# precalculus rotation problem?

Rotate the axes to eliminate the xy-term in the equation. Then write the equation in standard form.

65x^2 - 66xy + 65y^2 -1568 = 0

A. [(x`)^2/49] + [(y`)^2/16] = 1

B. [(x`)^2/36] + [(y`)^2/16] = 1

C. [(x`)^2/64] + [(y`)^2/25] = 1

D. [(x`)^2/36] + [(y`)^2/9] = 1

E. [(x`)^2/49] + [(y`)^2/25] = 1

Really having a hard time with these types of problems. Help would be appreciated

Relevance
• Rahil
Lv 5
8 years ago

If the x and y axis are rotated to give x' and y' then we can equally rotate x' and y' in the opposite direction to get x and y. Let t be the the angle of rotation.

So the substitution we will be using is

x = x' cos t - y' sin t

y = x' sin t + y' cos t

(this is the standard rotation transformation)

and we need to find the right choice of t to get rid of any x' y' terms.

Substituting it in gives

65x^2 - 66xy + 65y^2 -1568

= 65(x' cos t - y' sin t)^2

- 66(x' cos t - y' sin t)(x' sin t + y' cos t)

+ 65(x' sin t + y' cos t)^2

-1568

= 65(x' ^2 cos^2 t -2x'y' cos t sin t + y' ^2 sin^2 t)

- 66(x' ^2 cos t sin t + x'y' cos^2 t -x'y' sin^2 t - y' ^2 cos t sin t)

+ 65(x' ^2 sin^2 t + 2x'y' cos t sin t + y' ^2 cos^2 t)

-1568

Grouping terms together gives

= x' ^2 (65 cos^2 t - 66 cos t sin t + 65 sin^2 t)

+ y' ^2 (65 sin^2 t + 66 cos t sin t + 65 cos^2 t)

+ x' y' ( - 66 cos^2 t + 66 sin^2 t )

-1568

Using the trigonometry identity cos^2 t + sin^2 t = 1 gives

= x' ^2 (65 - 66 cos t sin t)

+ y' ^2 (65 + 66 cos t sin t)

+ x' y' ( - 66 cos^2 t + 66 sin^2 t )

-1568

Using the identity that 2 cos t sin t = sin 2t gives

= x' ^2 (65 - 33 sin 2t)

+ y' ^2 (65 + 33 sin 2t)

+ x' y' ( - 66 cos^2 t + 66 sin^2 t )

-1568

Using the identity that cos^2 t - sin^2 t = cos 2t gives

= x' ^2 (65 - 33 sin 2t) + y' ^2 (65 + 33 sin 2t) + x' y' ( - 66 cos 2t) -1568

We want to get rid of the x' y' term so we choose t so that cos 2t = 0

This forces sin 2t to be 1 or -1 it doesn't matter which one you choose (the only difference will be that x' and y' get swapped around).

Let's choose t = 45 degrees, so cos 2t = cos 90 = 0, and therefore sin 2t = sin 90 = 1

So now our equation becomes

x' ^2 (65 - 33) + y' ^2 (65 + 33) -1568 = 0

which rearranges to

x' ^2 (32/1568) + y' ^2 (98/1568) = 1

which is

x' ^2 /49 + y' ^2 /16 = 1