The equation of the given curve is x^2-4x+4y- | vedclass

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(D) The given equation is $x^2-4x+4y-8=0$.Rewriting the equation by completing the square:$x^2-4x+4 = -4y+8+4$$(x-2)^2 = -4(y-3)$Comparing this with $

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

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(D) The given equation is $x^2-4x+4y-8=0$.Rewriting the equation by completing the square:$x^2-4x+4 = -4y+8+4$$(x-2)^2 = -4(y-3)$Comparing this with $(x-h)^2 = 4a(y-k)$,we get $h=2, k=3$,and $4a = -4 \Rightarrow a = -1$.$(B)$ Vertex is $(h, k) = (2, 3)$.$(A)$ Focus is $(h, k+a) = (2, 3-1) = (2, 2)$.$(C)$ One end of the latus rectum is $(h+2a, k+a) = (2+2(-1), 3-1) = (0, 2)$ or $(h-2a, k+a) = (2-2(-1), 3-1) = (4, 2)$. Matching with options,we take $(4, 2)$.$(D)$ Point of intersection of the axis and directrix is $(h, k-a) = (2, 3-(-1)) = (2, 4)$.Thus,the correct matching is $A-V, B-III, C-I, D-IV$.

List-$I$	List-$II$
$(A)$ Focus	$(I)$ $(4,2)$
$(B)$ Vertex	$(II)$ $(3,2)$
$(C)$ One end of the latus rectum	$(III)$ $(2,3)$
$(D)$ Point of intersection of the axis and directrix	$(IV)$ $(2,4)$
	$(V)$ $(2,2)$

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The equation of the given curve is $x^2-4x+4y-8=0$. Match the following:List-$I$List-$II$$(A)$ Focus$(I)$ $(4,2)$$(B)$ Vertex$(II)$ $(3,2)$$(C)$ One end of the latus rectum$(III)$ $(2,3)$$(D)$ Point of intersection of the axis and directrix$(IV)$ $(2,4)$$(V)$ $(2,2)$The correct matching is: