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Question

If the focus of the parabola (yβ)2=4(xα) 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 C 0<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+y1=0 and same side as origin with respect to the line x+y3=0 Hence, α+β>0 and α+β<2

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