Ellipse Circumscribed In Triangle

_FlaMe

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

[tex] \frac{x^2}{100}+\frac{(y-5)^2}{25}=1 [/tex]

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

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

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

- I know f'(x) is the slope of the tangent line
- I have the equations for the top half of the parabola as well as the derivation
- I tried doing slope point form but for some reason it didn't work as I kept getting height = f(x)

dynamicsolo

The ellipse appears to be completely irrelevant to the proposition, except for the purpose of imposing a symmetry about the y-axis. I am going to make one small change and call the points on the ellipse (x_0, a) and (-x_0, a).

I am assuming that the base of the triangle RST is the x-axis, since nothing is given about that in the problem. The tangent lines to the point on the ellipse meet on the y-axis at a point (0, b). Since the slopes on the tangent lines will be +/- f' , the x-intercepts of those tangent lines will be (-b/f' , 0) and (b/f' , 0). Thus the area of the triangle is

A = (1/2) · (2b/f') · (b) = (b^2)/f' .

Now the slope of the tangent line from (0, b) to (x_0, a) is

f' = (a - b)/(x_0) ,

which is negative, which is also evident from the geometry. We can solve to find

b = a - (x_0)·f' = f - (x_0)·f' ,

since a = f(x_0) and I am transferring to the notation of the problem.

Substituting this into our area formula gives

A = { [ f - (x_0)·f' ]^2 }/f'

= { (f'^2)·[ (f/f') - (x_0) ]^2 }/f'

= f' · [ (f/f') - (x_0) ]^2 .

Since I have used the half of the triangle in the positive-x half-plane, f' < 0 , so taking the absolute value of this expression to obtain a positive area will introduce a minus sign.

Nowhere in this does it seem necessary to use any property of the ellipse other than its symmetry about the y-axis.