Question

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

Answer 1

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.

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

Damn, you're right.

Answer 7

@myininaya i don't know.. hmm

Answer 8

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

Answer 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?

Answer 10

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

Answer 11

@myininaya the second

Answer 12

And I wrote -7 instead of -17, my bad:\[y-7=\frac{ 20 }{ 9 }(x+17)\]\[y=\frac{ 20 }{ 9 }x+\frac{ 403 }{ 9 }\]

Answer 13

@naranjja there are 2 answers hehe

Answer 14

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

Answer 15

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

Answer 16

yes yes

Answer 17

"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. :)

Answer 18

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

Answer 19

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

Answer 20

find's not the term but to solve

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

yes

Answer 25

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.

Answer 26

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

Answer 27

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.

Answer 28

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

Answer 29

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

Answer 30

but i was just too stupid in not deriving the answers

Answer 31

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).

Answer 32

okay :)

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

With the other line that was given to you.

Answer 37

okay..

Answer 38

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.

Answer 39

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

Answer 40

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.

Answer 41

thank thank thank you so much @myininaya :))

Answer 42

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.

Answer 43

But okay. I understand. Have fun.