Tangent lines to ellipse from external point I'm given an ellipse with the equation
 $$\frac{x^2}{5^2} +\frac{y^2}{4^2} -1 = 0 $$

and I have to find the tangents to this ellipse  passing through a point $P(10, -8)$.
Use auxiliary coordinates $u:=\dfrac x5,v:=\dfrac y4$ to turn the ellipse to the unit circle. The external point is $(2,-2)$, which is at distance $\sqrt8$ of the origin, in direction $-45°$.

The point, the circle center and the tangency points form right triangles of small side $1$ and hypothenuses $\sqrt8$. Thus the small angle is given by $\tan\theta=\dfrac1{\sqrt{8-1}}$.

Hence, the slope of the tangents are given by

$$\tan(\pm\theta-45°)=\frac{\pm\sqrt7-1}{1\mp\sqrt7}.$$
After converting back to the $(x,y)$ coordinates,

$$m=\frac45\frac{\pm\sqrt7-1}{1\mp\sqrt7}.$$
The straight line has equation

$$y+8=m(x-10).$$ We plug that in the equation of the ellipse and get

$$\frac{x^2}{25}+\frac{(m(x-10)-8)^2}{16}-1=0.$$

This quadratic equation in $x$ has a double root (which ensures tangency) when its discriminant is zero, i.e.

$$75m^2+160m+48=0.$$

There are two solutions.