I on Finding the tangent of the ellipse

[Dave J]

Homework Statement

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

Homework 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 = 0But its this part where I can't go any further. See I only get X, so I can't solve for dy/dx without having y in the answer, and therefor I can't 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.

 

[courtrigrad]

it should be be 2x/9 + 2y/4 dy/dx = 0

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

solve for y.

 

[Dave J]

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.

 

[courtrigrad]

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

solve for y.

 

[Dave J]

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 couldn't be it.

 

[courtrigrad]

y = sqrt(20)/3 or -sqrt(20)/3dy/dx = -8x / 18y

and y = -sqrt(20)/3so it will be positive

 

[Dave J]

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

 

[courtrigrad]

$$\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$$

$$\frac{y^{2}}{4} = 1 - \frac{4}{9} = \frac{5}{9}$$

$$y^{2} = \frac{20}{9}$$

$$y = \pm \frac{\sqrt{20}}{3}$$

 

[HallsofIvy]

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

No, y^2= 4(0.55556)= 2.22222
so y= sqrt(2.22222)= 1.4907= sqrt(20)/3

Sqrt(0.138) = y

y = 0.37