If the vertex of the conic y^{2}-4y=4x-4a alw | vedclass

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(B) The given equation of the conic is $y^{2}-4y=4x-4a$.Completing the square for $y$,we get $y^{2}-4y+4=4x-4a+4$,which simplifies to $(y-2)^{2}=4(x-a

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$-\frac{1}{2} < a < 2$

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(B) The given equation of the conic is $y^{2}-4y=4x-4a$.Completing the square for $y$,we get $y^{2}-4y+4=4x-4a+4$,which simplifies to $(y-2)^{2}=4(x-a+1)$.Comparing this with the standard form $(y-k)^{2}=4A(x-h)$,the vertex is $(h, k) = (a-1, 2)$.Let the vertex be $(x_{0}, y_{0}) = (a-1, 2)$.The vertex lies between the lines $x+y-3=0$ and $2x+2y-1=0$.Rewriting the lines in terms of $x+y$:Line $1$: $x+y=3$.Line $2$: $2(x+y)=1 \implies x+y=\frac{1}{2}$.Since the vertex lies between these two parallel lines,the value of $x+y$ for the vertex must lie between the values of $x+y$ for the lines.Thus,$\frac{1}{2} Substituting $x_{0}+y_{0} = (a-1)+2 = a+1$:$\frac{1}{2} Subtracting $1$ from all parts:$\frac{1}{2}-1 < a < 3-1$,which gives $-\frac{1}{2} < a < 2$.

If the vertex of the conic $y^{2}-4y=4x-4a$ always lies between the straight lines $x+y=3$ and $2x+2y-1=0$,then

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