Question

If \[A(a{{t}^{2}},2at),B\left( \dfrac{a}{{{t}^{2}}},-\dfrac{2a}{t} \right)\] and \[C(a,0)\] then 2a is equal to:
(a) AM of CA and CB
(b) GM of CA and CB
(c) HM of CA and CB
(d) None of these

Answer 1

Hint: Solving this problem would require us the distance formula and some basic formulas of the progression theory. We first find the required distances and then we will find the AM, GM and HM of the distances and verify them.

Complete Step-by-Step solution:
Let us first calculate the distance between C and A which is
\[\begin{align}
  & CA=\sqrt{{{({{x}_{C}}-{{x}_{A}})}^{2}}+{{({{y}_{C}}-{{y}_{A}})}^{2}}} \\
 & CA=\sqrt{{{(a-a{{t}^{2}})}^{2}}+{{(0-2at)}^{2}}} \\
\end{align}\]
Now taking out the factor a as common from all the terms we get,
\[CA=a\sqrt{{{(1-{{t}^{2}})}^{2}}+{{(0-2t)}^{2}}}\]
Squaring both the terms inside we get,
\[\begin{align}
  & CA=a\sqrt{1-2{{t}^{2}}+{{t}^{4}}+4{{t}^{2}}} \\
 & CA=a\sqrt{1+2{{t}^{2}}+{{t}^{4}}} \\
\end{align}\]
Since the terms inside the square root are of the form \[{{a}^{2}}+2ab+{{b}^{2}}\] it becomes \[{{(a+b)}^{2}}\]
Thus,
\[CA=a\sqrt{{{(1+t)}^{2}}}\]
The square and square root mutually cancel each other. So finally we get,
\[CA=a(1+t)...........(i)\]
Now similarly we find CB.
\[\begin{align}
  & CB=\sqrt{{{({{x}_{C}}-{{x}_{B}})}^{2}}+{{({{y}_{C}}-{{y}_{B}})}^{2}}} \\
 & CB=\sqrt{{{\left( a-\dfrac{a}{{{t}^{2}}} \right)}^{2}}+{{\left( 0-\dfrac{2a}{t} \right)}^{2}}} \\
\end{align}\]
Now taking out the factor a as common from all the terms we get,
\[CB=a\sqrt{{{\left( 1-\dfrac{1}{{{t}^{2}}} \right)}^{2}}+{{\left( 0-\dfrac{2}{t} \right)}^{2}}}\]
Squaring both the terms inside we get,
\[\begin{align}
  & CB=a\sqrt{1-\dfrac{2}{t}+\dfrac{1}{{{t}^{2}}}+\dfrac{4}{t}} \\
 & CB=a\sqrt{1+\dfrac{2}{t}+\dfrac{1}{{{t}^{2}}}} \\
\end{align}\]
Since the terms inside the square root are of the form \[{{a}^{2}}+2ab+{{b}^{2}}\] it becomes \[{{(a+b)}^{2}}\]
Thus,
\[CB=a\sqrt{{{\left( 1+\dfrac{1}{t} \right)}^{2}}}\]
The square and square root mutually cancel each other. So finally we get,
\[\begin{align}
  & CB=a\left( 1+\dfrac{1}{t} \right) \\
 & CB=a\left( \dfrac{t+1}{t} \right)...........(ii) \\
\end{align}\]
Now we find the AM, GM and the HM of CA and CB.
\[\begin{align}
  & AM=\dfrac{CA+CB}{2} \\
 & AM=\dfrac{a(t+1)+\dfrac{a(t+1)}{t}}{2} \\
\end{align}\]
Taking out the a(t+1) out as a common factor we get,
\[\begin{align}
  & AM=a(t+1)\left( \dfrac{1+\dfrac{1}{t}}{2} \right) \\
 & AM=a(t+1)\left( \dfrac{1+t}{2} \right) \\
 & AM=\dfrac{a{{(t+1)}^{2}}}{2} \\
\end{align}\]
But according to option(a) and the question 2a is the AM.
So option(a) is wrong.
Now let us calculate the GM.
\[\begin{align}
  & GM={{(CA\times CB)}^{\dfrac{1}{2}}} \\
 & GM={{\left( a(t+1)\times a\dfrac{(t+1)}{t} \right)}^{\dfrac{1}{2}}} \\
\end{align}\]
Taking out a(t+1) out of the square out sign as it is appearing two times we get,
\[GM=\dfrac{a(t+1)}{\sqrt{t}}\]
But according to the question and option(b) 2a should be the GM.
So option(b) is wrong.
Now let’s calculate the HM.
\[\begin{align}
  & HM=\dfrac{2\times CA\times CB}{CA+CB} \\
 & HM=\dfrac{2\times a(t+1)\times \dfrac{a(t+1)}{t}}{a(t+1)+\dfrac{a(t+1)}{t}} \\
\end{align}\]
Now as we can see we can take the a(t+1) from the denominator and cancel one such term in the numerator. Also taking the LCM in the denominator we get,
\[HM=\dfrac{2\times \dfrac{a(t+1)}{t}}{\dfrac{1+t}{t}}\]
Cancelling the \[\dfrac{1+t}{t}\] term,
\[HM=2a\]
Thus according to option(c) and the question 2a is HM.
Thus option(c) is the correct option.

Note: You might get confused with the signs of the terms like a and (t+1) while taking them out of the root like what if they are negative and they will become complex while taking out of the root.
But you need not worry while taking them out of the root, because first of all the distance cannot be negative. But if you still have a doubt just include a modulus sign in that particular step and then remove it the next step. This will not confuse you…!!