How can we derive a Cartesian equation for the curve represented by the parametric equations r = 4 tan(θ) and r = sec(θ)?

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To derive a Cartesian equation from the given polar equations r = 4 tan(θ) and r = sec(θ), we start by utilizing the relationships between polar and Cartesian coordinates. In polar coordinates, r is the distance from the origin, and (x, y) are the Cartesian coordinates, which relate as follows:

  • x = r cos(θ)
  • y = r sin(θ)
  • r = √(x² + y²)

Step 1: Understanding the Given Equations

We have two polar equations to consider:

  • 1. r = 4 tan(θ)
  • 2. r = sec(θ)

First, we rewrite the tan(θ) and sec(θ) in terms of sine and cosine:

  • tan(θ) = sin(θ) / cos(θ)
  • sec(θ) = 1 / cos(θ)

Step 2: Expressing r in Terms of x and y

For the first equation, substituting the definition of tan(θ) gives us:

r = 4 (sin(θ) / cos(θ))

Multiplying both sides by cos(θ), results in:

r cos(θ) = 4 sin(θ)

In terms of x and y, this can be rewritten as:

x = 4 y / r

We have also r = √(x² + y²), thus substituting gives:

x = 4y / √(x² + y²)

Step 3: Working with the Second Equation

Next, for the second equation, rewriting sec gives:

r = 1 / cos(θ)

Multiplying by cos(θ) gives:

r cos(θ) = 1

This simplifies to:

x = 1

Hence, we now have two equations:

  • x = 4y / √(x² + y²)
  • x = 1

Step 4: Finding the Curve’s Cartesian Equation

Substituting x = 1 into the first equation: 

1 = 4y / √(1² + y²)

This leads to:

√(1 + y²) = 4y

Squaring both sides gives:

1 + y² = 16y²

Simplifying this results in:

15y² = 1

Thus:

y² = rac{1}{15}

This indicates:

y = rac{1}{	ext{√15}} 	ext{ or } -rac{1}{	ext{√15}}

Conclusion

In conclusion, the Cartesian equation for the curve described by the polar equations r = 4 tan(θ) and r = sec(θ) is comprised of two lines:

  • The vertical line x = 1
  • Horizontal lines at y = ± rac{1}{ ext{√15}}

This identifies the relationship between the curves represented in polar coordinates and their Cartesian counterpart.

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