Recognising a translated ellipse

Problem

Recognise and reduce to translated canonical form: $x^2 + 4y^2 + 2x - 8y + 1 = 0$.

Solution

I group the terms in $x$ and the terms in $y$, factoring out the $4$ in front of those in $y$: $(x^2 + 2x) + 4(y^2 - 2y) + 1 = 0.$

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

I divide by $4$ to obtain the translated canonical form: $\boxed{\dfrac{(x+1)^2}{4} + \dfrac{(y-1)^2}{1} = 1}.$

It is an ellipse with centre $O'(-1;1)$, semi-axes $a = 2$ (horizontal) and $b = 1$ (vertical). Since $a>b$, the foci are horizontal with $c = \sqrt{4-1} = \sqrt{3}$: $F_{1,2}(-1\pm\sqrt{3};\ 1)$.

Links

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