will someone please 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)
To check the work for finding two sets of parametric equations for the given rectangular equation x + y^2 = 4:
1. First set of parametric equations:
x = -y^2 + 4
y = -t^2
To check this, substitute the value of y from the second equation into the first equation:
x + (-t^2)^2 = 4
x + t^4 = 4
So, the first set of parametric equations x = -y^2 + 4 and y = -t^2 satisfies the original equation x + y^2 = 4.
2. Second set of parametric equations:
x = -t^3
y = -t^3 + 4
Again, substitute the value of y from the second equation into the first equation:
x + (-t^3 + 4)^2 = 4
x + t^6 - 8t^3 + 16 = 4
x + t^6 - 8t^3 + 12 = 0
So, the second set of parametric equations x = -t^3 and y = -t^3 + 4 also satisfies the original equation x + y^2 = 4.
Moving on to the second question, to find a polar equation of an ellipse with its focus at the pole, eccentricity e = 1/4, and directrix at y = 4:
The general polar equation for an ellipse with focus at the pole is given by:
r = r0 / (1 - e * cos θ)
1. Given that the eccentricity e = 1/4, we can substitute this value into the equation:
r = r0 / (1 - (1/4) * cos θ)
r = r0 / (1 - cos θ / 4)
2. Since the focus is at the pole, we know that the distance from the pole to the focus is equal to the distance from the pole to the directrix. In this case, the distance is 4 units. So, r0 = 4.
Substituting r0 = 4 into the equation we've derived:
r = 4 / (1 - cos θ / 4)
Simplifying,
r = 4 / (1 - (1/4) * cos θ)
r = 4 / (1 - cos θ / 4)
Therefore, the polar equation of the ellipse with its focus at the pole, an eccentricity of e = 1/4, and the directrix at y = 4, is r = 4 / (1 - cos θ / 4).