Question

There are two points $A(a,0)$ and $B(b,0)$ on positive $x$-axis, where $0<a<b$. Let $P(0,y)$ be a variable point on positive $y$-axis. Then the maximum value of $\angle APB$ occurs when $y$ is

• A. Arithmetic mean of $a\text{and}b$
• B. Geometric mean of $a\text{and}b$
• C. Harmonic mean of $a\text{and}b$
• D. $y^3=ab$

Answer 1

The correct option is B Geometric mean of $a\text{and}b$


From fig. we need to find the maximum value of $\theta$
$\tan (\alpha +\theta )=\frac{\tan \alpha +\tan \theta }{1-\tan \alpha \tan \theta }$
$\Rightarrow \frac{b}{y}=\frac{\frac{a}{y}+\tan \theta }{1-\frac{a}{y}\tan \theta }$
$\Rightarrow \frac{b}{y}-\frac{ab}{y^2}\tan \theta =\frac{a}{y}+\tan \theta$
$\Rightarrow (1+\frac{ab}{y^2})\tan \theta =\frac{b-a}{y}$
$\Rightarrow \tan \theta =\frac{(b-a)y}{y^2+ab}$

Let $f=\tan \theta =\frac{(b-a)y}{y^2+ab}$
For $\theta$ to be maximum, $f$ should also be maximum.
$\frac{df}{dy}=\frac{(y^2+ab)(b-a)-(b-a)y\left(2y\right)}{(y^2+ab{)}^2}=0$
$\Rightarrow y^2+ab-2y^2=0\Rightarrow y^2=ab...\left(i\right)$

$\frac{df}{dy}=\frac{(ab-y^2)(b-a)}{(y^2+ab{)}^2}$
$\frac{d^{2}f}{d{y}^2}{|}_{y^2=ab}=\frac{(y^2+ab{)}^2(b-a\left)\right(-2y)-(b-a\left)\right(ab-y^2\left)2\right(y^2+ab)2y}{(y^2+ab{)}^4}$
$=\frac{\left(2ab{)}^2\right(b-a\left)\right(-2\sqrt{ab})}{(2ab{)}^4}$
$=\frac{-(b-a)\left(2\sqrt{ab}\right)}{(2ab{)}^2}$

$\therefore \frac{d^{2}f}{d{y}^2}<0$
Hence, at $y^2=ab,\theta \to \text{maximum}$

$\therefore y\text{is G.M. of}a,b$.