Recognising an ellipse

Let us see how, starting from an expanded equation, completing the square reveals the hidden ellipse, with its centre and its semi-axes.

Example — Recognising an ellipse

$4x^2 + 9y^2 - 16x - 18y - 11 = 0$.

I factor out $4$ in front of $x^2, x$ and $9$ in front of $y^2, y$: $4(x^2 - 4x) + 9(y^2 - 2y) - 11 = 0.$

I complete the squares: $4\bigl[(x-2)^2 - 4\bigr] + 9\bigl[(y-1)^2 - 1\bigr] - 11 = 0$ $4(x-2)^2 + 9(y-1)^2 - 16 - 9 - 11 = 0$ $4(x-2)^2 + 9(y-1)^2 = 36.$

I divide by $36$: $\frac{(x-2)^2}{9} + \frac{(y-1)^2}{4} = 1.$

It is an ellipse with centre $O'(2;1)$, semi-axes $a = 3$ (horizontal), $b = 2$ (vertical), $c = \sqrt{9-4} = \sqrt{5}$. Foci $(2\pm\sqrt{5};\ 1)$.

Links

Topics: Ellipse
Concepts: Completing the square · Ellipse · Translated ellipse · Canonical equation
Skills: Analytic geometry · Simplifying