Question

If $x$ is very large as compare to $y$, then the value of $k$ in $\sqrt{\frac{x}{x+y}}\sqrt{\frac{x}{x-y}}=1+\frac{y^2}{k{x}^2}$ ?

Answer 1

$x>\ge y$

$y/a\to 0$

$LHS:\sqrt{\frac{x}{x+y}}\times \sqrt{\frac{x}{x-y}}-1+\frac{y^2}{k{x}^2}$

$=\sqrt{\frac{x62}{x^2-y^2}}$

$={\left(\frac{1}{1-y^2/x^2}\right)}^{1/2}$

$={(1-\frac{y^2}{x^2})}^{-1/2}$

$n$ is $-ve$ & $1\times 1<1$

$(1+x{)}^n=1+nx+\frac{n(n-1)}{x}\times x^{2}e....$

$\left|\frac{-y^2}{x^2}\right|<1$

$n\to -1/2$

$x\to -y^2/x^2$

${(1-\frac{y^2}{x^2})}^{-1/2}=1+-1/2\frac{-y^2}{x^2}$

$=1+\frac{y^2}{2x^2}$

$k=2$