A focal chord of the parabola y2=4ax meets it at P and Q. If S is the focus then 1SP+1SQ=
A
a
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B
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C
2a
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D
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Solution
The correct option is B Let P=(at21,2at1),Q=(at22,2at2) Now focus S=(a,0) Since PQ is a focal chord,t1t2=−1 SP=√(at21−a)2+(2at1−0)2=a√(t21−1)2+4t21=a√(t21−1)2=a(t21−1) SQ=√(at22−a)2+(2at2−0)2=a√(t22−1)2+4t21=a√(t22−1)2=a(t22−1) 1SP+1SQ=1a(t21+1)+1a(t22+1)=1a[1t21+1+1t22+1]=1a[t22+1+t22+1(t22+1)(t22+1)+]=1a(t21+t22+2t21+t21t22+t22+1) =1a[t21+t22+21+t21+t22+1]=1a