**1.) A parabola has a vertex at (-2,3) and its directrix is the line y=5. Write an equation of the parabola and find its focus and axis of symmetry.**

- I thought p = 3 , because its the difference between the y coordinate and the directrix.
- Then I plugged the vertex values for x and y into the equation of a hyperbola like so, (x+2)^2 = 4(-2)(y-3)
- I then expanded both sides out, x^2 + 4x + 4 = -8y + 24
- Then I solves for y, y = \(\displaystyle (-x^2 / 8) - (x / 2) + (5/2)\)

**2.) write an equation of the hyperbola whose foci are at (4,3) and (4,1) and whose eccentricity is (5/3). Find the vertices, center, the axis, and the asymptotes.**

- I know eccentricity is (c/a) or (c/b) depending on the axis. I'm not sure but I think c = 5 and b = 3
- Using the equation c^2 = a^2 + b^2 I solve for a ; a = 4
- Then I think the vertices are (0, 4) ( 0, 5)
- And I believe the axis is the X-axis
- Not sure how to get center, and asymptotes

**3.) For the conic 145x^2 + 120xy+ 180y^2 = 900, change the coordinate system using the rotation of axes formulas to the equation of a conic section in standard position. Indicate coordinates (from the original coordinate system) of all foci, vertices, and equations of all axes and asymptotes.**

- Okay I know this is an ellipsecot 2 (phee) = (145 - 180 /120) = ( -7 / 24)
- 90 degrees < 2(phee) < 180 degrees
- sin(phee) = sqrt(1-(-7/25) / 2) = (4/5)
- cos(phee) = sqrt(1+(-7/25) / 2) = (3/5)
- rotation of axes: x= (3/5)x' + (4/5)y' and y = (4/5)x' + (3/5)y'Not sure how to get foci, vertices, and equations of all axes and asymptotes.