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silverxx

find the tangents of the curve x^2 + y^2 - 6x + 4y = 0 from (-17, 7) the answers are 2x + 3y + 13 and 34x + 129y = 325

  • 6 months ago
  • 6 months ago

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  1. myininaya
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    You can find y'? I think you mean find the tangent of the curve at (-17,7), right? You can verify that point is on the circle by pluggin it in.

    • 6 months ago
  2. naranjja
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    \[x^2+y^2-6x+4y=0\]Deriving implicitly,\[2x+2y*y'-6+4*y'=0\]\[2y*y'+4*y'=-2x+6\]\[y'(2y+4)=-2x+6\]\[y'=\frac{ 2x+6 }{ 2y+4 }\]Evaluating at the point (-17,7),\[y'=\frac{ 2(-17)+6 }{2(7)+4 }=-\frac{ 28 }{ 18 }=\frac{ 14 }{ 9 }\]This is the slope. Now, using point-slope form,\[y-7=\frac{ 14 }{9 }(x+17)\]\[y=\frac{ 14 }{ 9 }x+\frac{ 301 }{ 9 }\]And that is the tangent that goes through that point.

    • 6 months ago
  3. silverxx
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    yes dy/dx = -2x+6/2y + 4

    • 6 months ago
  4. myininaya
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    hmmm...I don't think that point in on the circle.

    • 6 months ago
  5. silverxx
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    @naranjja you missed -2x + 6.. i think. the sign

    • 6 months ago
  6. naranjja
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    Damn, you're right.

    • 6 months ago
  7. silverxx
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    @myininaya i don't know.. hmm

    • 6 months ago
  8. naranjja
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    \[y'=\frac{ -2x+6 }{ 2y+4 }\]\[y'=\frac{ -2(-17)+6 }{ 2(7)+4 }=\frac{ 40 }{ 18 }=\frac{ 20 }{ 9 }\]Therefore,\[y-7=\frac{ 20 }{ 9 }(x+7)\]\[y=\frac{ 20 }{ 9 }x+\frac{ 203 }{ 9 }\]

    • 6 months ago
  9. myininaya
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    You want to find the tangent to the curve at (-17,7) That point isn't on the circle. Or do you want to find a tangent line to a point on the circle that goes through that point?

    • 6 months ago
  10. silverxx
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    you think it'll satisfy the answers?? it was written in the book hehe

    • 6 months ago
  11. silverxx
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    @myininaya the second

    • 6 months ago
  12. naranjja
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    And I wrote -7 instead of -17, my bad:\[y-7=\frac{ 20 }{ 9 }(x+17)\]\[y=\frac{ 20 }{ 9 }x+\frac{ 403 }{ 9 }\]

    • 6 months ago
  13. silverxx
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    @naranjja there are 2 answers hehe

    • 6 months ago
  14. myininaya
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    Is there anymore info given in the problem? Like the y-intercepts of the lines?

    • 6 months ago
  15. myininaya
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    Oh! Do you mean find the tangent points? And those are the lines that are given in the problem?

    • 6 months ago
  16. silverxx
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    yes yes

    • 6 months ago
  17. myininaya
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    "find the tangents of the curve x^2 + y^2 - 6x + 4y = 0 from (-17, 7) the answers are 2x + 3y =13 and 34x + 129y = 325" So we have 2x+3y=13 34x+129y=325 Put both into y=mx+b form. :)

    • 6 months ago
  18. myininaya
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    You guys already found the general slope m=y'=(-x+3)/(y+2)

    • 6 months ago
  19. silverxx
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    i need to find the process in order to reach the correct answers

    • 6 months ago
  20. silverxx
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    find's not the term but to solve

    • 6 months ago
  21. myininaya
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    Right. We are trying to find the tangent points. I'm telling you how to get there.

    • 6 months ago
  22. myininaya
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    You need to put both of those lines that were given into y=mx+b form.

    • 6 months ago
  23. myininaya
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    And is that one line 2x+3y=-13?

    • 6 months ago
  24. silverxx
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    yes

    • 6 months ago
  25. myininaya
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    I think it is because (-17,7) has to be on it. Ok so you find the slopes of both of those lines. Set it equal to the general slope of the curve you guys found. And solve both equations.

    • 6 months ago
  26. silverxx
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    wait.. my solution's -2x+6/ 2y+4 = y-7/ x- 17

    • 6 months ago
  27. myininaya
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    So we have 2x+3y=-13 34x+129y=325 I'm asking you to put both of those into y=mx+b form so we can identify the slopes of the tangent lines.

    • 6 months ago
  28. silverxx
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    got 2y^2 + 2x^2 - 40x - 10y + 79

    • 6 months ago
  29. silverxx
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    i think i have my own way to solve this..

    • 6 months ago
  30. silverxx
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    but i was just too stupid in not deriving the answers

    • 6 months ago
  31. myininaya
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    look I will do the first one for you it isn't hard We have 2x+3y=-13 Solve for y 3y=-2x-13 y=-2x/3-13 Slope is -2/3 So this means -2/3=(-x+3)/(y+2) Remember (-x+3)/(x+2) is what you guys found above. I just reduced the fraction. Solve for x and y. -2=-x+3 3=y+2 You should get point on a circle. More importantly the tangent point that is on the line 2x+3y=-13 that goes through (-17,7).

    • 6 months ago
  32. silverxx
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    okay :)

    • 6 months ago
  33. myininaya
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    But actually I could be wrong...ERRR...That result isn't on the circle. Let me think a little more.

    • 6 months ago
  34. myininaya
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    oh it is i just don't know how to add :)

    • 6 months ago
  35. myininaya
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    So that is one point. You can find the other using that same process.

    • 6 months ago
  36. myininaya
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    With the other line that was given to you.

    • 6 months ago
  37. silverxx
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    okay..

    • 6 months ago
  38. myininaya
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    34x + 129y = 325 So first step is to write this into y=mx+b form Identify the slope. Put y'=m and solve for x and y to find the other tangent point.

    • 6 months ago
  39. silverxx
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    okay. i'll try it later.. i think i need to digest this..

    • 6 months ago
  40. myininaya
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    You should get a result that isn't even on the circle. However you could find another tangent point whose line that goes through (-17,7) by looking at my workings from above.

    • 6 months ago
  41. silverxx
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    thank thank thank you so much @myininaya :))

    • 6 months ago
  42. myininaya
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    You can get another tangent point by looking at this: -2/3=(-x+3)/(y+2) I did -2=-x+3 , 3=y+2 You have also looked at it as 2/-3=(-x+3)/(y+2) which means you could solve 2=-x+3 and -3=y+2 to find the other point.

    • 6 months ago
  43. myininaya
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    But okay. I understand. Have fun.

    • 6 months ago
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