If {x^2} + 6x + 20y - 51 = 0,then the axis of | vedclass

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(A) Given equation of the parabola is ${x^2} + 6x + 20y - 51 = 0$. Rearranging the terms,we get: ${x^2} + 6x = -20y + 51$. Completing the square on th

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$x + 3 = 0$

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(A) Given equation of the parabola is ${x^2} + 6x + 20y - 51 = 0$. Rearranging the terms,we get: ${x^2} + 6x = -20y + 51$. Completing the square on the left side: ${x^2} + 6x + 9 = -20y + 51 + 9$. This simplifies to: ${(x + 3)^2} = -20y + 60$. Factoring out $-20$ on the right side: ${(x + 3)^2} = -20(y - 3)$. Comparing this with the standard form ${(x - h)^2} = 4a(y - k)$,the axis of the parabola is given by $x - h = 0$. Here,$h = -3$,so the axis is $x + 3 = 0$.

If ${x^2} + 6x + 20y - 51 = 0$,then the axis of the parabola is

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