Running path locus

An ellipse is the locus of points for which the sum of the distances to each focus is a constant amount. In this problem, the foci can be considered to be the location of the house and the location of the school. Then the sum of the distances from the house to $A$ and from $A$ to the school is a constant amount, $18-8=10$ . This constant amount is equal to $2a$ , the length of the major axis. Therefore, $a=5$ .

The distance between the foci of an ellipse is $2\sqrt{a^2-b^2}$ . In this problem, this distance is $8$ . Solving the equation $2\sqrt{a^2-b^2}=8$ for $b$ gives $b=3$ .

The center of the ellipse is halfway between the foci, $(4,0)$ . Therefore, $h=4$ and $k=0$ .

The general form of an ellipse is $\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$ . Making the substitutions for $a$ , $b$ , $h$ , and $k$ gives:

$\large\boxed{\frac{(x-4)^2}{25}+\frac{y^2}{9}=1}$

This solution assumes knowledge of an ellipse and its parameters. However, I'd be interested to see a solution that proves the same result assuming no previous knowledge of ellipses.

One point A where both of the first two legs are equal (i.e. 5 ) must be located on the vertical line that bisects (0,0) and (8,0). Therefore, A is at point (4,3) since 4^2 + 3^2 = 5^2. Testing x = 4 and y = 3 in each of the equations:

(4 - 12)^2 / 9 - 3^2 = 64/9 - 9 which is not equal to 1

(4-4)^2 / 25 + 3^2 / 9 = 0 + 1 which is equal to 1

(4 - 4)^2 + 3^2 = 0 + 9 which is not equal to 25

3^2 is not equal to -4 (4 - 9) = -4 (-5) = 20

Therefore 2nd equation must be the correct one.