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

Equation of parabola:
$y^2-4y=4x-4a$
$\Rightarrow y^2-4y+4=4x-4a+4$
$\Rightarrow (y-2{)}^2=4(x-(a-1))$

The vertex of the parabola,
$y-2=0$ and $x-(a-1)=0$
$\Rightarrow y=2$ and $x=a-1$

So $(a-1,2)$ should lie in $x+y=3$ and $x+y=\frac{1}{2}$
At $y=2,x=1,-\frac{3}{2}$
Hence,
$-\frac{3}{2}<a-1<1$
$\Rightarrow -\frac{1}{2}<a<2$