Question

The vertex of the parabola $(y-2{)}^2=16(x-1)$ is

• A. (1,2)
• B. (-1,2)
• C. (1,-2)
• D. (2,1)

Answer 1

The correct option is A

(1,2)

Given : $(y-2{)}^2=16(x-1)$

Let X=x-1,Y=y-2

$\therefore Y^2=16X$

Vertex (X=0,Y=0)=(x-1=0,y-2=0)=(x=1,y=2)

Hence,the vertex is at (1,2)