Ellipses word problem

• August 21st 2012, 01:39 PM
Greymalkin
Ellipses word problem
Attachment 24569

Problem above, from what I understand you are to find the vertical distance to an edge that is 6 feet from a vertex.
I'm trying to triangulate from the focus distances to the point to create an equation I can equate.
Using distance from the focus(c= $\sqrt{(25^2-14^2)}=20.71$
Setting the center at the origin, the distance from 2 foci = $2(\sqrt{20.71^2+14^2}=50$
Therefore the point 6 feet from the vertex, which is 19 ft from the origin, and is 39.71ft from the last focus, is equated by: $50-(39.71^2+b^2=c^2)$
This is where my approach halts, is my approach wrong? Or is there something I am just not seeing?

• August 21st 2012, 05:23 PM
GJA
Re: Ellipses word problem
Hi, Greymalkin.

I think your approach is good, nice work! I think a small hint will get things going in the right direction.

Since you're taking the origin to sit "in the middle" so-to-speak, we know that $x$ and $y$ must satisfy the equation

$\frac{x^{2}}{(25)^{2}}+\frac{y^{2}}{(14)^{2}}=1,$

where 25 is the major radius of the ellipse, and 14 is the minor radius. What you determined in your work was that the $x$ coordinate of the point we're interested is $x=-19$. Our goal is to know what the height $y$ is at this point; to determine this we can use the ellipse equation above since we know what $x$ is at the point of interest.

Does this help? Let me know if anything is unclear.

Good luck!

Edit: I was thinking of the bank on the left of the picture, that's why I have $x=-19$ above, but if you wanted to consider the right bank you could use $x=19$. It's all academic because we're going to square things in the equation of the ellipse anyways!