# Points on an ellipse

1. Mar 4, 2009

### Emethyst

Hello everyone, I am still relatively new to this site, so any mistakes I take full blame for.

My question is: At what point(s) on the ellipse x^2+4y^2=4 is the slope of the tangent line 1/2sqrt3?

I have found the derivative of the equation through implicit differentation (I came out with dy/dx=-2x/8y, if wrong please tell me) and thought I could solve the given equation for one of the variables, then plug it into the derivative while setting it = to the slope of the tangent line and solve for the variable, then simply use that point to find the other one, but it did not work.

Could someone be of assistance and show me how to go about this question, as it is 4 marks and I want to make sure I am prepared for the upcoming test. (For what its worth too I am in the beginning of grade 12 calculus so try and keep any explanation simple :tongue:)

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

2. Relevant equations

3. The attempt at a solution

2. Mar 4, 2009

### sutupidmath

ok.

Call the point at which the slope of the tg is as you have there $$(x_0,y_0)$$

then what this means is that you will have a system of two eq. in two unknowns:

$$\frac{1}{2\sqrt{3}}=-\frac{x_0}{4y_0}$$

and

$$x_0^2+4y_0^2=4$$

I assume you know how to solve for x_0 and y_o right?

[tex]

3. Mar 4, 2009

### Emethyst

Thanks a bunch, and I believe so, do you simply solve one of the equations for a variable (eg. y) and substitute it into the other one and solve for the one variable left, or is this the wrong way?

4. Mar 4, 2009

### sutupidmath

That is correct! so you will most probbably get two points.

5. Mar 4, 2009

### Emethyst

Thanks for all of that help sutupidmath, and you're right, it does come out two points :tongue: