I need help on Finding the tangent of the ellipse

1. The problem statement, all variables and given/known data
The problem is

Find the equation of the line with a positive slope that is tangent to the ellipse
(x^2)/9 + (y^2)/4 = 1
At x=2

2. Relevant equations
Now I know that to find the tangent, I find the derivative of the equation. So I got
2x/9 + 2y/4 dy/dx = 0

But its this part where I cant go any further. See I only get X, so I cant solve for dy/dx without having y in the answer, and therefor I cant solve for y.

I solved for dy/dx and got -16/18y, and just tp test I found what was Y when x = 2 for the original equation, it was 0.37.
So I adding 0.37 into -16/18y, i got -2.4 (which would be the slope of the line)

But it says find the line with the positive slope, after reading through the entire chapter in my textbook, and searching the internet, I am stuck and don't know what I'm doing wrong.

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 it should be be 2x/9 + 2y/4 dy/dx = 0 If x = 2, then (y^2)/4 = 1 - 4/9 solve for y.

 Quote by courtrigrad it should be be 2x/9 + 2y/4 dy/dx = 0
Sorry that was a typo, I originally had that, and typed it out wrong.

I need help on Finding the tangent of the ellipse

If x = 2, then (y^2)/4 = 1 - 4/9

solve for y.

 Quote by courtrigrad If x = 2, then (y^2)/4 = 1 - 4/9 solve for y.
I did solve for y, it was 0.37 but my answer was negative and therefor couldnt be it.

 y = sqrt(20)/3 or -sqrt(20)/3 dy/dx = -8x / 18y and y = -sqrt(20)/3 so it will be positive
 and y = -sqrt(20)/3 Where did you get that from. When I solved for y in the original equation. I get 0.4444 + (y^2)/4 = 1 0.55556 / 4 = y^2 Sqrt(0.138) = y y = 0.37
 $$\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$$ $$\frac{y^{2}}{4} = 1 - \frac{4}{9} = \frac{5}{9}$$ $$y^{2} = \frac{20}{9}$$ $$y = \pm \frac{\sqrt{20}}{3}$$

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