Need help urgently woth CALC 3

J

John45

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Sep 15, 2010
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22
hw.jpg
(click above to see questions)

Need help with questions 2 and 4
Please list step by step process this is due tomorrow morning and I don't have time to work it out, i've shown my work thus far if you want to tell me my errors and then post the solution with the steps you would really be saving my sanity! Thanks!

Work for #2



  • 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

Work for #4


  • Okay I know this is an ellipse cot 2Φ = (145 - 180 /120) = ( -7 / 24)
  • 90 degrees < 2Φ< 180 degrees
  • sinΦ = sqrt(1-(-7/25) / 2) = (4/5)
  • cosΦ = 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.
 
John45 said:
View attachment 2197
(click above to see questions)

Need help with questions 2 and 4
Please list step by step process this is due tomorrow morning and I don't have time to work it out, i've shown my work thus far if you want to tell me my errors and then post the solution with the steps you would really be saving my sanity! Thanks!

Work for #2



  • 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
Click to expand...
You seem to be simply guessing. You are told that the foci are at (4, 1) and (4, 3). You should have learned from the first examples of hyperbolae you saw that the foci lie on the axis so that the axis is x= 4. And the two foci, like the two vertices, are equally spaced from the center- the two foci are at (4, 1) and (4, 3) so the center is at (4, 2).

Work for #4


  • Okay I know this is an ellipse cot 2Φ = (145 - 180 /120) = ( -7 / 24)
  • 90 degrees < 2Φ< 180 degrees
  • sinΦ = sqrt(1-(-7/25) / 2) = (4/5)
  • cosΦ = 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.
Click to expand...
 
HallsofIvy said:
[/LIST]
You seem to be simply guessing. You are told that the foci are at (4, 1) and (4, 3). You should have learned from the first examples of hyperbolae you saw that the foci lie on the axis so that the axis is x= 4. And the two foci, like the two vertices, are equally spaced from the center- the two foci are at (4, 1) and (4, 3) so the center is at (4, 2).
Click to expand...

Halls,

Don't you see -s/he does not have time!!!

We have to spoon-feed the poster - don't just give clues!

Step-by-step solution needed - so that it can be "cut-and-paste"d.
 
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