Question

derivative question, please help!

Answer 1

Hi, this is the question I'm struggling with: Find equation of both the tangent lines to the ellipse \[x^2 +4y^2=36\] that pass through the point (12,3)

Answer 2

The derivative of the circle equation would be: \[2x+8yy'=0\] right?

Answer 3

correct so far

Answer 4

okay, so I'm confused about what to do next.

Answer 5

get y' all by itself

Answer 6

Okay. so that would give me\[y'= \frac{ -x }{ 4y }\] correct?

Answer 7

good, now this gives us the slope of the tangent line for any given point on the ellipse

use this to define a general equation in point slope form of a line

Answer 8

\[y-y_o=m(x-x_o)\]
\[y=mx-mx_o+y_o\]
\[m=y'~:~ (x_o,y_o)=given~point\]

Answer 9

so I should find out y'(12)? Since both lines need to pass through (12,3)?

Answer 10

the given point is most likely not on the ellipse; so no. At the moment leave y' as is and just insert it into the line equation

Answer 11

so then the line equation would be \[y=y'x-y'12+3?\]

Answer 12

yes, or at least thats the idea i have going thru my head at the moment; im also considering if using the tangent equation for any point on the ellipse would make life easier

Answer 13

http://facultypages.morris.umn.edu/~smukherj/HW4_calc1.pdf I'm not sure if that link will work, but question #5 is the one we're doing. It shows a picture of it and a hint.

Answer 14

nice :) yeah, my second thought was more in line with the "hint"

Answer 15

so, given a general point on the ellipse (a,b) and the y'

the equation of the tangent line at any given point is:\[y=\frac{-a}{4b}x+\frac{a}{4b}a+b\]

Answer 16

when x=12 and y=3; we should be able to restructure this setup

Answer 17

\[3=\frac{-a}{4b}12+\frac{a}{4b}a+b\]
\[3=\frac{-3a}{b}+\frac{a^2}{4b}+b\]
\[3=\frac{-3a}{b}+\frac{a^2}{4b}+b\]something along these lines. we can replace a and b with x and y prolly to help put it into a more usual format

Answer 18

That isn't the finale answer is it?

Answer 19

*final

Answer 20

no its not. but if we work it out we should get a set of points (prolly another ellipse) that we can compare to the original ellipse and see where they intersect

\[3b=-3a+\frac14a^2+b^2\]
\[0=\frac14a^2-3a+b^2-3b\]
\[0=\frac14(a^2-12a)+(b^2-3b)\]
\[0=\frac14(a^2-12a+36)-\frac{36}4+(b^2-3b+\frac94)-\frac94\]
\[0=\frac14(a-6)^2+(b-\frac94)^2-\frac{45}4\]gonna replace a and b now
\[1=\frac1{45}(a-6)^2+(b-\frac94)^2\]

i prolly missed some mathing in that .... but should give the general idea

Answer 21

lol, forgot to actually replace a and b with an x and y :/

Answer 22

I'm sorry, I'm a bit confused..

Answer 23

http://www.wolframalpha.com/input/?i=%28x-6%29%5E2%2F45%2B4%28y-9%2F4%29%5E2%2F45%3D1%2C+x%5E2%2B4y%5E2%3D36

the picture might help. and im sure there might be a simpler way to approach this.

Answer 24

So what's the equation of the second ellipse?

Answer 25

ideally it would be 3=12y'-xy'+y

Answer 26

so should I get y' by itself from that equation?

Answer 27

no, thats the general setup of the second ellipse given the information we found. y'=-x/4y

Answer 28

so I need to replace -x/4y into the equation wherever there's a y'?

Answer 29

yes

Answer 30

i think i might have made a divide by zero error thru my mathing

this, I believe, is the proper results

http://www.wolframalpha.com/input/?i=x%5E2%2B4y%5E2%3D36%2C12%28-x%2F4y%29%2Bx%28-x%2F4y%29%2By%3D3

Answer 31

so then \[3=\frac{ 12x }{ 4y}+\frac{ x^2 }{ 4y }+y\]

Answer 32

-12x ... but yes

Answer 33

\[3=\frac{ 3x }{ y }+\frac{ x^2 }{ 4y }\]

Answer 34

oh okay, so -3x

Answer 35

now equare that setup with the original ellipse

Answer 36

\[3=\frac{ -3x }{ y }+\frac{ x^2 }{ 4y }\]

Answer 37

\[x^2+4y^2-36=-\frac{ 3x }{ y }+\frac{ x^2 }{ 4y }-3\]
\[x^2+4y^2-36+\frac{ 3x }{ y }-\frac{ x^2 }{ 4y }+3=0\]

Answer 38

Okay. Should I solve for x or y?

Answer 39

you should prolly solve for a function of x; but mathically it doesnt really matter

Answer 40

Ugh, I can't figure out what either one is

Answer 41

\[x^2+4y^2-36+\frac{ 3x }{ y }-\frac{ x^2 }{ 4y }+3=0\]
\[x^2+4y^2-36+\frac{ 12x }{4 y }-\frac{ x^2 }{ 4y }+3=0\]
\[\frac{4yx^2+16y^3-36(4y)+12x-x^2+12y}{4y}=0\]
\[4x^2y+16y^3-144y+12x-x^2+12y=0\]

yeah, its being a pain

Answer 42

if we solve for x, we can use that in the original ellipse equation to determine our 2 points

Answer 43

Yeah, it's just really hard finding x. lol

Answer 44

gotta run

Answer 45

okay. thanks for the help