Question

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

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

Answer 1

Answer: (A)

The given equation $9 x^2-6 x+36 y+9=0$ can be written as $(3 x-1)^2=-4(9 y+2)$.
Hence, the vertex is $\left(\frac{1}{3},-\frac{2}{9}\right)$.