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

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

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

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(B) The given equation is $y^{2}-4y=4x-4a$. Completing the square for $y$, we get $(y-2)^{2}-4=4x-4a$. This simplifies to $(y-2)^{2}=4x-4a+4$, or $(y-2)^{2}=4(x-(a-1))$. Comparing this with the standard form $(y-k)^{2}=4A(x-h)$, the vertex is $(h, k) = (a-1, 2)$. The vertex lies between the lines $L_1: x+y-3=0$ and $L_2: 2x+2y-1=0$. For a point $(x_0, y_0)$ to lie between two parallel lines $ax+by+c_1=0$ and $ax+by+c_2=0$, the expressions $(ax_0+by_0+c_1)$ and $(ax_0+by_0+c_2)$ must have opposite signs, i.e.,$(ax_0+by_0+c_1)(ax_0+by_0+c_2) Substituting the vertex $(a-1, 2)$ into the lines: $L_1(a-1, 2) = (a-1)+2-3 = a-2$. $L_2(a-1, 2) = 2(a-1)+2(2)-1 = 2a-2+4-1 = 2a+1$. Thus, $(a-2)(2a+1) Solving this inequality, we find $a \in \left(-\frac{1}{2}, 2\right)$.

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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