For the parabola y^2+6y-2x+5=0:(I) The vertex | vedclass

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(B) Given equation of the parabola is $y^2+6y-2x+5=0$.Completing the square for $y$:$y^2+6y+9-9-2x+5=0$$(y+3)^2-4-2x=0$$(y+3)^2=2(x+2)$.Comparing this

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$I$ is true,$II$ is false

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(B) Given equation of the parabola is $y^2+6y-2x+5=0$.Completing the square for $y$:$y^2+6y+9-9-2x+5=0$$(y+3)^2-4-2x=0$$(y+3)^2=2(x+2)$.Comparing this with the standard form $(y-k)^2=4a(x-h)$,we get:Vertex $(h, k) = (-2, -3)$. Thus,statement $(I)$ is true.Here,$4a=2$,so $a=\frac{1}{2}$.The directrix of the parabola $(y-k)^2=4a(x-h)$ is given by $x=h-a$.$x = -2 - \frac{1}{2} = -\frac{5}{2}$.$2x+5=0$. Thus,statement $(II)$ is false.Therefore,$I$ is true and $II$ is false.

For the parabola $y^2+6y-2x+5=0$:
$(I)$ The vertex is $(-2,-3)$
$(II)$ The directrix is $y+3=0$
Which of the following is correct?

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For the parabola $y^2+6y-2x+5=0$:$(I)$ The vertex is $(-2,-3)$$(II)$ The directrix is $y+3=0$Which of the following is correct?