# Circumscribing an ellipse

1. Jun 20, 2005

### wisredz

Let P(x,a) and Q(-x,a) be two points on the upper half of the ellipse

$$\frac{x^2}{100}+\frac{(y-5)^2}{25}=1$$

centered at (0,5). A triangle RST is formed by using the tangent lines to the ellipse at Q and P.

Show that the area of the triangle is

$$A(x)=-f'(x)[x-\frac{f(x)}{f'(x)}]^2$$

where y=f(x) is the function representing the upper half of the ellipse.

Last edited: Jun 20, 2005
2. Jun 20, 2005

### StatusX

Have you been able to figure out where the third line of the triangle intersects the ellipse?

3. Jun 20, 2005

### wisredz

I never thought of that but that point isn't even related to f(x). But the point is (0,0). I don't understand how to use that point. The triangle's R corner is on the y axis. The other two are on the x axis.

4. Jun 20, 2005

### StatusX

Ok, so just find its base and height. Remember that f'(x) is the slope of the line tangent to the ellipse, so you can use it to find where that line intersects the x and y axes.

5. Jun 20, 2005

### wisredz

Well, it all get's messed up because I do not know what to do when trying to find an equation for one of the edges. That is because I have f'(x) in terms of x and I have the point Q(x,a) and when I try it everyting gets messed up. What should I do now? I had already tried until this point but I always get lost right here...

6. Jun 20, 2005

### StatusX

a=f(x), and point Q is at (x,f(x)). Did you misunderstand this part? Just draw everything. The line passes through Q and has a slope of f'(x). You need the x and y intercepts to get the base and height of the triangle.

7. Jun 20, 2005

### wisredz

The point-slope equation for the tangent passing through the point $Q(x_0,f(x_0))$ would be $y-f(x_0)=f'(x)(x-x_0)$ right? But when I give x the value of 0 I get $x_0$ as the y intersection. Am I doing something wrong here?

8. Jun 20, 2005

### StatusX

The y intercept is the y-value when x=0, so plug in 0 for x and solve for y.

9. Jun 21, 2005

### wisredz

Yeah, I know that. But when I plug in x=0, f'(x)=0. So the right hand side of the point slope equation becomes 0. from here $y=f(x_0)$, which is quite impossible by the figure drawn in the book.

Btw, I'll give f(x) and f'(x) in case that you may spot an error in the calculations.

$$f(x)= \frac {(\sqrt(100-x^2)}{2}+5$$
$$f'(x)= \frac {-x}{2*\sqrt(100-x^2)}$$

I actually graphed these functions and everything seems to be alright...

10. Jun 21, 2005

### StatusX

that's f'(x0), right?

11. Jun 22, 2005

### wisredz

Thanks a lot, I did it now. My mistake was not using f'(x_0) but instead f'(x). Thanks a lot again...