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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
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.
\[x^2+y^2-6x+4y=0\]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.
yes dy/dx = -2x+6/2y + 4
hmmm...I don't think that point in on the circle.
@naranjja you missed -2x + 6.. i think. the sign
@myininaya i don't know.. hmm
\[y'=\frac{ -2x+6 }{ 2y+4 }\]\[y'=\frac{ -2(-17)+6 }{ 2(7)+4 }=\frac{ 40 }{ 18 }=\frac{ 20 }{ 9 }\]Therefore,\[y-7=\frac{ 20 }{ 9 }(x+7)\]\[y=\frac{ 20 }{ 9 }x+\frac{ 203 }{ 9 }\]
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?
you think it'll satisfy the answers?? it was written in the book hehe
And I wrote -7 instead of -17, my bad:\[y-7=\frac{ 20 }{ 9 }(x+17)\]\[y=\frac{ 20 }{ 9 }x+\frac{ 403 }{ 9 }\]
@naranjja there are 2 answers hehe
Is there anymore info given in the problem? Like the y-intercepts of the lines?
Oh! Do you mean find the tangent points? And those are the lines that are given in the problem?
"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. :)
You guys already found the general slope m=y'=(-x+3)/(y+2)
i need to find the process in order to reach the correct answers
find's not the term but to solve
Right. We are trying to find the tangent points. I'm telling you how to get there.
You need to put both of those lines that were given into y=mx+b form.
And is that one line 2x+3y=-13?
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.
wait.. my solution's -2x+6/ 2y+4 = y-7/ x- 17
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.
got 2y^2 + 2x^2 - 40x - 10y + 79
i think i have my own way to solve this..
but i was just too stupid in not deriving the answers
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).
But actually I could be wrong...ERRR...That result isn't on the circle. Let me think a little more.
oh it is i just don't know how to add :)
So that is one point. You can find the other using that same process.
With the other line that was given to you.
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.
okay. i'll try it later.. i think i need to digest this..
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.
thank thank thank you so much @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.
But okay. I understand. Have fun.