The vertex of the parabola 9x^2 - 6x + 36y + | vedclass

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Question

(A) Given the equation of the parabola: $9x^2 - 6x + 36y + 9 = 0$.Rearranging the terms to complete the square for $x$: $9x^2 - 6x = -36y - 9$.Divide

Vedclass · Accepted Answer

$(1/3, -2/9)$

Vedclass · Answer

(A) Given the equation of the parabola: $9x^2 - 6x + 36y + 9 = 0$.Rearranging the terms to complete the square for $x$: $9x^2 - 6x = -36y - 9$.Divide by $9$: $x^2 - \frac{2}{3}x = -4y - 1$.Add $(\frac{1}{3})^2 = \frac{1}{9}$ to both sides: $x^2 - \frac{2}{3}x + \frac{1}{9} = -4y - 1 + \frac{1}{9}$.$(x - \frac{1}{3})^2 = -4y - \frac{8}{9}$.$(x - \frac{1}{3})^2 = -4(y + \frac{2}{9})$.Comparing this with the standard form $(x - h)^2 = 4a(y - k)$,the vertex $(h, k)$ is $(\frac{1}{3}, -\frac{2}{9})$.

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

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