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.

x² + y² = 1

Step-by-step 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

(0,,,0)

step-by-step explanation:

the table of x and y values is shown in the attached pic(part of question)

we have to decide according that graph

given points are (0, 0) (–1, 0), (2, 0)

when we graph these points on graph they show an increasing line as shown in the attached pic