Question

find the tangents of the curve x^2+ y^2 - 6x + 4y = 0 from (-17, 7). the answers are 2x+3y+13=0 and 34x+129y = 325.. already have the process but i didnt derived to the answer. please help.

Answer 1

what i did first was i get the slope through dy/dx

Answer 2

what is your derivative?

Answer 3

okay wait

Answer 4

-2x+6/ 2y+4

Answer 5

there is no need to work it all the way to that step; sinc e y' is what we are after, just input the point values directly.

x^2+ y^2 - 6x + 4y = 0

2x+ 2y y' - 6 + 4 y' = 0

2(-17)+ 2(7) y' - 6 + 4 y' = 0

-34 + 14y' - 6 + 4y' = 0

-40 + 18y' = 0

-20 + 9y' = 0

etc

Answer 6

nothing wrong with working thru all the algebra for y', its just not needed is all

Answer 7

okay.. so, whats the next step? :)

Answer 8

our slope is 9/20, so lets see if we get one of those results at least; the next is finding another suitable x,y that fits the bill

Answer 9

\[y=\frac 9{20}(x+17)+7\]
\[20y=9(x+17)+140\]
\[20y=9x+294\]
\[9x-20y=-294\]
now we see if im reading it correctly to begin with :)

is -17,7 a solution to the given equation?

Answer 10

yes yes

Answer 11

-17,7 is not a solution to: x^2 +y^2 -6x +4y

so what it is saying is that that is a point of in the distance that tangents pass thru

Answer 12

so we do have to use the general format :)

Answer 13

2x+ 2y y' - 6 + 4 y' = 0

x+ y y' - 3 + 2y' = 0

y' = (3-x)/(2+y)

Answer 14

im sorry. i messed up. it is the point on the curve x^2 + y^2 blabla

Answer 15

for some point(s); (a,b) the follwing system holds

\[7 = \frac{3-a}{2+b}(-17-a)+b\]
\[a^2+b^2-6a+4b=0\]

Answer 16

does that make sense? for some tangent slopes coming off of the circle as (a,b), passes thru the point (-17,7) has to satisfy that system

Answer 17

yes it is..

Answer 18

\[7 = \frac{3-a}{2+b}(-17-a)+b\]
\[7(2+b) = (3-a)(-17-a)+b(2+b)\]
\[14+7b = a^2+14a-51+2b+b^2\]hope i saw that last part correctly :)

\[65 = a^2+b^2+14a-5b\]
you have 2 circle equations that intersect at the "tangent" points needed

Answer 19

\[a^2+b^2+14a-5b-65=a^2+b^2-6a+4b\]
\[14a-5b-65=-6a+4b\]
\[20a=9b+65\]
\[a=\frac9{20}b+\frac{13}{4}\]
now we can produce a quadratic in b, to solve for b

Answer 20

im following, okay..

Answer 21

\[a^2+b^2-6a+4b=0\]
\[(\frac9{20}b+\frac{13}{4})^2+b^2-6(\frac9{20}b+\frac{13}{4})+4b=0\]
solve for b, so that you can solve for a; then use those point to define the lines between (a,b) and (-17,7)

I have to get to class for now

Answer 22

yes! thank you!! so much!! Take care :) have to digest this!! :)