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

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

answer is A.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Given, points are

A = (t^{2}, 2 t), B = (\frac{1}{t^{2}}, - \frac{2}{t}), S = (1, 0)

Distance between $' S' and' A' = \sqrt{{(t^{2} - 1)}^{2} + {(2 t)}^{2}}$

$S A = \sqrt{t^{4} + 1 - 2 t^{2} 4 t^{2}}$

$S A = \sqrt{{(t^{2} + 1)}^{2}}$

$S A = t^{2} + 1$

Distance between $' S' and' B' = \sqrt{{(1 - \frac{1}{t^{2}})}^{2} + {(\frac{2}{t})}^{2}}$

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

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

$S B = \frac{1}{t^{2}} \sqrt{{(t^{2} + 1)}^{2}}$

$S B = \frac{t^{2} + 1}{t^{2}}$

$∴ \frac{1}{S A} + \frac{1}{S B} = \frac{1}{t^{2} + 1} + \frac{1}{\frac{t^{2} + 1}{t^{2}}}$

$= \frac{1}{t^{2} + 1} + \frac{t^{2}}{t^{2} + 1} = 1$

Watch 3-min video & get full concept clarity