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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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Verified Answer
The correct answer is:
$-\frac{1}{2} < a < 2$
Equation of the vertex of conic
$$
y^{2}-4 y=4 x-4 a
$$
$\Rightarrow \quad y^{2}-4 y+4=4 x-4 a+4$
$\Rightarrow \quad(y-2)^{2}=4[x-(a-1)]$
$\therefore$
Vertex $=(a-12)$
From figure,
clearly. $-\frac{3}{2} < a-1 < 1$
$\therefore$
$-\frac{1}{2} < a < 2$
$$
y^{2}-4 y=4 x-4 a
$$
$\Rightarrow \quad y^{2}-4 y+4=4 x-4 a+4$
$\Rightarrow \quad(y-2)^{2}=4[x-(a-1)]$
$\therefore$
Vertex $=(a-12)$
From figure,
clearly. $-\frac{3}{2} < a-1 < 1$
$\therefore$
$-\frac{1}{2} < a < 2$
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