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

- anonymous

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- schrodinger

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- 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

\[x^2+y^2-6x+4y=0\]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

yes dy/dx = -2x+6/2y + 4

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## More answers

- myininaya

hmmm...I don't think that point in on the circle.

- anonymous

@naranjja you missed -2x + 6.. i think. the sign

- anonymous

Damn, you're right.

- anonymous

@myininaya i don't know.. hmm

- 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

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

you think it'll satisfy the answers?? it was written in the book hehe

- anonymous

@myininaya the second

- 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

@naranjja there are 2 answers hehe

- myininaya

Is there anymore info given in the problem?
Like the y-intercepts of the lines?

- myininaya

Oh! Do you mean find the tangent points? And those are the lines that are given in the problem?

- anonymous

yes yes

- 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

You guys already found the general slope m=y'=(-x+3)/(y+2)

- anonymous

i need to find the process in order to reach the correct answers

- anonymous

find's not the term but to solve

- myininaya

Right. We are trying to find the tangent points.
I'm telling you how to get there.

- myininaya

You need to put both of those lines that were given into y=mx+b form.

- myininaya

And is that one line 2x+3y=-13?

- anonymous

yes

- 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

wait.. my solution's -2x+6/ 2y+4 = y-7/ x- 17

- 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

got 2y^2 + 2x^2 - 40x - 10y + 79

- anonymous

i think i have my own way to solve this..

- anonymous

but i was just too stupid in not deriving the answers

- 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

okay :)

- myininaya

But actually I could be wrong...ERRR...That result isn't on the circle. Let me think a little more.

- myininaya

oh it is
i just don't know how to add :)

- myininaya

So that is one point. You can find the other using that same process.

- myininaya

With the other line that was given to you.

- anonymous

okay..

- 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

okay. i'll try it later.. i think i need to digest this..

- 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

thank thank thank you so much @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

But okay. I understand. Have fun.

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