Implicit equation of a sheared ellipse

Question

I have the following implicit equation of an ellipse

$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1.$

What will be the equation once it is sheared in the $x$ direction by an angle $\phi$ ?

In other words, suppose I wanted to apply the following shear transformation to the points of an ellipse

$\begin{bmatrix} x'\\ y' \end{bmatrix} = \begin{bmatrix} 1 & \tan(\phi)\\ 0 & 1 \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix}$

How would this transformation be represented in an implicit ellipse equation?

Answer 1

\begin{align*} x' &= x+y\tan \phi \\ y' &= y \\ x &= x'-y'\tan \phi \\ \frac{x^2}{a^2}+\frac{y^2}{b^2} &= \frac{(x'-y'\tan \phi)^2}{a^2}+\frac{y'^2}{b^2} \\ \frac{x'^2-2x'y'\tan \phi+y'^2\tan^2 \phi}{a^2}+\frac{y'^2}{b^2} &= 1 \\ \end{align*}

$$b^2x'^2\cos^2 \phi-b^2x'y'\sin 2\phi+(a^2\cos^2 \phi+b^2\sin^2 \phi)y'^2-a^2 b^2 \cos^2 \phi=0$$