If the focus of parabola $(y - k)^{2} = 4 (x - h)$ always lies between $x + y = 1$ and $x + y = 3$ then :

0 < h + k < 2

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0 < h + k < 1

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1 < h + k < 2

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1 < h + k < 3

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Solution

The correct option is B $0 < h + k < 2$
We know that the focus of the parabola is $(h + 1, k)$
It is given that focus lies between $x + y = 1$ and $x + y = 3$
that is
$1 < x + y < 3$
$1 < h + 1 + k < 3$
$0 < h + k < 2$

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