Convert to Rectangular: r*tanΘ/secΘ=2
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y=2
y=½
x=2
x=½
To convert a polar equation to rectangular form, we need to use the following conversions:
1. r = √(x^2 + y^2) (distance from the origin to the point)
2. Θ = arctan(y / x) (angle made with the positive x-axis)
Let's go step by step and convert each of the given polar coordinates to rectangular form:
1. r * tan(Θ) / sec(Θ) = 2
First, let's simplify the equation using trigonometric identities:
tan(Θ) = sin(Θ) / cos(Θ)
sec(Θ) = 1 / cos(Θ)
Substituting these values into the equation:
r * (sin(Θ) / cos(Θ)) / (1 / cos(Θ)) = 2
Simplifying further:
r * sin(Θ) = 2
Now, using the conversion formulas:
r = √(x^2 + y^2)
Θ = arctan(y / x)
We can rewrite the equation as:
√(x^2 + y^2) * sin(arctan(y / x)) = 2
To solve for x and y, we need to square both sides of the equation to eliminate the square root:
(x^2 + y^2) * sin^2(arctan(y / x)) = 4
Simplifying further, using the trigonometric identity:
sin^2(arctan(y / x)) = (y / x)^2 / ((y / x)^2 + 1)
We can rewrite the equation as:
(x^2 + y^2) * (y / x)^2 / ((y / x)^2 + 1) = 4
Next, let's simplify the equation by multiplying both sides by (y / x)^2 + 1:
(x^2 + y^2) * (y^2) = 4 * ((y / x)^2 + 1) * (x^2)
Expanding the equation:
x^2 * y^2 + y^4 = 4 * (y^2 + x^2)
We can rearrange the equation to get it in the standard quadratic form:
x^2 * y^2 - 4 * x^2 - 3 * y^2 = 0
Now, we have a quadratic equation in terms of x^2 and y^2. To solve for x and y, we can substitute u = x^2 and v = y^2:
u * v - 4 * u - 3 * v = 0
This equation represents a hyperbola.
To find the rectangular form of this equation, we need more information about the variables x and y. Is there any additional context or equations provided?
since tanθ/secθ = sinθ, can you handle
r sinθ = 2
??