Question

If ${\text{x}}$ varies as ${\text{y}}$ , prove that ${{\text{x}}^2} + {\text{ }}{{\text{y}}^2}$ varies as ${{\text{x}}^2}{\text{ - }}{{\text{y}}^2}$

Answer 1

In this question we have that prove the expression given. For that we are going to use the binomial expansion to prove the problem.
Here binomial expansion of powers of a binomial. According to the theorem, it is possible to expand the polynomial ${\left( {{\text{x + y}}} \right)^{\text{n}}}$ into a sum involving terms of the form ${\text{a}}{{\text{x}}^{\text{b}}}{{\text{y}}^{\text{c}}}$, where the exponents ${\text{b}}$ and ${\text{c}}$ are nonnegative integers with ${\text{b + c = n}}$, and the coefficients a of each term is a specific positive integer depending on ${\text{n}}$ and ${\text{b}}$.
Here we use some constant that belongs to the next step by dividing the terms, we prove it.

Complete step-by-step answer:
Given that ${\text{x}}$ varies as ${\text{y}}$,
To prove that ${{\text{x}}^2} + {\text{ }}{{\text{y}}^2}$ varies as ${{\text{x}}^2}{\text{ - }}{{\text{y}}^2}$
Now, Let us consider that ${\text{x = ky}}$ where ${\text{k}}$ is a constant. Then
Now we consider that ${{\text{x}}^2} + {\text{ }}{{\text{y}}^2}$
Here substitute the value of ${\text{x}}$ that we have taken as,
Now, ${{\text{x}}^2}{\text{ + }}{{\text{y}}^2} = {{\text{k}}^2}{{\text{y}}^2} + {{\text{y}}^2}$
Taking common terms of this equation, we get
$= {{\text{y}}^2}\left( {{{\text{k}}^2} + 1} \right)$
Now we consider from given equation is ${{\text{x}}^2}{\text{ - }}{{\text{y}}^2}$
Here substitute the value of ${\text{x}}$ that we have taken as,
Now, ${{\text{x}}^2}{\text{ - }}{{\text{y}}^2} = {{\text{k}}^2}{{\text{y}}^2}{\text{ - }}{{\text{y}}^2}$
Taking common terms of this equation, we get
$= {{\text{y}}^2}\left( {{{\text{k}}^2} - 1} \right)$
Next, we divide the term of given equation as ${{\text{x}}^2} + {\text{ }}{{\text{y}}^2}$ varies as ${{\text{x}}^2}{\text{ - }}{{\text{y}}^2}$,
$\dfrac{{{{\text{x}}^2}{\text{ + }}{{\text{y}}^2}{\text{ }}}}{{{{\text{x}}^2}{\text{ - }}{{\text{y}}^2}}}{\text{ }} = {\text{ }}\dfrac{{\left( {{{\text{k}}^2} + {\text{ }}1} \right)}}{{\left( {{{\text{k}}^2} - {\text{ }}1} \right)}}$
Now, let us take that ${\text{l}} = {\text{ }}\dfrac{{\left( {{{\text{k}}^2} + {\text{ }}1} \right)}}{{\left( {{{\text{k}}^2} - {\text{ }}1} \right)}}$, a constant
That implies, $\dfrac{{{{\text{x}}^2}{\text{ + }}{{\text{y}}^2}{\text{ }}}}{{{{\text{x}}^2}{\text{ - }}{{\text{y}}^2}}}{\text{ }} = {\text{ l}}$
By cross multiplication of this equation, we get that
${{\text{x}}^2} + {{\text{y}}^2} = {\text{l}}\left( {{{\text{x}}^2} - {{\text{y}}^2}} \right)$ where $\dfrac{{{{\text{x}}^2}{\text{ + }}{{\text{y}}^2}{\text{ }}}}{{{{\text{x}}^2}{\text{ - }}{{\text{y}}^2}}}{\text{ }} = {\text{ }}\dfrac{{\left( {{{\text{k}}^2} + {\text{ }}1} \right)}}{{\left( {{{\text{k}}^2} - {\text{ }}1} \right)}}$ a constant
Hence proved.

Note: A binomial is a polynomial with two variables. It describes the algebraic expansion of the powers. The theorem and its generalizations can be used to prove results and solve problems in combinatorics, algebra, calculus, and many other areas of mathematics.