Question

Ax+by=1 bx+ay=2ab/a2+b2 then (x*2+y*2) (a*2+b*2)=?

Answer 1

$\textbf{Given:}$

$ax+by=1\\\\bx+ay=\frac{2ab}{a^2+b^2}$

$ax+by-1=0\\\\bx+ay-\frac{2ab}{a^2+b^2}=0$

$\text{By cross multiplication rule, we get}$

$\displaystyle\frac{x}{\frac{-2ab^2+a}{a^2+b^2}}=\frac{y}{-b+\frac{2a^2b}{a^2+b^2}}=\frac{1}{a^2-b^2}$

$\implies\displaystyle\frac{x}{\frac{-2ab^2+a^3+ab^2}{a^2+b^2}}=\frac{y}{\frac{-a^2b-b^3+2a^2b}{a^2+b^2}}=\frac{1}{a^2-b^2}$

$\implies\displaystyle\frac{x}{\frac{a^3-ab^2}{a^2+b^2}}=\frac{y}{\frac{a^2b-b^3}{a^2+b^2}}=\frac{1}{a^2-b^2}$

$\implies\displaystyle\frac{x}{\frac{a(a^2-b^2)}{a^2+b^2}}=\frac{y}{\frac{b(a^2-b^2)}{a^2+b^2}}=\frac{1}{a^2-b^2}$

$\implies\displaystyle\frac{x}{\frac{a}{a^2+b^2}}=1$

$\implies\displaystyle\bf\,x=\frac{a}{a^2+b^2}$

$\text{and}$

$\implies\displaystyle\frac{y}{\frac{b}{a^2+b^2}}=1$

$\implies\displaystyle\bf\,y=\frac{b}{a^2+b^2}$

$\text{Now,}$

$\displaystyle\,x^2+y^2=(\frac{a}{a^2+b^2})^2+(\frac{b}{a^2+b^2})^2$

$\displaystyle\,x^2+y^2=\frac{a^2}{(a^2+b^2)^2}+\frac{b^2}{(a^2+b^2)^2}$

$\displaystyle\,x^2+y^2=\frac{a^2+b^2}{(a^2+b^2)^2}$

$\displaystyle\,x^2+y^2=\frac{1}{a^2+b^2}$

$\implies\boxed{\bf(x^2+y^2)(a^2+b^2)=1}$

Answer 2

ax+by=1-----equation 1

bx +ay=2ab/a2 +b2

Let x=0 and a=0 put in equ 1

From equ 1.

y=1/b

(0²+1/b2)(0²+b2)=1/b2 *b2=1

=1. Ans