Use Rotation of Axes Formulas
-$$\mathrm { xy } + 16 = 0 ; \quad \theta = 45 ^ { \circ }$$
A) $\frac { y ^ { \prime 2 } } { 32 } - \frac { x ^ { \prime 2 } } { 32 } = 1$
B) $y ^ { \prime 2 } = - 32 x ^ { \prime }$
C) $\frac { y ^ { \prime 2 } } { 32 } + \frac { x ^ { \prime 2 } } { 32 } = 1$
D) $\frac { x ^ { 2 } } { 4 } + \frac { y ^ { 2 } } { 2 } = 1$

Question

Quizplus · Accepted Answer

The rotation of axes formulas for transforming $x$ and $y$ into $x'$ and $y'$ when the axes are rotated by an angle $\theta$ are given by:$$x = x' \cos \theta - y' \sin \theta$$$$y = x' \sin \theta + y' \cos \theta$$Given $\theta = 45^\circ$, $\cos 45^\circ = \sin 45^\circ = \frac{\sqrt{2}}{2}$. Substituting these into the given equation $xy + 16 = 0$ and using the rotation formulas:$$x'y'(\frac{\sqrt{2}}{2})^2 - x'y'(\frac{\sqrt{2}}{2})^2 + 16 = 0$$This simplifies to:$$0 + 16 = 0$$To find the correct form, we need to apply the rotation correctly. The mistake in the explanation shows a misunderstanding of how to apply the rotation formulas to the given equation. Let's correct this:Given equation is $xy + 16 = 0$. After rotation by $45^\circ$, we use the rotation formulas correctly:$$x = \frac{x' - y'}{\sqrt{2}}, \quad y = \frac{x' + y'}{\sqrt{2}}$$Substituting these into the equation $xy + 16 = 0$ gives:$$\left(\frac{x' - y'}{\sqrt{2}}\right)\left(\frac{x' + y'}{\sqrt{2}}\right) + 16 = 0$$$$ \Rightarrow \frac{x'^2 - y'^2}{2} + 16 = 0$$$$ \Rightarrow x'^2 - y'^2 + 32 = 0$$$$ \Rightarrow \frac{y'^2}{32} - \frac{x'^2}{32} = 1$$This matches choice A, which is the correct transformation of the given equation under a rotation of $45^\circ$.

Use Rotation of Axes Formulas - xy+16=0;θ=45∘\mathrm { xy } + 16 = 0 ; \quad \theta = 45 ^ { \circ }xy+16=0;θ=45∘

Use Rotation of Axes Formulas
- $\mathrm { xy } + 16 = 0 ; \quad \theta = 45 ^ { \circ }$