can you check my work?
find two sets of parametric equations for the given rectangular equation
x+y^2=4
x=-y^2+4
y=-t^2
x=-t^2+4
x=-t^3
y=-t^3+4
Find a polar equation of an ellipse with its focus at the pole an eccentricity of e=1/4 and directrix at y=4.
answer: √(x^2 + y^2) / (y - 4) = 1/4
4√(x^2 + y^2) = y - 4
4r = r sin θ - 4
(sin θ - 4) r = 4
r = 4/(sin θ - 4)
For the first question, you have provided two sets of parametric equations for the given rectangular equation x + y^2 = 4. Both sets are correct:
1. x = -y^2 + 4
y = -t^2
2. x = -t^2 + 4
y = -t^2
These parametric equations represent curves in the xy-plane that satisfy the original rectangular equation.
Now, for the second question, you need to find a polar equation for an ellipse with its focus at the pole, an eccentricity of e = 1/4, and the directrix at y = 4.
To find the polar equation, we can use the definition of an ellipse in polar coordinates. The definition states that the distance from any point on the ellipse to the focus divided by the distance from the same point to the directrix is equal to the eccentricity (e).
Here's how to solve it:
1. Given that the focus is at the pole (r = 0), the polar equation will have the form r = A / (1 - e * cos θ), where r is the distance from the pole, θ is the angle, and A is a constant.
2. Since the focus is at the pole, the distance from the focus to any point on the ellipse is simply r.
3. The distance from any point on the ellipse to the directrix, which is a horizontal line y = 4, can be determined using the formula: d = r * (1 - e^2) / (1 - e * cos θ), where d is the distance.
4. The eccentricity e is given as 1/4.
5. Plugging in the values, we get the equation: r = A / (1 - (1/4) * cos θ).
So, the polar equation of the ellipse is r = A / (1 - (1/4) * cos θ).
I apologize for the incorrect answer I initially provided.