Write the appropriate rotation formulas so that in a rotated system the equation has no x'y'-term.
-$$14 x ^ { 2 } - 24 x y + 7 y ^ { 2 } + 30 x - 40 y - 45 = 0$$
A) $x = \frac { 3 } { 5 } x ^ { \prime } - \frac { 4 } { 5 } y ^ { \prime } ; y = \frac { 4 } { 5 } x ^ { \prime } + \frac { 3 } { 5 } y ^ { \prime }$
B) $x = \frac { 7 } { 25 } x ^ { \prime } - \frac { 24 } { 25 } y ^ { \prime } ; y = \frac { 24 } { 25 } x ^ { \prime } + \frac { 7 } { 25 } y ^ { \prime }$
C) $x = \frac { 89 } { 100 } x ^ { \prime } - \frac { 9 } { 20 } y ^ { \prime } ; y = \frac { 9 } { 20 } x ^ { \prime } + \frac { 89 } { 100 } y ^ { \prime }$
D) $x = \frac { 1 } { 2 } x ^ { \prime } - \frac { \sqrt { 3 } } { 2 } y ^ { \prime } ; y = \frac { \sqrt { 3 } } { 2 } x ^ { \prime } + \frac { 1 } { 2 } y ^ { \prime }$

Question

Quizplus · Accepted Answer

The rotation formula to eliminate the $x'y'$-term is derived from the angle $\theta$ that satisfies $\tan(2\theta) = \frac{2B}{A-C}$, where $A$, $B$, and $C$ are coefficients from the general conic section equation $Ax^2 + 2Bxy + Cy^2 + Dx + Ey + F = 0$. For the given equation, $A = 14$, $B = -12$ (since $2B = -24$), and $C = 7$. Plugging these into the formula gives $\tan(2\theta) = \frac{-24}{14-7} = -\frac{24}{7}$. The angle $\theta$ that satisfies this condition corresponds to the rotation that will eliminate the $x'y'$-term. The coefficients $\frac{3}{5}$ and $\frac{4}{5}$ in option A correspond to the cosine and sine of the angle of rotation, respectively, indicating that this rotation will align the axes with the principal axes of the conic section, thereby eliminating the $x'y'$-term.

Write the Appropriate Rotation Formulas So That in a Rotated 14x2−24xy+7y2+30x−40y−45=014 x ^ { 2 } - 24 x y + 7 y ^ { 2 } + 30 x - 40 y - 45 = 014x2−24xy+7y2+30x−40y−45=0

Write the Appropriate Rotation Formulas So That in a Rotated $14 x ^ { 2 } - 24 x y + 7 y ^ { 2 } + 30 x - 40 y - 45 = 0$