Identify and put conic in standard form: $4x^2 - 16x + 3y^2 + 24y + 52 = 0$

Question

How do I put this conic in standard form and identify it?

$$4x^2 - 16x + 3y^2 + 24y + 52 = 0$$

Answer 1

Start by completing the square.

$$\begin{align} 4x^2 - 16 x + 3y^2 + 24 y + 52 & = (4x^2 - 16x \color{blue}{\bf + 16}) + (3y^2 + 24y \color{blue}{\bf + 48}) + 52 \color{blue}{\bf - 16 - 48} \\ \\ & = 4(x^2 -4x + 4) + 3(y^2 + 8 y + 16) - 12 = 0\\ \\ & = \cdots \end{align}$$

Added:

Factoring gives us: $$\begin{align} 4(x^2 -4x + 4) + 3(y^2 + 8 y + 16) - 12 & = 0\\ \\ 4(x - 2)^2 + 3(y + 4)^2 &= 12\end{align}$$

Now divide through by $12$.

$$\dfrac{(x-2)^2}{3} + \frac{(y+4)^2}{4} = 1 \iff \left(\dfrac{x-2}{\sqrt 3}\right)^2 + \left(\frac{y+4}{2}\right)^2 = 1$$ This equation describes an ellipse with center $(2, -4)$.