Finding angle of rotation $\theta$ that transforms a standard ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ into $2x^2+2xy+2y^2-5=0$

Question

The ellipse with equation $2x^2+2xy+2y^2-5=0$ is obtained from the standard ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ by rotating through angle $\theta$.

Determine the angle of rotation $\theta$.

I used the standard method of the coefficient of $xy$~ (can't get ~ signs) being $0$. I worked out $\theta = \frac{\pi}{4}$.

Is this the correct angle of rotation ?

Answer 1

This is how the rotated ellipse looks like, according to Desmos:

Apparently the ellipse is rotated by $-\frac \pi4$, if the long axis was horizontal. However, if the long axis was vertical then the ellipse is rotated by $\frac \pi4$.

The coordinates of a point on the rotated ellipse are $$(x_r,y_r)=(x\cos \theta-y\sin \theta,x\sin \theta+y\cos \theta)$$

The given equation of the rotated ellipse is: $$\begin{align}&2x^2_r+2y^2_r+2x_ry_r=5\\ &\Leftrightarrow 2(x\cos \theta-y\sin \theta)^2+2(x\sin \theta+y\cos \theta)^2+2(x\cos \theta-y\sin \theta)(x\sin \theta+y\cos \theta)=5\\&\Leftrightarrow 2x^2(1+\cos \theta\sin\theta)+2y^2(1-\cos \theta \sin\theta)+2xy\underbrace{(\cos^2\theta-\sin^2\theta)}_{=0}=5\\&\Leftrightarrow\boxed{\frac{x^2}{\frac 5{2(1+\frac{\sin 2\theta}2)}}+ \frac{y^2}{\frac 5{2(1-\frac{\sin 2\theta}2)}}=1, \cos^2\theta=\sin^2\theta}\end{align}$$

Two solutions are possible depending on the solution of $\cos^2\theta=\sin^2\theta$:

$$\begin{align}\cos^2\theta=\sin^2\theta&\Rightarrow \cos\theta=\sin\theta\\&\Rightarrow \theta=\frac \pi4+k\pi, k\in \Bbb Z\\&\Rightarrow a^2=\frac 53, b^2=5\text{ : long axis is vertical}\end{align}$$

I would not consider an ellipse with long axis vertical because the rotation is presumed to come off a standard position where the original ellipse was centered in origin with the long axis horizontal.

$$\begin{align}\cos^2\theta=\sin^2\theta&\Rightarrow \cos\theta=-\sin\theta\\&\Rightarrow \theta=\frac {3\pi}4+k\pi, k\in \Bbb Z\\&\Rightarrow a^2=5, b^2=\frac 53\text{ : long axis is horizontal}\end{align}$$

Answer 2

My working. I can solve $\cot(2 \theta) = \frac{A-C}{B} = 0$, so $2\theta = \frac{\pi}{2} + k \pi$, so $\theta = \frac{\pi}{4} + \frac{k \pi}{2}$. Does that mean any angles of this form work ? I choose $\theta = \frac{\pi}{4}$.

Transformation: $x = X \cos(\theta) - Y\sin(\theta)$ and $y = X\sin(\theta)+Y\cos(\theta)$.

I subbed this in to obtain $\frac{X^2}{5/3} + \frac{Y^2}{5} = 1$