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If the vertex of the conic $y^{2}-4 y=4 x-4 a$ always lies between the straight lines $x+y=3$ and $2 x+2 y-1=0 .$ then
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The correct answer is:
$-\frac{1}{2} < a < 2$
We have, $y^{2}-4 y=4 x-4 a$ $\Rightarrow \quad(y-2)^{2}-4=4 x-4 a$
$\Rightarrow \quad(y-2)^{2}=4 x-4 a+4$
$\Rightarrow \quad(y-2)^{2}=4|x-(a-1)|$
Hence, vertex is $(a-1,2)$
Since, vertex lies between the lines $x+y=3$ and $2 x+2 y-1=0,$ then
$(a-1+2-3)(2 a-2+4-1) < 0$
$\Rightarrow(a-2)(2 a+1) < 0 \Rightarrow a \in\left(-\frac{1}{2}, 2\right)$
$\Rightarrow \quad(y-2)^{2}=4 x-4 a+4$
$\Rightarrow \quad(y-2)^{2}=4|x-(a-1)|$
Hence, vertex is $(a-1,2)$
Since, vertex lies between the lines $x+y=3$ and $2 x+2 y-1=0,$ then
$(a-1+2-3)(2 a-2+4-1) < 0$
$\Rightarrow(a-2)(2 a+1) < 0 \Rightarrow a \in\left(-\frac{1}{2}, 2\right)$
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