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. $1<\alpha +\beta <3$
• B. $0<\alpha +\beta <3$
• C. $0<\alpha +\beta <2$
• D. $-1<\alpha +\beta <2$

Answer 1

The correct option is C $0<\alpha +\beta <2$

For the given parabola the value of $a=1$ and vertex at $(\alpha ,\beta )$.
Hence focus of the parabola will be $(\alpha +1,\beta ).$


Clearly, focus must lie to the opposite side of the origin w.r.t. the line $x+y-1=0$ and same side as origin with respect to the line $x+y-3=0$ Hence, $\alpha +\beta >0$ and $\alpha +\beta <2$