Can $y$ be made the subject in such equation?

Question

I came across the equation:

$${x^2} + 2xy + {3y^2} = 1$$

I tried making $\space y \space$ the subject, but the furthest I got was:

$$ 2xy + {3y^2} = 1 - {x^2} $$

$$ y(2x + 3y) = 1 - {x^2} $$

$$ y = {{ 1 - {x^2}}\over { 2x + 3y }}$$

But I can't seem to be able to get $\space y \space$ on its own.

I even tried taking $\space \log \space$ of both sides, but that failed.

Does anyone know a way to do it?

Answer 1

You can just use the quadratic equation

$$y = \frac{-2x \pm \sqrt{4x^2 - 4\times(3)\times(x^2-1)}}{2\times3}$$

$$y = \frac{-2x \pm \sqrt{4x^2 - 12 x^2 + 12}}{6}$$

$$y = \frac{-2x \pm \sqrt{12 - 8 x^2}}{6}$$

$$y = \frac{-x \pm \sqrt{3 - 2 x^2}}{3}$$

Answer 2

Your equation is a quadratic form which can be written as $$ 1 = u^t A u = (x, y) \begin{pmatrix} 1 & 1 \\ 1 & 3 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} $$

This a symmetric matrix and we can calculate the eigenvectors $v_1$, $v_2$ and eigenvalues $\lambda_1$, $\lambda_2$:

>> A = [1,1;1,3]
A =

   1   1
   1   3

>> [B,C]=eig(A)
B =

  -0.92388   0.38268
   0.38268   0.92388

C =

Diagonal Matrix

   0.58579         0
         0   3.41421

We have $$ \DeclareMathOperator{diag}{diag} (v_1 v_2)^{-1} A(v_1 v_2) = (v_1 v_2)^{-1} (\lambda_1 v_1, \lambda_2 v_2) = \diag(\lambda_1, \lambda_2) \iff \\ A = (v_1, v_2) \diag(\lambda_1, \lambda_2) (v_1, v_2)^{-1} $$ This can be interpreted such that the matrix $T = (v_1, v_2)^{-1}$ transforms the coordinates $u$ into $u' = T u$, and $A$ is diagonal in that system, so that your equation transforms into \begin{align} 1 &= u^t T^{-1} T A T^{-1} T u \\ &= (u')^t \diag(\lambda_1, \lambda_2) u' \\ &= \lambda_1 (x')^2 + \lambda_2 (y')^2 \\ &= \left( \frac{x'}{1/\sqrt{\lambda_1}}\right)^2 + \left( \frac{y'}{1/\sqrt{\lambda_2}}\right)^2 \\ &= \left( \frac{x'}{a}\right)^2 + \left( \frac{y'}{b}\right)^2 \quad (*) \end{align} which is an ellipse with semi axes $a = 1/\sqrt{\lambda_1}=1.31$ and $b = 1/\sqrt{\lambda_2}=0.54$.

(Large version)

The idea now is to use some parametrization of $(*)$ and use $T^{-1}$ to transform into the original system.

We might use $$ y' = \pm b \sqrt{1 - \left( \frac{x'}{a}\right)^2} $$ which gives $$ u = T^{-1} u' \iff \\ \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -0.92388 & 0.38268 \\ 0.38268 & 0.92388 \end{pmatrix} \begin{pmatrix} x' \\ \pm b \sqrt{1 - \left( \frac{x'}{a}\right)^2} \end{pmatrix} $$