How can we factorise a general second degree expression?

Question

The expression:

$$13x^2 - 18xy +37y^2 + 2x +14y -2 = 0$$

What I tried was making factors like $(x-a)(y-b)+ $ other terms.

$$\left(x-\dfrac{7}{9}\right)\left(y-\dfrac{1}{9}\right)+\dfrac{2}{81}-\dfrac{13x^2}{18}-\dfrac{37y^2}{18}=0$$

Actually, I do not want to completely factorise the expression, but only make it into groups of factors so that I can easily identify the conic that it represents.

The method that we were taught was by calculating $\Delta$, given by:

$$\Delta = abc +2fgh-af^2-bg^2-ch^2$$

and cases on when it represents which conic. But the problem asks for elements of conic too, like focus, directrix etc. These can be identified only when we have it in proper factorised form, which I am unable to do.

Please suggest ways of factorisation as a general rule.

Answer 1

This book seems to do conic sections in the first twenty pages or so, although the online preview shows only the end of the chapter. It would appear they do not require linear algebra before this material, so that might be an advantage. They do a hyperbola on page E-24, an ellipse pages E-27 to E-29, then exercise E-29 to E-30, just the final answers pages E-30 to E-32. This book is Geometry & Vector Calculus by A. R. Vasishtha.

Here is the treatment of a different ellipse, pages E-27 to E-29. If not clear, you can go to the web preview and compare.

THIS SEEMS TO BE a good book for this, extra practice problems and solutions, to go with an existing textbook in engineering mathematics. To get anywhere with this problem, basic linear algebra and simple multivariable calculus (partial derivatives) are required, unless you are able to memorize a large number of formulas. An edition of the main textbook seems available here. Maybe safe to use, maybe not. I cannot seem to find conic sections in this book, though. They jump to three dimensional stuff.

Te center of the ellipse is at $(1/4, 1/4).$ If we take translated coordinates $(p,q)$ with $$ p = x + \frac{1}{4}, \; \; q = y + \frac{1}{4}, $$ or $$ x = p - \frac{1}{4}, \; \; y = q - \frac{1}{4}, $$ we get $$ 13 p^2 - 18 pq + 37 q^2 = 4. $$

For the next part, a rotation, you need to deal with eigenvalues and eigenvectors. Not bad, they deliberately made the eigenvalues integers. However, square roots do appear.

Let $$ u = \frac{3}{\sqrt {10}} p + \frac{1}{\sqrt {10}} q, $$ $$ v = - \frac{1}{\sqrt {10}} p + \frac{3}{\sqrt {10}} q, $$ or $$ p = \frac{3}{\sqrt {10}} u - \frac{1}{\sqrt {10}} v, $$ $$ q = \frac{1}{\sqrt {10}} u + \frac{3}{\sqrt {10}} v. $$

Then $$ 10 u^2 + 40 v^2 = 4. $$

Answer 2

We have the conic as $$Q(x,y) = 13x^2-18xy+37y^2+2x+14y-2=0$$ Comparing with the standard form $$Q(x,y) = Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$ we get, $A=13, B=-18, C=37, D=2, E=14, F=-2$. To determine the type of conic, we create a matrix $$ A_Q =\begin{bmatrix} A & B/2 & D/2\\ B/2 & C & E/2\\ D/2 & E/2 & F \end{bmatrix}$$ and then classify the conic as follows:

$(1)$ If $\det A_Q = 0$, the conic is degenerate.
$(2)$ If $\det A_Q \neq 0$, then we can see what type of conic section it is by computing the minor, $\det A_{{33}}$:
$(a)$ $Q$ is a hyperbola if $\det A_{33} <0$.
$(b)$ $Q$ is a parabola if $\det A_{33}=0$, and
$(c)$ $Q$ is an ellipse if $\det A_{33}>0$.

---

For our conic, we have $$\det A_{33} = \begin{vmatrix} 13 & -9\\ -9 & 37\end{vmatrix} = (13)(37)-(-9)(-9) = 400>0$$ So, it is an ellipse. Hope it helps.