Question

The vertex of the parabola x² + 8x + 12y + 4 = 0 is
(a) (−4, 1)
(b) (4, −1)
(c) (−4, −1)
(d) (4, 1)

Answer 1

(a) (−4, 1)

Given:
x² + 8x + 12y + 4 = 0
$$\begin{gathered} \Rightarrow {\left(x+4\right)}^2-16+12y+4=0 \\ \Rightarrow {\left(x+4\right)}^2+12y-12=0 \\ \Rightarrow {\left(x+4\right)}^2=-12\left(y-1\right) \end{gathered}$$

Let $X=x+4,Y=y-1$

∴ $X^2=-12Y$

Vertex = $\left(X=0,Y=0\right)=\left(x+4=0,y-1=0\right)=\left(x=-4,y=1\right)$

Hence, the vertex is at (−4, 1).