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

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Question

(A) Given the equation of the parabola: $9y^2 - 12y - 16x - 57 = 0$. Rearranging the terms: $9y^2 - 12y = 16x + 57$. Completing the square for the $y$

Vedclass · Accepted Answer

$3y = 2$

Vedclass · Answer

(A) Given the equation of the parabola: $9y^2 - 12y - 16x - 57 = 0$. Rearranging the terms: $9y^2 - 12y = 16x + 57$. Completing the square for the $y$ terms: $(3y)^2 - 2(3y)(2) + 2^2 = 16x + 57 + 4$. $(3y - 2)^2 = 16x + 61$. $9(y - 2/3)^2 = 16(x + 61/16)$. $(y - 2/3)^2 = \frac{16}{9}(x + 61/16)$. The axis of the parabola is given by the linear term inside the square: $3y - 2 = 0$,which simplifies to $3y = 2$.

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

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