Finding, without derivatives, the line through $(9,6.125)$ that is tangent to the parabola $y=-\frac18x^2+8$ 
What part I am looking at right now is f). My parabola equation for the front edge of the roof is $y=-\frac18x^2+8$. All the laser lights have an equation that is tangent to the parabola except the red laser beam. The point that I have for the red laser beam source’s location for the tower is at $(9,6.125)$. How can I determine an equation of the lines that are tangent to the parabola and pass through $(9,6.125)$, without using derivatives? 
Thank you for answering ahead of time, I had a lot of trouble with this question!
I tried to find an equation that goes through the vertex and got the equation -5/24x+8, then found the midpoint of the x on the two intersections, (0+5/3)/2= 5/6. After that, I plugged it into the equation and recieved 2279/288. I thought the line passing through (5/6, 2279/288) and (9,6.125) would be tangent to the parabola. Unfortunately this was incorrect, and I do not know how to solve this without derivatives...!
 A: Let $m$ be the gradient of the red beam.  You want to solve simultaneously
$$y=-\tfrac18x^2+8\quad\hbox{and}\quad y-6.125=m(x-9)\ .$$
This gives $-\tfrac18x^2+8=6.125+m(x-9)$, or simplifying,
$$x^2+8mx-(15+72m)=0\ .$$
Now the beam will miss the parabola if this equation has no solutions; it will "hit" the parabola if there are two solutions; and it will be tangent to the parabola if there is one solution.  A quadratic has one solution if and only if the discriminant is zero, that is,
$$64m^2+4(15+72m)=0\ .$$
This will give you two values of $m$ from which you can calculate two values of $x$ for the point of tangency.  But one of these will have $x>9$ which for the present problem must be rejected.
I'll leave you to fill in the details.  Good luck!
A: Your lines have to pass through $(9, 6.125)$
All you have to do is to find the slope of such lines.
If the slope is $m$, the equation of such a line is
$$y-6.125 = m(x-9)$$
This line intersects the parabola at the point where $$y=-(1/8)x^2+8$$
That is $$-(1/8)x^2+8 -6.125=m(x-9)$$
In order to have tangency you need this quadratic equation to have double roots. 
That is $b^2-4ac=0$ where $b=-m$ and $a=-(1/8)$ and $c=9m+1.875$
We get $$m^2-4(-1/8)(9m+1.675)=0$$
Solve for $m$ and you will have your tangent lines.
