Question

Let ${}^{\prime }{a}^{\prime }$ be the variance of the first $25$ natural numbers. If a focal chord of $y^2=4ax$ makes an angle $\alpha \in (0,\frac{\pi }{4}]$ with the positive direction of $x$ axis, then the minimum length of the focal chord is

Answer 1

variance ${\sigma }^2=\frac{\sum (x_i{)}^2}{n}-{\left(\frac{\sum x_i}{n}\right)}^2$
For first $n$ natural numbers
$\sum x_i^2=\frac{n(n+1)(2n+1)}{6}$
$\sum x_i=\frac{n(n+1)}{2}$
${\sigma }^2=\frac{(n+1)(2n+1)}{6}-\frac{(n+1{)}^2}{4}$
$=\frac{(n+1)}{12}[4n+2-3n-3]$
$=\frac{(n^2-1)}{12}$
$a=\frac{(25{)}^2-1}{12}=52$ ($\because n=25$)

Length of the focal chord $=4a{cosec}^2\alpha$
$\Rightarrow 4\times 52{cosec}^2\alpha$
$\Rightarrow 208{cosec}^2\alpha$
Now $0<\alpha \le \frac{\pi }{4}$
$\Rightarrow \sqrt{2}\le cosec\alpha <\infty$
$\Rightarrow 2\le {cosec}^2\alpha <\infty$
Minimum Length $=208\times 2=416$