I need help with some Trig problems...please! Mostly deals with Conic Sections.?

Question

1)Find the center, vertices, and foci for the ellipse: 4x^2+36y^2=144

2) Find the equation of a parabola with its vertex at the origin if its directrix is y = -5
[A] y=20x^2
[B] y= (1/20)x^2
[C] x= (1/20)y^2
[D] x= -5y^2

3) Identify the conic defined by the following equation: 36x^2 = 25-25y^2

4) Solve: x^2 + y^2=16,
((x^2)/4) - ((y^2)/9) = 1

Thank you!!! Any help at all is greatly appreciated!

Answer 1

1)Find the center, vertices, and foci for the ellipse: 4x^2+36y^2=144
x^2/36 + y^2/4 = 1
Center is at (0,0). Vertices at (-6,0 and (6,0)
Foci at (-2sqrt(5),0) and (2sqrt(5), 0)

2) Find the equation of a parabola with its vertex at the origin if its directrix is y = -5. Its focus is at (0,5) So 2p = 20.
y= 20x^2 is answer.

3) Identify the conic defined by the following equation: 36x^2 = 25-25y^2 --> 36x^2 +25y^2 = 25
So x^2/(5/6)^2+y^2 = 1, so its an ellipse.

4) Solve: x^2 + y^2=16,
This is a circle with center at (0,0) and radius = 4

Answer 2

(2x - 4)² = 8y - 16 4x² - 16x + 16 = 8y - 16 8y = 4x² - 16x + 16 + 16 8y = 4x² - 16x + 32 y = 1/2 x² - 2x + 4 y = (1/2 x² - 2x) + 4 y = 1/2(x² - 4x) + 4 y = 1/2(x² - 4x + 4) + 4 - 1/2(4) y = 1/2(x - 2)² + 4 - 2 y = 1/2(x - 2)² + 2 Vertex (2, 2) ¯¯¯¯¯¯¯¯¯¯ h = 2 k = 2 a = 1/2 p = 1 / 4a p = 1 / 4(1/2) p = 1 / (4/2) p = 1/2 Focus (h, k + p) Focus (2, 2 + 1/2) Focus (2, 4/2 + 1/2) Focus (2, 5/2) ¯¯¯¯¯¯¯¯¯¯¯¯ Directrix: y = k - p y = 2 - 1/2 y = 4/2 - 1/2 y = 3/2 ¯¯¯¯¯¯¯