If, for non-zero real variables $x,y$, and real parameter $a>1$,
$x:y=(a+1):(a-1)$,
then, the ratio $(x^2\text{–}{y}^2):(x^2+y^2)$ is

Problem Analysis:

- We are given the ratio of two non-zero real variables, $x$ and $y$, in terms of a parameter $a>1$. The given ratio is $x:y=(a+1):(a-1)$.
- The objective is to find the ratio $(x^2\text{–}{y}^2):(x^2+y^2)$ in terms of $a$.

Deriving the Ratio

From the given ratio, we can write:

$\frac{x}{y}=\frac{a+1}{a-1}$

Squaring both sides, we get the ratio of the squares:

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

Let $R=\frac{x^2}{y^2}$. We need to find the ratio $\frac{x^2-y^2}{x^2+y^2}$. To simplify this expression, we can divide both the numerator and the denominator by $y^2$:

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

Substituting and Simplifying

Now, substitute the expression for $R$:

$R-1=\frac{(a+1{)}^2}{(a-1{)}^2}-1=\frac{(a+1{)}^2-(a-1{)}^2}{(a-1{)}^2}$

Using the identity $(A+B{)}^2-(A-B{)}^2=4AB$, where $A=a$ and $B=1$:

$(a+1{)}^2-(a-1{)}^2=4a(1)=4a$

So, $R-1=\frac{4a}{(a-1{)}^2}$.

Similarly, calculate $R+1$:

$R+1=\frac{(a+1{)}^2}{(a-1{)}^2}+1=\frac{(a+1{)}^2+(a-1{)}^2}{(a-1{)}^2}$

Using the identity $(A+B{)}^2+(A-B{)}^2=2(A^2+B^2)$, where $A=a$ and $B=1$:

$(a+1{)}^2+(a-1{)}^2=2(a^2+1^2)=2(a^2+1)$

So, $R+1=\frac{2(a^2+1)}{(a-1{)}^2}$.

Final Ratio Calculation

Now, compute the required ratio $\frac{R-1}{R+1}$:

$\frac{R-1}{R+1}=\frac{\frac{4a}{(a-1{)}^2}}{\frac{2(a^2+1)}{(a-1{)}^2}}$

Cancel out the common term $\frac{1}{(a-1{)}^2}$ (since $a\ne 1$):

$\frac{R-1}{R+1}=\frac{4a}{2(a^2+1)}$

Simplify the expression:

$\frac{R-1}{R+1}=\frac{2a}{a^2+1}$

Therefore, the ratio $(x^2\text{–}{y}^2):(x^2+y^2)$ is $2a:(a^2+1)$.