Question

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

Answer 1

find y'

Answer 2

2x+8yy'=0

Answer 3

8yy'=-2x

y'=-2x/(8y)

y'=-x/(4y)

Answer 4

now plug in (12,3) to figure out the slope of the tangent line

Answer 5

I distinctly remember doing this earlier today . Obviously gogetta you're in the same class as someone else! Or maybe same text book!

Answer 6

then use (12,3) and y' evaluated at (12,3) to find the equation of the tangent line

Answer 7

y'=-12/(4 *3)=-1 is slope

Answer 8

yeah i made a mistake!

Answer 9

\[y=-1x+b \]
\[3=-1(12)+b => b=15 \]

Answer 10

y=-x+15

Answer 11

@myininaya you made a mistake as well

Answer 12

\[(12,3)\] is not on the graph. you have to find two lines tangent to the graph that pass through
\[(12,3)\]

Answer 13

omg that point is not even on the graph lol

Answer 14

yes i just checked lol

Answer 15

No -- you've solved assuming that (12,3) lies on the ellipse. But it most definitely does not, as that point doesn't satisfy the ellipse equation.

So, the question is a little more subtle.

Answer 16

i didn't read the question lol

Answer 17

i think you need to reason that if the point (x,y) is on the graph then
\[\frac{y-3}{x-12}=-1\] and also that
\[x^2+4y^2=36\] and solve from there

Answer 18

@ jamesj apparently you solved this already, so instead of me messing up the algebra maybe you can link to it

Answer 19

Hmm ... it was about 8 hours ago. Let's just do it again.

Note first that it is not necessarily true that slope of the line will be -1

Answer 20

Draw a diagram and convince yourself of that.

Answer 21

oh right the whole thing is wrong. in fact the slope cannot be -1

Answer 22

What we need to do is write down the general equation for a tangent line and find those that pass through (12,3)

Answer 23

should be
\[\frac{y-3}{x-12}=-\frac{x}{4y}\]

Answer 24

Right AND we know the x and y satisfy the ellipse equation.

Now solve.

Answer 25

i'll leave that as an exercise...

Answer 26

lol

Answer 27

i must have messed this up somewhere. think my algebra is gone

Answer 28

Let's just do it.

Multiplying out
4y^2 - 12y = -x^2 + 12x
x^2 + 4y^2 = 12(x + y)

Now LHS = 36 so x + y = 3

Answer 29

yeah i got that far...

Answer 30

ok now i see . we can write
\[y=3-x\] and put
\[x^2+4(3-x)^2=36\] so get y

Answer 31

Something's wrong.

Answer 32

While you guys figure out the hard tangent, the easy one is y= 3

Answer 33

There's a mistake somewhereI can't put my finger on it.

Answer 34

what i am wondering is when you solve
\[\frac{y-3}{x-12}=-\frac{x}{4y}\] and
\[x^2+4y^2=36\] don't you get an infinite number of solutions?
it just comes out to
\[x+y=3\]

Answer 35

oh wait...

Answer 36

\[x^2+4(3-x)^2=36\]
\[5x^2-24x=0\]
\[x=0,x=\frac{24}{5}\] whew

Answer 37

Yes, but that's problematic, because the tangents at x = 0 are the lines y = 2, y =-2

Answer 38

So what about those at x = 24/5 ... ah yes, those work.