Consider the parabola y^2+2x+2y-3=0 and match | vedclass

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(A) The given equation of the parabola is $y^2+2x+2y-3=0$.Rewriting the equation by completing the square for $y$:$(y^2+2y+1)-1+2x-3=0$$(y+1)^2+2x-4=0

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$A-III, B-II, C-IV, D-I$

Vedclass · Answer

(A) The given equation of the parabola is $y^2+2x+2y-3=0$.Rewriting the equation by completing the square for $y$:$(y^2+2y+1)-1+2x-3=0$$(y+1)^2+2x-4=0$$(y+1)^2 = -2(x-2)$Comparing this with the standard form $(y-k)^2 = -4a(x-h)$,we get:Vertex $(h, k) = (2, -1)$$-4a = -2 \Rightarrow a = \frac{1}{2}$Axis: $y-k=0 \Rightarrow y+1=0$Focus: $(h-a, k) = (2-\frac{1}{2}, -1) = (\frac{3}{2}, -1)$Directrix: $x = h+a$ $\Rightarrow x = 2+\frac{1}{2}$ $\Rightarrow x = \frac{5}{2}$ $\Rightarrow 2x-5=0$Thus,the matches are:$A \rightarrow III$ (Equation of directrix is $2x-5=0$)$B \rightarrow II$ (Focus is $(\frac{3}{2}, -1)$)$C \rightarrow IV$ (Equation of axis is $y+1=0$)$D \rightarrow I$ (Vertex is $(2, -1)$)Therefore,the correct match is $A-III, B-II, C-IV, D-I$.

$A. \ 2x-5=0$	$I. \ \text{Vertex}$
$B. \ (\frac{3}{2}, -1)$	$II. \ \text{Focus}$
$C. \ y+1=0$	$III. \ \text{Equation of directrix}$
$D. \ (2, -1)$	$IV. \ \text{Equation of the axis}$
	$V. \ \text{Equation of the Latus rectum}$

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