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

• A

  $2 < a < 4$
• Correct answer
  B

  $-\frac{1}{2} < a < 2$
• C

  $0 < a < 2$
• D

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

Solution

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. $-\frac{3}{2} < a-1 < 1$ $\therefore$ $-\frac{1}{2} < a < 2$

Hence, the correct answer is option: 'B'.