Consider the parabola x2+4y=0. Let P(a,b) be any fixed point inside the parabola and let S be the focus of parabola. Then the minimum value of SQ+PQ as point Q moves on parabola is:
A
|1−a|
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B
|ab|+1
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C
√a2+b2
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D
1−b
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Solution
The correct option is D1−b Let the foot of perpendicular from Q to the directrix be N. Now, since focal distance of Q=QN So, for SQ+PQ to be minimum QN+PQ should also be minimum since SQ=QN Here, S,P are fixed points and Q,N are variable points. ∴ For QN+PQ to be minimum, P,Q,N should be collinear. So minimum value of SQ+PQ=PQ+QN=PN=1−b