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

The correct option is B $-\frac{1}{2}<a<2$

Vertex of $y^2-4y=4x-4a$ is $(a-1,2)$
So, $(a-1+2-3)(2a-2+4-1)<0$
$(a-2)(2a+1)<0$
$-\frac{1}{2}<a<2$