Find $k$ such that the area is an integer 
For some positive integers k, the parabola with equation $y = \frac{x^2}{k} - 5$ intersects
  the circle with equation $x^2 + y^2 = 25$ at exactly three distinct points A, B and C.
  Determine all such positive integers k for which the area of $\triangle ABC$ is an integer.

Point $A = (0, -5)$ and also, I found that the area is:
$\frac{1}{2} BH$ the area of half the triangle $= \frac{1}{2} xy$ so just multiply by $2$ in the end.
In the end, I ended with the coordinate for $B$ as 
$B = (\sqrt{10k + k^2}, 5 + k)$.
So the area of the WHOLE Triangle (ABC) is:
$A = \sqrt{k}\sqrt{10 + k}(10 + k)$
Which value of $k$ that will make this an integer is the hard part.
Then I found that:
$A = (10 + k)\sqrt{(k + 5)^2 - 25}$ but I still cannot proceed.

Simplifying $x^2+(\frac{x^2}{k}-5)^2=25$ gives
$$x^2(x^2+k^2-10k)=0.$$
In order for this equation to have three distinct real roots, we need to have $10k-k^2\gt 0,$ i.e. $0\lt k\lt 10$.
Then, $$x=0,\pm\sqrt{10k\color{red}{-}k^2}.$$
The area of $\triangle{ABC}$ is given as
$$[\triangle{ABC}]=\frac 12\times 2\sqrt{10k-k^2}\times (5-k+5)=(10-k)\sqrt{k(10-k)}.$$
Checking each of $k=1,2,\cdots, 9$ gives that the answer is $k=\color{red}{1,2,5,8,9}$.
