Equations of lines tangent to an ellipse

You need a line that passes through the point $(4,6)$ and that touches the ellipse at just one point. The vertical line does that and you've already found it. Obviously there is exactly one other tangent line (and if that's not obvious to you, then draw the picture and look at it!).

Nonvertical lines passing through the point $(4,6)$ have equation $y-6=m(x-4)$.

That implies $y=mx-4m+6$, so we can put $mx-4m+6$ in place of $y$:

$$ \frac{x^2}{16} + \frac{(mx-4m+6)^2}{4} = 1. $$ This is equivalent to $$ \underbrace{(1+4m)}x^2 + \underbrace{-8m(4m-6)}\ x + \underbrace{4(4m-6)^2 -16} = 0. $$

This equation is quadratic in $x$. We therefore want a quadratic equation with exactly one solution. A quadratic equation $ax^2+bx+c=0$ has exactly one solution precisely if its discriminant $b^2-4ac$ is $0$. So we have $$ b^2-4ac = \underbrace{64m^2(4m-6)^2 - 4(1+4m)(4(4m-6)^2-16) = 0}. $$ Now we only need to solve this last equation for $m$.

Try changing the variables, linear transf, x=4X, y=2Y. This turns the ellipse into the unit circle, and (4,6) goes to (1,3). Now draw it and use elementary trigonometry to get the slope and the equation, in X and Y. Convert back to x and y.

We have that $x^2+4y^2=16$, so using implicit differentiation gives $2x+8yy^{\prime}=0$ and therefore $\;\;\;\displaystyle y ^{\prime}=-\frac{x}{4y}$.

Since the slope of the line between $(x,y)$ and $(4,6)$ is given by $\frac{y-6}{x-4}$, you have that $\displaystyle -\frac{x}{4y}=\frac{y-6}{x-4}$.

This gives $-x^2+4x=4y^2-24y$, so now you can use this equation and the equation $x^2+4y^2=16$ to find the other point of tangency.