Finding a positive definite matrix to satisfy the general equation of an ellipse

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This discussion focuses on deriving a positive definite matrix A that satisfies the equation of an ellipse represented as \(v^TAv=1\), where \(v=(x, y)^T\). The equation is given by \(\left(\frac{x}{a_1}\right)^2 + \left(\frac{y}{a_2}\right)^2 - 2\left(\frac{xy}{a_1a_2}\right)\cos(\delta)=\sin^2(\delta)\). The user successfully identifies relationships between the coefficients \(a\), \(b\), \(c\), and \(d\) in the matrix A and the parameters \(a_1\), \(a_2\), and \(\delta\), concluding that \(b\) and \(c\) remain free variables unless additional constraints are applied. The discussion emphasizes the importance of ensuring that the matrix A is positive definite.

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kalish1
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I am trying to find a matrix A such that $(1)$ can be written as $v^TAv=1$ where $v=(x, y)^T$.

$(1)$: $$\left(\frac{x}{a_1}\right)^2 + \left(\frac{y}{a_2}\right)^2 - 2\left(\frac{xy}{a_1a_2}\right)\cos(\delta)=\sin^2(\delta)$$ $$a_1, a_2, \sin(\delta)\neq 0.$$

I am positive that $\cos(\delta)$ should not be $\cos^2(\delta)$, as it is not even indicated in my textbook's errata.

**Here is my attempt:**

$v^TAv=1 \iff (x,y)A(x,y)^T=1 \iff A [=] 2 $x $2$ $$\iff 1 = (x \ y) \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \iff 1 = (x \ y) \begin{pmatrix} ax+by \\ cx+dy\end{pmatrix} \iff ax^2 + bxy + cxy + dy^2 = 1 \iff a\sin^2(\delta)x^2 + b\sin^2(\delta)xy + c\sin^2(\delta)xy + d\sin^2(\delta)y^2 = \sin^2(\delta)$$

So, $a\sin^2(\delta)=\frac{1}{{a_1}^2}, d\sin^2(\delta)=\frac{1}{{a_2}^2}, (b+c)\sin^2(\delta)=\frac{-2\cos(\delta)}{a_1a_2}$.

**Where I'm stuck:** Beyond this, I can't seem to separate $b$ and $c$! Is one of them just going to be a free variable?

Thanks.

By the way, I have cross-posted this question on Math Stack Exchange but did not get a satisfactory answer.

Moderator Edit: Link to the Thread at Stack Exchange: matrices - Finding a positive definite matrix to satisfy the general equation of an ellipse - Mathematics Stack Exchange
 
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An first step to this is to divide through by sin(x) to get
sin^2(x) so that it actually is equal to 1.

write
\begin{pmatrix}x & y \end{pmatrix}\begin{pmatrix} a & b \\ c & d\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix}= \begin{pmatrix}ax+ cy & cx+ dy\end{pmatrix}\begin{pmatrix}x \\ y \end{pmatrix}= ax^2+ bxy+ cxy+ dy^2= ax^2+ (b+ c)xy+ dy^2 which we compare to \frac{1}{a_1^2sin^2(\delta)}x^2- \frac{2cos(\delta)}{a_1a_2sin^2(\delta)}xy+ \frac{1}{a_2^2sin^2(\delta)}y^2= 1.

So a= \frac{1}{a_1^2sin^2(\delta)}, b+ c= \frac{2cos(\delta)}{a_1a_2asin^2(\delta)}, and d= \frac{1}{a_2^2sin^2(\delta)}. That has a single equation for b and c so there can be many such matrices unless you have some other condition such the matrix being orthogonal- or positive definite!
 
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