Tangent Lines to an Ellipse Passing Through an Outside Point

[horacescope]

Homework Statement

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

Homework Equations

The equation of an ellipse is x²/a² + y²/b² = 1.
I converted the given equation to
x²/36 + y²/9 = 1 by dividing each value by 36.

The Attempt at a Solution

My online homework website is telling me I'm half right (I found the equation of ONE of the tangent lines to the ellipse passing through (12,3)). Here's what I have:

I took the derivative of the equation of the ellipse and got y' = -x/4y.
I set this value equal to the slope of a line passing through the point (12,3), (y-3)/(x-12). I then solved for x² and y² to find

x²=-4y² + 12y + 12x
y²=-x²/4 + 3x + 3y

I then substituted these equations, one at a time, for the values of x² and y² in the original equation of the ellipse and solved for the unknown value. I think I took an unnecessary step doing that, but I need the practice! I found that

x=0
y=±3

I then determined the slopes of the lines passing through (0,3)-->(12,3) and (0,-3)-->(12,3). Clearly, the line running from (0,3)-->(12,3) has a slope of 0, and the equation of the line is y=3. This is the part that I got correct.
The slope of the line running through (0,-3)-->(12,3) I found to be 1/2. However, my homework website is telling me the equation y=1/2x-3 is incorrect.

Any assistance you can give me would be very appreciated! Honestly, we haven't gone over ANY problems like this in class, so the fact I'm even half right has been enormously satisfying. But I'd like to know what I'm doing wrong... I have a feeling I probably did something stupid, like dropped a negative sign somewhere. There are only 6 questions on this section's homework, so as it is I have an 83.3% and can't make an A unless I get this last piece of the last question correct. So frustrating!

Thanks in advance!

 

[Dick]

Homework Statement

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

Homework Equations

The equation of an ellipse is x²/a² + y²/b² = 1.
I converted the given equation to
x²/36 + y²/9 = 1 by dividing each value by 36.

The Attempt at a Solution

My online homework website is telling me I'm half right (I found the equation of ONE of the tangent lines to the ellipse passing through (12,3)). Here's what I have:

I took the derivative of the equation of the ellipse and got y' = -x/4y.
I set this value equal to the slope of a line passing through the point (12,3), (y-3)/(x-12). I then solved for x² and y² to find

x²=-4y² + 12y + 12x
y²=-x²/4 + 3x + 3y

I then substituted these equations, one at a time, for the values of x² and y² in the original equation of the ellipse and solved for the unknown value. I think I took an unnecessary step doing that, but I need the practice! I found that

x=0
y=±3

I then determined the slopes of the lines passing through (0,3)-->(12,3) and (0,-3)-->(12,3). Clearly, the line running from (0,3)-->(12,3) has a slope of 0, and the equation of the line is y=3. This is the part that I got correct.
The slope of the line running through (0,-3)-->(12,3) I found to be 1/2. However, my homework website is telling me the equation y=1/2x-3 is incorrect.

Any assistance you can give me would be very appreciated! Honestly, we haven't gone over ANY problems like this in class, so the fact I'm even half right has been enormously satisfying. But I'd like to know what I'm doing wrong... I have a feeling I probably did something stupid, like dropped a negative sign somewhere. There are only 6 questions on this section's homework, so as it is I have an 83.3% and can't make an A unless I get this last piece of the last question correct. So frustrating!

Thanks in advance!

Homework Statement

Homework Equations

The Attempt at a Solution

If you got x^2=-4y^2 + 12y + 12x from your slope relation, then that tells you x^2+4y^2=12y+12x. x^2+4y^2=36. So 36=12y+12x. Solve that for y and substitute back into the ellipse equation. You should get two values of x that satisfy it. Not just x=0.

 

[horacescope]

Thanks! Of course it was right in front of me!