If Pat2,2at,Qat2,−2at and S(a,0) be any three points then 1SP+1SQ=

SP=at2−a2+4a2t2=a2t2−12+4t2=a2t2+12=at2+1SQ=at2−a2+4a2t2=a21d2+12=a1t2+11SP+1SQ=1a(1+t2)+t2a(1+t2)=1a

If Pat2,2at,Qat2,−2at and S(a,0) be any three points then 1SP+1SQ=

SP=at2−a2+4a2t2=a2t2−12+4t2=a2t2+12=at2+1SQ=at2−a2+4a2t2=a21d2+12=a1t2+11SP+1SQ=1a(1+t2)+t2a(1+t2)=1a

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$If P (a t^{2}, 2 a t), Q (\frac{a}{t^{2}}, - \frac{2 a}{t}) and S (a, 0) be any three points then \frac{1}{S P} + \frac{1}{S Q} =$

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$\frac{3}{a}$

$\frac{1}{a}$

$\frac{2}{a}$

answer is C.

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Detailed Solution

$S P = \sqrt{{(a t^{2} - a)}^{2} + 4 a^{2} t^{2}}$

$= \sqrt{a^{2} {(t^{2} - 1)}^{2} + 4 t^{2}} = \sqrt{a^{2} {(t^{2} + 1)}^{2}} = a (t^{2} + 1)$

$S Q = \sqrt{{(\frac{a}{t^{2}} - a)}^{2} + \frac{4 a^{2}}{t^{2}}} = \sqrt{a^{2} {(\frac{1}{d^{2}} + 1)}^{2}} = a (\frac{1}{t^{2}} + 1)$

$\frac{1}{S P} + \frac{1}{S Q} = \frac{1}{a (1 + t^{2})} + \frac{t^{2}}{a (1 + t^{2})} = \frac{1}{a}$

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If Pat2,2at,Qat2,−2at and S(a,0) be any three points then 1SP+1SQ=

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$If P (a t^{2}, 2 a t), Q (\frac{a}{t^{2}}, - \frac{2 a}{t}) and S (a, 0) be any three points then \frac{1}{S P} + \frac{1}{S Q} =$