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

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

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(\frac{- 1}{3}, \frac{- 1}{2})

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(\frac{- 1}{3}, \frac{1}{2})

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(\frac{1}{3}, \frac{1}{2})

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Solution

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

The equation can be written as $(3 x - 1)^{2} = - 4 (9 y + 2)$
This is of the form $x^{2} = 4 a y$ . To get the vertex, we will equate 3x-1 = 0 and 9y+2 = 0
Hence the vertex is $(\frac{1}{3}, - \frac{2}{9})$ .

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