Minimum value of slope of line tangent to ellipse I would like to solve the following problem.

Given that the point $p$ lies on the ellipse $$\frac{x^2}{100} + \frac{(y - 12)^2}{75} = 1$$ and the line $y = ax$ passes through $p$, let $n$ be the smallest positive value of $a$. Compute $n$.

What first came to my mind was finding the slope of the line tangent to the ellipse. This would require me to set the discriminant of the resulting quadratic in $x$ to be set to $0$.
However, the numbers in the denominator make the calculations a bit annoying. Is there a better way to solve the problem? Thanks!

Not the best in math way answer because it is not general solution but it is based on my comment:

My idea is (for easy solution without hard work) to find the line
  $y=αx$ that is tangent in the circle that contains the ellipse and
  then test next few values with the real ellipse. [I know it is a hack
  but also know that the numbers seems to be a little bit silly for this
  question... Sure the solution is close to $1$ if not $1$]

The circle that contains the ellipse (don't know English name) is:
$x^2+(y-12)^2=10^2$
The tangent on this circle that can pass through $(0,0)$ is a line of function $y=\alpha \cdot x$ that has distance d=10 from the point $(0,12)$.
line is: 
$\alpha \cdot x -y=0$ and point (0,12) have a distance:
$d=\dfrac{|\alpha \cdot 0 -12|}{\sqrt{\alpha^2+1}}=\dfrac{12}{\sqrt{\alpha^2+1}}=10$
Thus:
$100\alpha^2+100=144\Longrightarrow \alpha^2=\frac{44}{100}$.
We are looking for integer values of $\alpha$, and so we will start from value $1$ and test a few integers with the real ellipse [Possibly $1$ could already passes through].
Test value $\alpha=1$
$\begin{cases}y=x\\
\dfrac{x^2}{100}+\dfrac{(x-12)^2}{75}=1
\end{cases}\hspace{10pt}
\begin{cases}y=x\\
x^2+\dfrac{4}{3}(x-12)^2=100
\end{cases}\hspace{10pt}
\begin{cases}y=x\\
x^2+\dfrac{4}{3}x^2 +\dfrac{4}{3}144-\dfrac{4}{3}\cdot 24 \cdot x=100 
\end{cases}$
$\begin{cases}\dfrac{7}{3}x^2-32\cdot x+16\cdot 12-100=0\\\end{cases}$
The last one becomes:
$7x^2- 96 x +(192-100)\cdot 3=0$
$7x^2-96 x+276=0$ with
$\Delta=(-96)^2-4\cdot 7\cdot 276\approx 10,000-7000 >0$
Solution =1
Edit: Visualizing why I found the numbers  silly:

