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

2 < a < 4

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

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0 < a < 2

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- \frac{1}{2} < a < \frac{3}{2}

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Solution

The correct option is B $- \frac{1}{2} < a < 2$
$y^{2} - 4 y = 4 x - 4 a \Rightarrow y^{2} - 4 y + 4 = 4 x - 4 a + 4 \Rightarrow (y - 2)^{2} = 4 (x - a + 1)$

On comparing the above equation with the standard equation of the parabola $(y - k)^{2} = 4 (x - h)$ with $(h, k)$ as their vertex, we get the vertex of parabola as $(a - 1, 2)$ .
Since, it is given that the vertex lies between straight lines, $x + y = 3$ and $2 x + 2 y - 1 = 0$

$\Rightarrow (a - 1 + 2 - 3) (2 a - 2 + 4 - 1) < 0$
$\Rightarrow (a - 2) (2 a + 1) < 0$
$\Rightarrow - \frac{1}{2} < a < 2$

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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