Find the inradius of a triangle situated within an ellipse

Question

Question: $∆ABC$ is situated within an ellipse whose major axis is of length $10$ and whose minor axis is of length $8$. Point $A$ is a focus of the ellipse, point $B$ is an endpoint of the minor axis and point $C$ is on the ellipse such that the other focus lies on $BC$. Compute the inradius of ∆ABC.

I am given the hint of the area formula involving the inradius $A = rs$.

First I call the other focus $D$. I know that $P=20$ and therefore $s=10$ because $AB+BD=10$ and $AC+CD=10$. So I have $A = rs = 10r$. If I can get $A$, I immediately have $r$ but I'm not sure how to get there.

I've figured out that the focal length ($AD$) is 3 and that $AB = BD = 5$, but it hasn't helped. Can someone point me in the right direction?

Answer 1

Line $BC$ has equation $$\frac{x}{3}+\frac{y}{4}=1$$ Ellipse has equation $$\frac{x^2}{25}+\frac{y^2}{16}=1$$ Intersection points are $$B(0,4);\;C\left( \frac{75}{17},-\frac{32}{17}\right)$$ Perimeter of $\triangle ABC=20$. Half perimeter $p=10$

Area of $\triangle ABC;\;\mathcal{A}=\frac{300}{17}$

radius of inscribed circle is $$r=\frac{A}{p}=\frac{30}{17}$$

Answer 2

I'll be using coordinate geometry to arrive at my hint. First off, we now that A is situated right at the focus of the ellipse. For our convenience, let us assume that it is the positive focus, (5,0). Also, we know that B is situated at the endpoint of the minor axis. Again, for our convenience, let us assume that it is the positive endpoint, and so we fix B as (4,0).

Well, first of all, we know that the other focus lies on the line connecting B and C. BUT, we also know that the other focus is (-5,0).

Find the equation of the line joining these points, and then substitute y into the equation of the ellipse. You'll get a quadratic equation in x. Solve it, however tedious it is, get the value of (x,y).

Henceforth, use the distance formula to evaluate s,a,b,c.

Substitute them into your formula and solve the rest.