anonymous
  • anonymous
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
Mathematics
schrodinger
  • schrodinger
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myininaya
  • myininaya
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.
anonymous
  • anonymous
\[x^2+y^2-6x+4y=0\]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.
anonymous
  • anonymous
yes dy/dx = -2x+6/2y + 4

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myininaya
  • myininaya
hmmm...I don't think that point in on the circle.
anonymous
  • anonymous
@naranjja you missed -2x + 6.. i think. the sign
anonymous
  • anonymous
Damn, you're right.
anonymous
  • anonymous
@myininaya i don't know.. hmm
anonymous
  • anonymous
\[y'=\frac{ -2x+6 }{ 2y+4 }\]\[y'=\frac{ -2(-17)+6 }{ 2(7)+4 }=\frac{ 40 }{ 18 }=\frac{ 20 }{ 9 }\]Therefore,\[y-7=\frac{ 20 }{ 9 }(x+7)\]\[y=\frac{ 20 }{ 9 }x+\frac{ 203 }{ 9 }\]
myininaya
  • myininaya
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?
anonymous
  • anonymous
you think it'll satisfy the answers?? it was written in the book hehe
anonymous
  • anonymous
@myininaya the second
anonymous
  • anonymous
And I wrote -7 instead of -17, my bad:\[y-7=\frac{ 20 }{ 9 }(x+17)\]\[y=\frac{ 20 }{ 9 }x+\frac{ 403 }{ 9 }\]
anonymous
  • anonymous
@naranjja there are 2 answers hehe
myininaya
  • myininaya
Is there anymore info given in the problem? Like the y-intercepts of the lines?
myininaya
  • myininaya
Oh! Do you mean find the tangent points? And those are the lines that are given in the problem?
anonymous
  • anonymous
yes yes
myininaya
  • myininaya
"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. :)
myininaya
  • myininaya
You guys already found the general slope m=y'=(-x+3)/(y+2)
anonymous
  • anonymous
i need to find the process in order to reach the correct answers
anonymous
  • anonymous
find's not the term but to solve
myininaya
  • myininaya
Right. We are trying to find the tangent points. I'm telling you how to get there.
myininaya
  • myininaya
You need to put both of those lines that were given into y=mx+b form.
myininaya
  • myininaya
And is that one line 2x+3y=-13?
anonymous
  • anonymous
yes
myininaya
  • myininaya
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.
anonymous
  • anonymous
wait.. my solution's -2x+6/ 2y+4 = y-7/ x- 17
myininaya
  • myininaya
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.
anonymous
  • anonymous
got 2y^2 + 2x^2 - 40x - 10y + 79
anonymous
  • anonymous
i think i have my own way to solve this..
anonymous
  • anonymous
but i was just too stupid in not deriving the answers
myininaya
  • myininaya
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).
anonymous
  • anonymous
okay :)
myininaya
  • myininaya
But actually I could be wrong...ERRR...That result isn't on the circle. Let me think a little more.
myininaya
  • myininaya
oh it is i just don't know how to add :)
myininaya
  • myininaya
So that is one point. You can find the other using that same process.
myininaya
  • myininaya
With the other line that was given to you.
anonymous
  • anonymous
okay..
myininaya
  • myininaya
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.
anonymous
  • anonymous
okay. i'll try it later.. i think i need to digest this..
myininaya
  • myininaya
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.
anonymous
  • anonymous
thank thank thank you so much @myininaya :))
myininaya
  • myininaya
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.
myininaya
  • myininaya
But okay. I understand. Have fun.

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