Question

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

• A. $2<a<4$
• B. $-\frac{1}{2}<a<2$
• C. $0<a<2$
• D. $-\frac{1}{2}<a<\frac{3}{2}$

Answer 1

Answer: (B)

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

We have, $y^2-4y=4x-4a$
$\Rightarrow (y-2{)}^2-4=4x-4a$
$\Rightarrow (y-2{)}^2-4=4x-4a$
$\Rightarrow (y-2{)}^2=4[x-(a-1)]$
Hence, vertex is $(a-1,2)$.
Since, vertex lies between the lines $x+y=3$ and $2x+2y-1=0$, then
$(a-1+2-3)(2a-2+4-1)<0$
$\Rightarrow (a-2)(2a+1)<0\Rightarrow a\in (-\frac{1}{2},2)$