Question

Find equations of both the tangent lines to the ellipse
x2 + 4y2 = 36
that pass through the point (12, 3).

Answer 1

might want to develop the derivative of the ellipse for a any general point

Answer 2

can you derive the ellipse using implicit differentation?

Answer 3

wouldn't it just be 2x+8y=0?

Answer 4

yes, but with a y' added in for good measure

Answer 5

2x+8yy'=0

y'=-x/y for any (x,y) on the ellipse

Answer 6

deleted me 4

Answer 7

y'=-x/4y

Answer 8

so now I just plug in the point(12,3) to get the slop for the tangent equation?

Answer 9

slope**

Answer 10

so lets take a general point (a,b) along the ellipse to define its slope

y' = -a/4b also has to be the slope of the line that goes thru (12,3)
\[y=-\frac{a}{4b}(x-12)+3\]but (a,b) also on that line
\[b=-\frac{a}{4b}(a-12)+3\]
\[b=-\frac{a^2}{4b}-\frac{12}{4b}+3\]
\[4b^2=-a^2+12+12b\]
\[a^2+4b^2-12b=12\]
\[a^2+4(b^2-3b)=12\]
\[a^2+4(b^2-3b+\frac94-\frac94)=12\]
\[a^2+4(b-3)^2-9=12\]
\[a^2+4(b-3)^2=21\]
\[\frac{a^2}{21}+\frac{4(b-3)^2}{21}=1\]

make sure i did those right

Answer 11

essentially, we create another ellipse; where this ellipse intersects the original ellipse will define the 2 points that are needed

Answer 12

i see one error, that doesnt really change things to much\[(b-\frac32)^2\]

Answer 13

wouldn't it be b=−a2/4b+12a/4b+3?

Answer 14

and then we solve from there? (this is near the very top)

Answer 15

yes, then the last term becomes 3*4b/4b = 12b/4b

and we can collect all the numberators into one place; cross multiply the 4b over; and simplify

Answer 16

i fixed the typo in the next line :)

Answer 17

oh, that makes sense. Okay, I think I understand it for the most part.

Answer 18

its fairly simple, it just making sure you dont trip over te algebra along the way :)

Answer 19

so I understand all the algebra but I am unsure on how to find the exact equations of the tangent lines... can you help me on that?

Answer 20

well, to see what it is graphically

http://www.wolframalpha.com/input/?i=y%3D-x%2F%284y%29%28x-12%29%2B3%2C+x%5E2+%2B+4y%5E2+-+36+%3D+0

Answer 21

set the equations of the ellipses equal to each other; im not sure that i typed out the process correctly tho ...

Answer 22

\[b=-\frac{a}{4b}(a-12)+3\]
\[4b^2=-a(a-12)+12b\]
\[4b^2-12b=-(a^2-12a)\]
\[4(b^2-3b+\frac94)-9=-(a^2-12a+36)+36\]
\[4(b-\frac32)^2-9=-(a-6)^2+36\]
\[(a-6)^2+4(b-\frac32)^2=45\]
thats better

Answer 23

that makes sense :) so how do I get from there to the equations? I am lost :/

Answer 24

I am confused on how to find the equations of the tangent lines