If A=t2, 2t, B=1t2,-2t, S=(1, 0) then 1SA+1SB=

A=t2, 2t, B=1t2, -2t, S=(1, 0) SA=(1-t2)2+4t2 =(1+t2)2=1+t2 SB=1-1t22+4t2 =1+1t22 =1+1t2 =t2+1t2 Now 1SA+1SB=11+t2+t2t2+1 =1+t21+t2 =1

If A=t2, 2t, B=1t2,-2t, S=(1, 0) then 1SA+1SB=

A=t2, 2t, B=1t2, -2t, S=(1, 0) SA=(1-t2)2+4t2 =(1+t2)2=1+t2 SB=1-1t22+4t2 =1+1t22 =1+1t2 =t2+1t2 Now 1SA+1SB=11+t2+t2t2+1 =1+t21+t2 =1

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If $A = (t^{2}, 2 t), B = (\frac{1}{t^{2}}, - \frac{2}{t}), S = (1, 0)$ then $\frac{1}{SA} + \frac{1}{SB} =$

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Detailed Solution

$A = (t^{2}, 2 t), B = (\frac{1}{t^{2}}, \frac{- 2}{t}), S = (1, 0) SA = \sqrt{{(1 - t^{2})}^{2} + 4 t^{2}} = \sqrt{{(1 + t^{2})}^{2}} = 1 + t^{2} SB = \sqrt{{(1 - \frac{1}{t^{2}})}^{2} + \frac{4}{t^{2}}} = \sqrt{{(1 + \frac{1}{t^{2}})}^{2}} = 1 + \frac{1}{t^{2}} = \frac{t^{2} + 1}{t^{2}} Now \frac{1}{SA} + \frac{1}{SB} = \frac{1}{1 + t^{2}} + \frac{t^{2}}{t^{2} + 1} = \frac{1 + t^{2}}{1 + t^{2}} = 1$