Question

$A=(a{t}^2,2at)\text{and}B=(\frac{a}{t^2},\frac{-2a}{t}),\text{them}AB=?$

• A. $a{(t-\frac{1}{t})}^2$
• B. $a{(1+\frac{1}{t})}^2$
• C. $a{(t+\frac{1}{t})}^2$
• D. $a(t^2-\frac{1}{t^2})$

Answer 1

The correct option is C $a{(t+\frac{1}{t})}^2$

$Formulaused$
$distanceformula=\sqrt{{(y_2-y_1)}^2+{(x_2-x_1)}^2}$
$(x^2-y^2)=(x+y)(x-y)$
$Given$
$A=(a{t}^2,2at),B=(\frac{a}{t^2},\frac{-2a}{t})$
$AB=\sqrt{{(\frac{-2a}{t}-2at)}^2+{(\frac{a}{t^2}-a{t}^2)}^2}$
$=\sqrt{4a^2{(\frac{-1}{t}-t)}^2+a^2{(\frac{1}{t^2}-t^2)}^2}$
$=a\sqrt{4{(\frac{1}{t}+t)}^2+{\left((\frac{1}{t}+t)(\frac{1}{t}-t)\right)}^2}$
$=a\sqrt{{(\frac{1}{t}+t)}^2[4+{(\frac{1}{t}-t)}^2]}$
$=a(\frac{1}{t}+t)\sqrt{\frac{1}{t^2}+t^2+2}$
$=a(\frac{1}{t}+t)\sqrt{{(\frac{1}{t}+t)}^2}$
$=a{(\frac{1}{t}+t)}^2$