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silverxx

  • one year ago

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

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  1. myininaya
    • one year ago
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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.

  2. naranjja
    • one year ago
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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.

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

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

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

  6. naranjja
    • one year ago
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    Damn, you're right.

  7. silverxx
    • one year ago
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    @myininaya i don't know.. hmm

  8. naranjja
    • one year ago
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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,\[y-7=\frac{ 20 }{ 9 }(x+7)\]\[y=\frac{ 20 }{ 9 }x+\frac{ 203 }{ 9 }\]

  9. myininaya
    • one year ago
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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?

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

  11. silverxx
    • one year ago
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    @myininaya the second

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

  13. silverxx
    • one year ago
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    @naranjja there are 2 answers hehe

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

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

  16. silverxx
    • one year ago
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    yes yes

  17. myininaya
    • one year ago
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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. :)

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

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

  20. silverxx
    • one year ago
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    find's not the term but to solve

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

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

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

  24. silverxx
    • one year ago
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    yes

  25. myininaya
    • one year ago
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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.

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

  27. myininaya
    • one year ago
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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.

  28. silverxx
    • one year ago
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    got 2y^2 + 2x^2 - 40x - 10y + 79

  29. silverxx
    • one year ago
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    i think i have my own way to solve this..

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

  31. myininaya
    • one year ago
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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).

  32. silverxx
    • one year ago
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    okay :)

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

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

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

  36. myininaya
    • one year ago
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    With the other line that was given to you.

  37. silverxx
    • one year ago
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    okay..

  38. myininaya
    • one year ago
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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.

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

  40. myininaya
    • one year ago
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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.

  41. silverxx
    • one year ago
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    thank thank thank you so much @myininaya :))

  42. myininaya
    • one year ago
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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.

  43. myininaya
    • one year ago
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    But okay. I understand. Have fun.

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