Answer: \(a=-1\).
Use gradients for straight-line work: parallel lines have equal gradients and perpendicular lines have gradients whose product is \(-1\).
At the points of intersection, \(y=3x-11\) and \(xy=4-3x-2x^2\).
Substitute \(y=3x-11\) into the curve:
\(x(3x-11)=4-3x-2x^2.\)
So
\(3x^2-11x=4-3x-2x^2.\)
Hence
\(5x^2-8x-4=0.\)
Factorising or solving gives
\(x=2 \quad\text{or}\quad x=-\frac25.\)
The corresponding \(y\)-coordinates are
\(y=3(2)-11=-5\)
and
\(y=3\left(-\frac25\right)-11=-\frac{61}{5}.\)
So
\(A=\left(-\frac25,-\frac{61}{5}\right), \qquad B=(2,-5).\)
The midpoint of \(AB\) is
\(\left(\frac{-\frac25+2}{2},\frac{-\frac{61}{5}-5}{2}\right) =\left(\frac45,-\frac{43}{5}\right).\)
The gradient of \(AB\) is the gradient of the line \(y=3x-11\), which is \(3\). Therefore the gradient of the perpendicular bisector is
\(-\frac13.\)
The perpendicular bisector is
\(y+\frac{43}{5}=-\frac13\left(x-\frac45\right).\)
The point \(C=(a,-8)\) lies on this line. Substitute \(y=-8\):
\(-8+\frac{43}{5}=-\frac13\left(a-\frac45\right).\)
So
\(\frac35=-\frac13\left(a-\frac45\right).\)
Therefore
\(a-\frac45=-\frac95,\)
and hence
\(a=-1.\)
