If the focus of parabola $(y-k)^2=4(x-h)$ always lies between the lines $x+y=1$ and $x+y$ $=3$ then
If the focus of parabola $(y-k)^2=4(x-h)$ always lies between the lines $x+y=1$ and $x+y$ $=3$ then
$0 \lt h+k \lt 2$
$0 \lt h+k \lt 1$
$1 \lt h+k \lt 2$
$1 \lt h+k \lt 3$
Solution
Coordinate of focus will be $(h+1, k)$
Now focus should lie to the opposite side of origin with respect to line $x+y-1=0$ and same side as origin with respect to line $x+y-3=0$
Hence $h+k\gt0$ and $h+k \lt 2$.
Hence $h+k\gt0$ and $h+k \lt 2$.