Question

Vertex of the parabola $9x^2-6x+36y+9=0$ is

• A. $(\frac{1}{3},\frac{-2}{9})$
• B. $(\frac{-1}{3},\frac{1}{2})$
• C. $(\frac{-1}{3},\frac{-1}{2})$
• D. $(\frac{1}{3},\frac{1}{2})$

Answer 1

The correct option is A $(\frac{1}{3},\frac{-2}{9})$

Given,

$9x^2-6x+36y+9=0$

$(3x{)}^2-2(3x)=-36y-9$

$(3x{)}^2-2(3x)+1=-36y-9+1$

$(3x-1{)}^2=-36y-8$

$(3x-1{)}^2=-36(y+\frac{2}{9})$

$(x-a{)}^2=4c(y-b)$

${(x-\frac{1}{3})}^2=-4(y+\frac{2}{9})$

$x-\frac{1}{3}=0$

$\Rightarrow x=\frac{1}{3}$

$y+\frac{2}{9}=0$

$\Rightarrow y=-\frac{2}{9}$

$\therefore (\frac{1}{3},-\frac{2}{9})$