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Brut [27]
3 years ago
9

R = sec(θ) − 2cos(θ), where -π/2 < θ < π/2

Mathematics
1 answer:
Alex3 years ago
4 0

Answer:

  y = (x/(1-x))√(1-x²)

Step-by-step explanation:

The equation can be translated to rectangular coordinates by using the relationships between polar and rectangular coordinates:

  x = r·cos(θ)

  y = r·sin(θ)

  x² +y² = r²

__

  r = sec(θ) -2cos(θ)

  r·cos(θ) = 1 -2cos(θ)² . . . . . . . . multiply by cos(θ)

  r²·r·cos(θ) = r² -2r²·cos(θ)² . . . multiply by r²

  (x² +y²)x = x² +y² -2x² . . . . . . . substitute rectangular relations

  x²(x +1) = y²(1 -x) . . . . . . . . . . . subtract xy²-x², factor

  y² = x²(1 +x)/(1 -x) = x²(1 -x²)/(1 -x)² . . . . multiply by (1-x)/(1-x)

  \boxed{y=\dfrac{x\sqrt{1-x^2}}{1-x}}

__

The attached graph shows the equivalence of the polar and rectangular forms.

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