MCQ
If $A(a{t^2},\,2at),\;B(a/{t^2},\, - 2a/t)$ and $C(a,\,0)$, then $2a$ is equal to
- A$A.M$. of $CA$ and $CB$
- B$G.M$. of $CA$ and $CB$
- ✓$H.M$. of $CA$ and $CB$
- DNone of these
$ = a\sqrt {({t^2} + 1 + 2{t^2})} = a\,(1 + {t^2})$
$CB = \sqrt {{{\left( {\frac{a}{{{t^2}}} - a} \right)}^2} + {{\left( {\frac{{ - 2a}}{t}} \right)}^2}} = a\,\left( {1 + \frac{1}{{{t^2}}}} \right)$
$H.M$. of $CA$ and $CB$ $ = \frac{{2{a^2}\,(1 + {t^2})\,\left( {1 + \frac{1}{{{t^2}}}} \right)}}{{a\,\left[ {1 + {t^2} + 1 + \frac{1}{{{t^2}}}} \right]}} = 2a$.
$\left[ {\because \,\,{\text{H}}{\text{.M}}.\,{\text{of }}x {\text{and }}y = \frac{{2xy}}{{x + y}}} \right]$
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