Question

Suppose that x=2t/(1+t²) and y=(1-t²)/(1+t²).Show that the value of x²+y² does not depend on the value of t.

Answer 1

x² + y² = 1

Explanation:

Hi there!

We two functions of t

x = 2t/(1 + t²)

y = (1 - t²) / (1 + t²)

First, let´s square x:

x² = [2t/(1 + t²)]²

Apply distributive property

x² = (2t)²/(1+t²)²

x² = 4t²/(1+t²)(1+t²)

Apply distributive property

x² = 4t² /(1 + 2t² + t⁴)

Now let´s square y:

y² = [(1 - t²) / (1 + t²)]²

Apply distributive property

y² = (1 - t²)² / (1 + t²)²

y² = (1 - t²)(1 - t²) / (1 + t²)(1 + t²)

Apply distributive property

y² = (1 - 2t² + t⁴) / ( 1 + 2t² + t⁴)

x² + y² = [4t² /(1 + 2t² + t⁴)] + [(1 - 2t² + t⁴) / ( 1 + 2t² + t⁴)]

Since the denominators are equal, we have to sum the numerators

x² + y² = (4t² + 1 - 2t² + t⁴)/ (1 + 2t² + t⁴)

x² + y² = (1 + 2t² + t⁴) / (1 + 2t² + t⁴)

x² + y² = 1

Then, x² + y² = 1 and therefore does not depend on the value of t