Find the vertex,focus,axis,length of latus re | vedclass

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(B) Given parabola: $y^{2} = 4x + 4y$Rearranging the terms: $y^{2} - 4y = 4x$Completing the square: $y^{2} - 4y + 4 = 4x + 4$$(y - 2)^{2} = 4(x + 1)$C

Vedclass · Accepted Answer

$(-1, 2), (0, 2), y = 2, 4, x = -2$

Vedclass · Answer

(B) Given parabola: $y^{2} = 4x + 4y$Rearranging the terms: $y^{2} - 4y = 4x$Completing the square: $y^{2} - 4y + 4 = 4x + 4$$(y - 2)^{2} = 4(x + 1)$Comparing with the standard form $Y^{2} = 4AX$,where $Y = y - 2$,$X = x + 1$,and $A = 1$.Vertex: $(X = 0, Y = 0) \implies (x + 1 = 0, y - 2 = 0) \implies (-1, 2)$.Focus: $(X = A, Y = 0) \implies (x + 1 = 1, y - 2 = 0) \implies (0, 2)$.Axis: $Y = 0 \implies y - 2 = 0 \implies y = 2$.Length of latus rectum: $4A = 4(1) = 4$.Equation of directrix: $X = -A \implies x + 1 = -1 \implies x = -2$.

Find the vertex,focus,axis,length of latus rectum,and the equation of the directrix for the parabola $y^{2} = 4x + 4y$.

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