The ratio of the $A.M.$ and $G.M.$ of two positive numbers $a$ and $b,$ is $m: n .$ Show that $a: b=(m+\sqrt{m^{2}-n^{2}}):(m-\sqrt{m^{2}-n^{2}})$

Let the two numbers be $a$ and $b$

$A.M.=\frac{a+b}{2}$ and $G.M.=\sqrt{a b}$

According to the given condition,

$\frac{a+b}{2 \sqrt{a b}}=\frac{m}{n}$

$\Rightarrow \frac{(a+b)^{2}}{4(a b)}=\frac{m^{2}}{n^{2}}$

$\Rightarrow(a+b)^{2}=\frac{4 a b m^{2}}{n^{2}}$

$\Rightarrow(a+b)=\frac{2 \sqrt{a b} m}{n}$         ........$(1)$

Using this in the identity $(a-b)^{2}=(a+b)^{2}-4 a b,$ we obtain

$(a-b)^{2}=\frac{4 a b m^{2}}{n^{2}}-4 a b=\frac{4 a b\left(m^{2}-n^{2}\right)}{n^{2}}$

$\Rightarrow(a-b)=\frac{2 \sqrt{a b} \sqrt{m^{2}-n^{2}}}{n}$       .........$(2)$

Adding $(1)$ and $(2),$ we obtain

$2 a=\frac{2 \sqrt{a b}}{n}(m+\sqrt{m^{2}-n^{2}})$

$\Rightarrow a=\frac{\sqrt{a b}}{n}(m+\sqrt{m^{2}-n^{2}})$

Substituting the value of $a$ in $(1),$ we obtain

$b=\frac{2 \sqrt{a b}}{n} m-\frac{\sqrt{a b}}{n}(m+\sqrt{m^{2}-n^{2}})$

$=\frac{\sqrt{a b}}{n} m-\frac{\sqrt{a b}}{n} \sqrt{m^{2}-n^{2}}$

$=\frac{\sqrt{a b}}{n}(m-\sqrt{m^{2}-n^{2}})$

$\therefore a:b = \frac{a}{b} = \frac{{\frac{{\sqrt {ab} }}{n}(m + \sqrt {{m^2} - {n^2}} )}}{{\frac{{\sqrt {ab} }}{n}(m - \sqrt {{m^2} - {n^2}} )}} = \frac{{(m + \sqrt {{m^2} - {n^2}} )}}{{(m - \sqrt {{m^2} - {n^2}} )}}$

Thus, $a: b=(m+\sqrt{m^{2}-n^{2}}):(m-\sqrt{m^{2}-n^{2}})$