CALC 3 Hyperbolas, Parabolas, and elipse

Apr 2010
29
0
Need help with 3 questions.

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.
 
Jan 2008
484
109
UK
1) looks fine. since the directrix is horizontal the axis of symmetry is vertical, through the vertex.

2) Haven't looked at it.

3) You have a slip somewhere.

I have

x = 4/5 x' + 3/5 y'

and

y= -3/5 x' + 4/5 y'

I lazily subbed these into 145x^2 + 120 xy + 180y^2 =900 using a CAS and it simplified to 100x'^2 + 225 y'^2 = 900.

This can be further simplified to (x'/3)^2 + (y'/2)^2 = 1.