The distance between (at^22at) and ( a t^2 , -2a t ) - Vidyadip

MCQ

The distance between $(\text{at}^2\text{2at}) $ and $ (\frac{\text{a}}{\text{t}^2},\frac{-2\text{a}}{\text{t}})$

✓
$\text{a}(\text{t}+\frac{1}{\text{t}})^2$
B
$\text{a}(\text{t}+\frac{1}{\text{t}})\text{units}$
C
$(\text{t}+\frac{1}{\text{t}})^2\text{units}$
D
$\text{a}(\text{t}-\frac{1}{\text{t}})^2\text{units}$

✓

Answer

Correct option: A.

$\text{a}(\text{t}+\frac{1}{\text{t}})^2$

The ditance between $(\text{at}^2\text{2at}) $ and $ (\frac{\text{a}}{\text{t}^2},\frac{-2\text{a}}{\text{t}})$
$=\sqrt{\Bigg(\frac{\text{a}}{\text{t}^2}{\text{-at}^2\Bigg)}^2\Bigg(\frac{\text{-2a}}{\text{t}}{\text{-at}^2\Bigg)}^2}$
$=\text{a}\sqrt{\frac{1}{\text{t}^4}+\text{t}^4-2+\frac{4}{\text{t}^2}+4\text{t}^2+8}$
$=\text{a}\sqrt{\frac{1}{\text{t}^4}+\text{t}^4+\frac{4}{\text{t}^2}+4\text{t}^2+6}$
$=\text{a}\sqrt{\frac{1}{\text{t}^4}+\text{t}^4-4+2+\frac{4}{\text{t}^2}+4\text{t}^2}$
$=\sqrt{(\text{t}^2+\frac{1}{\text{t}^2}+2)}^2$
$=\text{a}(\text{t}^2+\frac{1}{\text{t}^2}+2)^2$
$=\text{a}(\text{t}+\frac{1}{\text{t}})^2\text{units}$

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