The area of the triangle formed by the points of intersection of the parabola $y=a(x+5)(x-1)$ with the coordinate axes is 12 . Find $a$ if the parabola opens upward.

Step 1 ：The given parabola is \(y=a(x+5)(x-1)\).

Step 2 ：The points of intersection of the parabola with the coordinate axes are the roots of the equation \(y=0\), which are \(x=-5\) and \(x=1\).

Step 3 ：So, the triangle formed by the points of intersection of the parabola with the coordinate axes has vertices at \((-5,0)\), \((1,0)\), and \((0,5a)\).

Step 4 ：The base of the triangle is the line segment between \((-5,0)\) and \((1,0)\), which has length \(1-(-5)=6\).

Step 5 ：The height of the triangle is the y-coordinate of the point \((0,5a)\), which is \(5a\).

Step 6 ：The area of the triangle is given by the formula \(\frac{1}{2} \times \text{base} \times \text{height}\), so we have \(12 = \frac{1}{2} \times 6 \times 5a\).

Step 7 ：Solving for \(a\) gives \(a = \frac{12}{15} = \frac{4}{5}\).

Step 8 ：\(\boxed{a = \frac{4}{5}}\) is the final answer.