cotangent function. We know that B will 1 AC 2cot never be zero since the whole point is to B eliminate the xy term. 1 1 A C Also, the range of the inverse cotangent is cot 2 B between 0E and 180E, so 02180 , which means that 090 .
While it is possible to get an angle using this formula, it is usually easier to draw a triangle with 2θ as the angle and use the half-angle identities to find the values for sin θ and cos θ. 1cos2 1cos2 cos2 and sin2 2 2 The discriminant, BAC2 4 , can be used to determine the type of conic section. The discriminant is not affected by the rotation, so you can use the original coefficients or the new ones after the rotation. The discriminant does not detect the degenerate cases. If BAC2 40 , the conic is a circle or an ellipse. If BAC2 40 , the conic is a parabola. If BAC2 40 , the conic is a hyperbola.
Example Problem 27xxyy22 270 57 282 25 x 285 22 y 287 0 2 The discriminant is BAC2 4 2 70 4 2 7 5 7 280 280 0 Since the discriminant is 0, this is going to be a parabola.
AC27 57 37 3 cot 2 B 270 270210 1210 If you need to know the angle, then tan1 32.31153324 23 But you don't need to know the angle to get rid of the xy term. 3 Draw a triangle so that cot 2 and use it to determine the 210 values for the cosine and sine using the half angle identities. Use the positive values for cos and sin since is an acute angle. 1cos2 13/75 5 cos2 cos 227 7 1cos2 13/72 2 sin2 sin 227 7
Make the substitutions into the formulas above. A 2755 270221 57 1020100 7777 7 310210 B 270 777 27 57 6 6 0 C 2 7221 2 7055 5 7 4 20 25 49 7777 77 5 2 D 28 2 2 5 77 28 5 2 2 2810 10 10 4 392 77 E 28 2 2 52 28 5 2 2 5 77 282 2 10 5 2 10 196 77 F 28 7 The new equation in the xy plane is 00xxyy22 49 392 x 196 y 2870 . 777 7 Multiplying everything by gives yxy2 8440 . 49 Notice that the discriminant is still zero. Graph before rotation. BAC22404010
If you put general equation into standard form by completing the square, you get yyx2 484 yy2 44844 x yx282 1 2 yx2421 Same graph after rotation by 32.3 . This is a parabola with a vertex at 1, 2 , opening along the positive x axis, with a focal length of 2. That puts the focus at 1, 2 and the directrix at x 3 .
sin 2 7 2 Use tan to get the correct cos 57 5 angle on the rotated axes. tan1 2 5 32.3