Consider the parabola $\mathcal{P}$ of equation $\;y = x^2,$ and the line $L$ of equation $y = x+6.$ Let $P(x_P,y_P)$ be a point on the arc of the parabola $\mathcal{P}$ below $L.$ Let $A$ and $B$ be the points of intersection of $\mathcal{P}$ and $L.$
a) Find an equation of the line $L_P$ passing through $P$ which is perpendicular to $L.$
b) Let $Q_P$ be the point of intersection of $L$ and $L_P.$ Express the distance between $P$ and $Q_P$ using only $x_P$ and $y_P.$
c) Find the point $P$ on the arc of $\mathcal{P}$ below $L$ such that the area of the triangle $APB$ is maximal.
I had no problem doing part a) and got $y=8-x,$ however, I cannot continue further with part b) and c) 
without hesitation. Please help me!
Hint: You can easily find out the coordinates of $Q_P$ by solving the system of linear equations \begin{align} y-x=6\\ x+y=8 \end{align} Knowing $Q_P$ you can express the distance between $P$ and $Q_P$ in terms of $(x_P,y_P)$.
For the third part, first find out the distance between $A$ and $B$. Let that distance be $b$. Let the distance between $P$ and $Q_P$ be $h$. Then the area of triangle is $\frac{bh}{2}$. Note that the only variables are $x_P$ and $y_P$. Now use the relation $y=x^2$ and take the derivative of area with respect to either $x$ or $y$. (Depends on your substitution.) I hope you know how to proceed.
P.S:My answer is based on the assumption that your first part is correct.