If the focus of the parabola (y−β)2=4(x−α) always lies between the lines x+y=1 and x+y=3 then
A
1<α+β<3
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B
0<α+β<3
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C
0<α+β<2
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D
−1<α+β<2
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Solution
The correct option is C0<α+β<2 For the given parabola the value of a=1 and vertex at (α,β). Hence focus of the parabola will be (α+1,β).
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, α+β>0 and α+β<2