Question

If the focus of the parabola ${(y-\beta )}^2=4(x-\alpha )$ always lies between the lines $x+y=1$ and $x+y=3$, then

• A. $1<\alpha +\beta <2$
• B. $0<\alpha +\beta <1$
• C. $0<\alpha +\beta <2$
• D. None of these

Answer 1

Answer: (C)

$0<\alpha +\beta <2$

We know that focus of the parabola $Y^2=4aX$ is $(X=a,Y=0)$
Thus focus of the parabola $(y-\beta {)}^2=4(x-\alpha )$ is $(x-\alpha =a,y-\beta =0)$
which is $(\alpha +1,\beta )$
Now it is given that this point lies between the lines $x+y=1$ and $x+y=3$
Thus $(\alpha +1)+\beta >1$ and $(\alpha +1)+\beta <3$
$\alpha +\beta >0$ and $\alpha +\beta <2$
Hence $0<\alpha +\beta <2$