Find the centers of all ellipses with a covertex at (0,6), a focus at (8,0), & integral center coordinates.?

Question

There should be six of them (plus two degenerate circles.)

sorry-not six of them-10 I think besides (the points themselves)

Answer 1

Let me try this question.

There is a covertex, V at (0,6) and a focus, F at (8,0).
Let C be the center with cords (p,q), where p and q are supposed to be integers.
So from the geometry, CV must be perpendicular to CF.
Also let CV = b, and CF = c.
So b^2 + c^2 = 6^2 + 8^2 = 100

Now b^2 = p^2 + (q-6)^2 = p^2 + q^2 - 12q + 36
c^2 = (8-p)^2 + q^2 = p^2 + q^2 - 16p + 64

From the above we get
p^2 + q^2 = 6q + 8p
or p^2 - 8p + (q^2 + 6q) = 0
solving the quadratic for p,
p = 4 +/- sqrt(16 + 6q - q^2)

So we wish f(q) = 16 + 6q - q^2 to be a perfect square.
since f(q) >= 0 for -2 <= q <= 8
we can obtain by trial and error that the following allow for perfect squares:
q = -2, -1, 0, 3, 6, 7, 8
So the possibilities for the center, C are:
(4,-2); (1,-1); (7,-1); (0,0); (8,6); (7,7); (1,7); (4,8); (-1,3); (9,3)

The points (8,0) and (0,6) were ignored because they coincided with F and V respectively.

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Alternatively
since CV and CF are at right angles, C must lie on a circle of radius 5 drawn from the midpoint of F and V. ie C must lie on
(x-4)^2 + (y-3)^2 = 5^2
This will result in the same integral values as above.