Question

Question(1)
2
1
If A = (t2, 2t), B =
S = (1, 0) then
+
2
t2
t
SA
SB
1
B 2.
с 3
D
4​

Answer 1

SOLUTION

GIVEN

$\displaystyle \sf{A = ( {t}^{2} , 2t) \: \: , \: B = \bigg( \frac{1}{ {t}^{2} } , - \frac{2}{t} \bigg) \:, \: S = (1,0) }$

TO CHOOSE THE CORRECT OPTION

$\displaystyle \sf{ \frac{1}{SA} + \frac{1}{SB} }$

A. 1

B. 2

C. 3

D. 4

EVALUATION

Here it is given that

$\displaystyle \sf{A = ( {t}^{2} , 2t) \: \: , \: B = \bigg( \frac{1}{ {t}^{2} } , - \frac{2}{t} \bigg) \:, \: S = (1,0) }$

∴ SA

$\displaystyle \sf{ = \sqrt{{( {t}^{2} - 1)}^{2} + {(2t - 0)}^{2} } }$

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

$\displaystyle \sf{ = \sqrt{{( {t}^{2} + 1)}^{2} } }$

$\displaystyle \sf{ = ( {t}^{2} + 1) }$

Again

SB

$\displaystyle \sf{ = {\sqrt{ {\bigg( \frac{1}{ {t}^{2} } - 1 \bigg)}^{2} + { \bigg( - \frac{2}{t} - 0 \bigg)}^{2}} } }$

$\displaystyle \sf{ = {\sqrt{ {\bigg( \frac{1}{ {t}^{2} } - 1 \bigg)}^{2} + \frac{4}{ {t}^{2} } } } }$

$\displaystyle \sf{ = {\sqrt{ {\bigg( \frac{1}{ {t}^{2} } + 1 \bigg)}^{2} } } }$

$\displaystyle \sf{ = \bigg( \frac{1}{ {t}^{2} } + 1 \bigg) }$

$\displaystyle \sf{ = \bigg( \frac{1 + {t}^{2} }{ {t}^{2} } \bigg) }$

$\displaystyle \sf{ \therefore \: \: \frac{1}{SA} + \frac{1}{SB} }$

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

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

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

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

$= 1$

FINAL ANSWER

The correct option is A. 1

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