Question

The vertex of the parabola $x^2+8x+12y+4=0$ is

• A. (-4,1)
• B. (4,-1)
• C. (-4,-1)
• D. (4,1)

Answer 1

The correct option is A

(-4,1)

Given: $x^2+8x+12y+4=0$

$\Rightarrow (x+4{)}^2-16+12y+4=0$

$\Rightarrow (x+4{)}^2+12y-2=0$

$\Rightarrow (x+4{)}^2=-12(y-1)$

$LetX=x+4,Y=y-1$

$\therefore X^2=-12Y$

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

Hence,the vertex is at (-4,1).