The axis of the parabola 9y^2 - 16x - 12y - 5 | vedclass

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Question

(A) Given the equation of the parabola: $9y^2 - 16x - 12y - 57 = 0$Rearranging the terms to complete the square for $y$:$9(y^2 - \frac{4}{3}y) = 16x +

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$3y = 2$

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(A) Given the equation of the parabola: $9y^2 - 16x - 12y - 57 = 0$Rearranging the terms to complete the square for $y$:$9(y^2 - \frac{4}{3}y) = 16x + 57$$9(y^2 - \frac{4}{3}y + \frac{4}{9}) = 16x + 57 + 4$$9(y - \frac{2}{3})^2 = 16x + 61$$(y - \frac{2}{3})^2 = \frac{16}{9}(x + \frac{61}{16})$This is of the form $(y - k)^2 = 4a(x - h)$,where the axis of the parabola is given by $y - k = 0$.Here,$k = \frac{2}{3}$,so the axis is $y - \frac{2}{3} = 0$,which simplifies to $3y = 2$.

The axis of the parabola $9y^2 - 16x - 12y - 57 = 0$ is

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