For the parabola y^2+6y-2x+5=0,match the item | vedclass

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

Vedclass · Accepted Answer

$I$-$F$,$II$-$A$,$III$-$C$,$IV$-$E$

Vedclass · Answer

(B) Given equation: $y^2+6y-2x+5=0$Completing the square for $y$: $(y^2+6y+9) - 9 - 2x + 5 = 0$$(y+3)^2 = 2x + 4$$(y+3)^2 = 2(x+2)$Comparing with $(y-k)^2 = 4a(x-h)$,we get vertex $V(h, k) = (-2, -3)$. This matches $F$.Here,$4a = 2$,so $a = \frac{1}{2}$.Focus is $(h+a, k) = (-2 + \frac{1}{2}, -3) = (-\frac{3}{2}, -3)$. This matches $A$.Equation of directrix is $x = h - a = -2 - \frac{1}{2} = -\frac{5}{2}$,which implies $2x + 5 = 0$. This matches $C$.Equation of the axis is $y = k$,so $y = -3$,which implies $y + 3 = 0$. This matches $E$.Thus,the correct matching is $I-F, II-A, III-C, IV-E$.

List-$I$ (Geometric Property)	List-$II$ (Coordinates/Equations)
$I$. Vertex	$A$. $\left(-\frac{3}{2}, -3\right)$
$II$. Focus	$B$. $\left(\frac{3}{2}, -3\right)$
$III$. Equation of the directrix	$C$. $2x + 5 = 0$
$IV$. Equation of the axis	$D$. $2x + y + 3 = 0$
	$E$. $y + 3 = 0$
	$F$. $(-2, -3)$

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